Chapter 2
1. We use Eq. 2-2 and Eq. 2-3. During a time tc when the velocity remains a positive constant, speed is
equivalent to velocity, and distance is equivalent to displacement, with ∆x = v tc.
(a) During the first part of the motion, the displacement is ∆x1 = 40 km and the time interval is
(40 km)
t1 =
= 133
. h.
(30 km / h)
During the second part the displacement is ∆x2 = 40 km and the time interval is
t2 =
(40 km)
= 0.67 h.
(60 km / h)
Both displacements are in the same direction, so the total displacement is
∆x = ∆x1 + ∆x2 = 40 km + 40 km = 80 km.
The total time for the trip is t = t1 + t2 = 2.00 h. Consequently, the average velocity is
vavg =
(80 km)
= 40 km / h.
(2.0 h)
(b) In this example, the numerical result for the average speed is the same as the average velocity 40 km/h.
(c) As shown below, the graph consists of two contiguous line segments, the first having a slope of 30 km/h and
connecting the origin to (t1, x1) = (1.33 h, 40 km) and the second having a slope of 60 km/h and connecting (t1,
x1) to (t, x) = (2.00 h, 80 km). From the graphical point of view, the slope of the dashed line drawn from the
origin to (t, x) represents the average velocity.
3. The speed (assumed constant) is v = (90 km/h)(1000 m/km) ⁄ (3600 s/h) = 25 m/s. Thus, in 0.50 s, the car
travels (0.50 s)(25 m/s) ≈ 13 m.
5. Using x = 3t – 4t2 + t3 with SI units understood is efficient (and is the approach we will use), but if we wished
to make the units explicit we would write
x = (3 m/s)t – (4 m/s2)t2 + (1 m/s3)t3.
We will quote our answers to one or two significant figures, and not try to follow the significant figure rules
rigorously.
(a) Plugging in t = 1 s yields x = 3 – 4 + 1 = 0.
21
22
(b) With t = 2 s we get x = 3(2) – 4(2)2+(2)3 = –2 m.
(c) With t = 3 s we have x = 0 m.
(d) Plugging in t = 4 s gives x = 12 m.
For later reference, we also note that the position at t = 0 is x = 0.
(e) The position at t = 0 is subtracted from the position at t = 4 s to find the displacement ∆x = 12 m.
(f) The position at t = 2 s is subtracted from the position at t = 4 s to give the displacement ∆x = 14 m. Eq. 2-2,
then, leads to
vavg =
∆x 14 m
=
= 7 m/s.
∆t
2s
(g) The horizontal axis is 0 ≤ t ≤ 4 with SI units understood.
Not shown is a straight line drawn from the point at (t, x) = (2, –2) to the highest point shown (at t = 4 s) which
would represent the answer for part (f).
9. The values used in the problem statement make it easy to see that the first part of the trip (at 100 km/h) takes
1 hour, and the second part (at 40 km/h) also takes 1 hour. Expressed in decimal form, the time left is 1.25
hour, and the distance that remains is 160 km. Thus, a speed v = (160 km)/(1.25 h) = 128 km/h is needed.
13. (a) Denoting the travel time and distance from San Antonio to Houston as T and D, respectively, the average
speed is
savg1 =
D (55 km/h)(T/2) + (90 km/h)(T / 2)
=
= 72.5 km/h
T
T
which should be rounded to 73 km/h.
(b) Using the fact that time = distance/speed while the speed is constant, we find
savg2 =
which should be rounded to 68 km/h.
D
=
T
D/2
55 km/h
D
= 68.3 km/h
/2
+ 90Dkm/h
23
(c) The total distance traveled (2D) must not be confused with the net displacement (zero). We obtain for the
two-way trip
savg =
2D
= 70 km/h.
+ 68.3Dkm/h
D
72.5 km/h
(d) Since the net displacement vanishes, the average velocity for the trip in its entirety is zero.
(e) In asking for a sketch, the problem is allowing the student to arbitrarily set the distance D (the intent is not
to make the student go to an Atlas to look it up); the student can just as easily arbitrarily set T instead of D, as
will be clear in the following discussion. We briefly describe the graph (with kilometers-per-hour understood
for the slopes): two contiguous line segments, the first having a slope of 55 and connecting the origin to (t1, x1)
= (T/2, 55T/2) and the second having a slope of 90 and connecting (t1, x1) to (T, D) where D = (55 + 90)T/2. The
average velocity, from the graphical point of view, is the slope of a line drawn from the origin to (T, D). The
graph (not drawn to scale) is depicted below:
15. We use Eq. 2-4. to solve the problem.
(a) The velocity of the particle is
v=
dx d
=
(4 − 12t + 3t 2 ) = −12 + 6t .
dt dt
Thus, at t = 1 s, the velocity is v = (–12 + (6)(1)) = –6 m/s.
(b) Since v < 0, it is moving in the negative x direction at t = 1 s.
(c) At t = 1 s, the speed is |v| = 6 m/s.
(d) For 0 < t < 2 s, |v| decreases until it vanishes. For 2 < t < 3 s, |v| increases from zero to the value it had in
part (c). Then, |v| is larger than that value for t > 3 s.
(e) Yes, since v smoothly changes from negative values (consider the t = 1 result) to positive (note that as t → +
∞, we have v → + ∞). One can check that v = 0 when t = 2 s.
(f) No. In fact, from v = –12 + 6t, we know that v > 0 for t > 2 s.
17. We use Eq. 2-2 for average velocity and Eq. 2-4 for instantaneous velocity, and work with distances in
centimeters and times in seconds.
24
(a) We plug into the given equation for x for t = 2.00 s and t = 3.00 s and obtain x2 = 21.75 cm and x3 = 50.25
cm, respectively. The average velocity during the time interval 2.00 ≤ t ≤ 3.00 s is
vavg =
∆x 50.25 cm − 2175
. cm
=
∆t
3.00 s − 2.00 s
which yields vavg = 28.5 cm/s.
(b) The instantaneous velocity is v =
dx
dt
= 4.5t 2 , which, at time t = 2.00 s, yields v = (4.5)(2.00)2 = 18.0 cm/s.
(c) At t = 3.00 s, the instantaneous velocity is v = (4.5)(3.00)2 = 40.5 cm/s.
(d) At t = 2.50 s, the instantaneous velocity is v = (4.5)(2.50)2 = 28.1 cm/s.
(e) Let tm stand for the moment when the particle is midway between x2 and x3 (that is, when the particle is at xm
= (x2 + x3)/2 = 36 cm). Therefore,
xm = 9.75 + 15
. tm3
⇒
tm = 2.596
in seconds. Thus, the instantaneous speed at this time is v = 4.5(2.596)2 = 30.3 cm/s.
(f) The answer to part (a) is given by the slope of the straight line between t = 2 and t = 3 in this x-vs-t plot. The
answers to parts (b), (c), (d) and (e) correspond to the slopes of tangent lines (not shown but easily imagined) to
the curve at the appropriate points.
19. We represent its initial direction of motion as the +x direction, so that v0 = +18 m/s and v = –30 m/s (when t
= 2.4 s). Using Eq. 2-7 (or Eq. 2-11, suitably interpreted) we find
(−30 m/s) − (+1 m/s)
aavg =
= − 20 m/s 2
2.4 s
which indicates that the average acceleration has magnitude 20 m/s2 and is in the opposite direction to the
particle’s initial velocity.
22. We use Eq. 2-2 (average velocity) and Eq. 2-7 (average acceleration). Regarding our coordinate choices, the
initial position of the man is taken as the origin and his direction of motion during 5 min ≤ t ≤ 10 min is taken
to be the positive x direction. We also use the fact that ∆x = v∆t ' when the velocity is constant during a time
interval ∆t' .
(a) The entire interval considered is ∆t = 8 – 2 = 6 min which is equivalent to 360 s, whereas the sub-interval in
which he is moving is only ∆t' = 8 − 5 = 3min = 180 s. His position at t = 2 min is x = 0 and his position at t =
8 min is x = v∆t' = (2.2)(180) = 396 m . Therefore,
25
vavg
396 m − 0
=
= 110
. m / s.
360 s
(b) The man is at rest at t = 2 min and has velocity v = +2.2 m/s at t = 8 min. Thus, keeping the answer to 3
significant figures,
aavg =
2.2 m / s − 0
= 0.00611 m / s2 .
360 s
(c) Now, the entire interval considered is ∆t = 9 – 3 = 6 min (360 s again), whereas the sub-interval in which he
is moving is ∆t' = 9 − 5 = 4 min = 240 s ). His position at t = 3 min is x = 0 and his position at t = 9 min
is x = v∆t ' = (2.2)(240) = 528 m . Therefore,
vavg =
528 m − 0
= 1.47 m / s.
360 s
(d) The man is at rest at t = 3 min and has velocity v = +2.2 m/s at t = 9 min. Consequently, aavg = 2.2/360 =
0.00611 m/s2 just as in part (b).
(e) The horizontal line near the bottom of this x-vs-t graph represents the man standing at x = 0 for 0 ≤ t < 300 s
and the linearly rising line for 300 ≤ t ≤ 600 s represents his constant-velocity motion. The dotted lines
represent the answers to part (a) and (c) in the sense that their slopes yield those results.
The graph of v-vs-t is not shown here, but would consist of two horizontal “steps” (one at v = 0 for 0 ≤ t < 300 s
and the next at v = 2.2 m/s for 300 ≤ t ≤ 600 s). The indications of the average accelerations found in parts (b)
and (d) would be dotted lines connecting the “steps” at the appropriate t values (the slopes of the dotted lines
representing the values of aavg).
23. We use v = v0 + at, with t = 0 as the instant when the velocity equals +9.6 m/s.
(a) Since we wish to calculate the velocity for a time before t = 0, we set t = –2.5 s. Thus, Eq. 2-11 gives
c
h
c
h
v = (9.6 m / s) + 3.2 m / s2 ( −2.5 s) = 16
. m / s.
(b) Now, t = +2.5 s and we find
v = (9.6 m / s) + 3.2 m / s2 (2.5 s) = 18 m / s.
26
25. The constant acceleration stated in the problem permits the use of the equations in Table 2-1.
(a) We solve v = v0 + at for the time:
t=
v − v0
=
a
1
10
(3.0 × 10 8 m / s)
= 31
. × 10 6 s
9.8 m / s 2
which is equivalent to 1.2 months.
(b) We evaluate x = x0 + v0 t + 12 at 2 , with x0 = 0. The result is
x=
1
( 9.8 m/s 2 ) (3.1×106 s)2 = 4.6 ×1013 m .
2
26. We take +x in the direction of motion, so v0 = +24.6 m/s and a = – 4.92 m/s2. We also take x0 = 0.
(a) The time to come to a halt is found using Eq. 2-11:
0 = v0 + at ⇒ t =
24.6 m/s
= 5.00 s .
− 4.92 m/s 2
(b) Although several of the equations in Table 2-1 will yield the result, we choose Eq. 2-16 (since it does not
depend on our answer to part (a)).
0 = v02 + 2ax ⇒ x = −
(24.6 m/s)2
= 61.5 m .
2 ( − 4.92 m/s2 )
(c) Using these results, we plot v0t + 12 at 2 (the x graph, shown next, on the left) and v0 + at (the v graph, on the
right) over 0 ≤ t ≤ 5 s, with SI units understood.
27. Assuming constant acceleration permits the use of the equations in Table 2-1. We solve v 2 = v02 + 2a ( x − x0 )
with x0 = 0 and x = 0.010 m. Thus,
a=
v 2 − v02 (5.7 ×105 m/s) 2 − (1.5 ×105 m/s)2
=
= 1.62 ×1015 m/s2 .
2x
2(0.010 m)
29. We separate the motion into two parts, and take the direction of motion to be positive. In part 1, the vehicle
accelerates from rest to its highest speed; we are given v0 = 0; v = 20 m/s and a = 2.0 m/s2. In part 2, the
27
vehicle decelerates from its highest speed to a halt; we are given v0 = 20 m/s; v = 0 and a = –1.0 m/s2
(negative because the acceleration vector points opposite to the direction of motion).
(a) From Table 2-1, we find t1 (the duration of part 1) from v = v0 + at. In this way, 20 = 0 + 2.0t1 yields t1 = 10 s.
We obtain the duration t2 of part 2 from the same
equation. Thus, 0 = 20 + (–1.0)t2 leads to t2 = 20 s, and the
total is t = t1 + t2 = 30 s.
(b) For part 1, taking x0 = 0, we use the equation v2 = v2,0 + 2a(x – x0) from Table 2-1 and find
x=
v 2 − v02 (20 m/s) 2 − (0)2
=
= 100 m .
2a
2(2.0 m/s 2 )
This position is then the initial position for part 2, so that when the same equation is
x − 100 m =
used in part 2 we obtain
v 2 − v02 (0) 2 − (20 m/s) 2
.
=
2a
2(−1.0 m/s 2 )
Thus, the final position is x = 300 m. That this is also the total distance traveled should be evident (the vehicle
did not "backtrack" or reverse its direction of motion).
30. The acceleration is found from Eq. 2-11 (or, suitably interpreted, Eq. 2-7).
a=
∆v
=
∆t
1000 m / km I
b1020 km / hgF
G
H3600 s / h JK
1.4 s
= 202.4 m / s 2 .
In terms of the gravitational acceleration g, this is expressed as a multiple of 9.8 m/s2 as follows:
 202.4 m/s 2 
a=
g = 21g .
2 
 9.8 m/s 
31. We assume the periods of acceleration (duration t1) and deceleration (duration t2) are periods of constant a
so that Table 2-1 can be used. Taking the direction of motion to be +x then a1 = +1.22 m/s2 and a2 = –1.22 m/s2.
We use SI units so the velocity at t = t1 is v = 305/60 = 5.08 m/s.
(a) We denote ∆x as the distance moved during t1, and use Eq. 2-16:
(5.08 m/s) 2
v = v + 2a1∆x ⇒ ∆x =
= 10.59 m ≈ 10.6 m.
2(1.22 m/s 2 )
2
2
0
(b) Using Eq. 2-11, we have
t1 =
v − v0 5.08 m/s
=
= 4.17 s.
a1
1.22 m/s 2
The deceleration time t2 turns out to be the same so that t1 + t2 = 8.33 s. The distances traveled during t1 and t2
are the same so that they total to 2(10.59 m) = 21.18 m. This implies that for a distance of 190 m – 21.18 m =
168.82 m, the elevator is traveling at constant velocity. This time of constant velocity motion is
28
t3 =
168.82 m
= 33.21 s.
5.08 m / s
Therefore, the total time is 8.33 s + 33.21 s ≈ 41.5 s.
33. The problem statement (see part (a)) indicates that a = constant, which allows us to use Table 2-1.
(a) We take x0 = 0, and solve x = v0t + 12 at2 (Eq. 2-15) for the acceleration: a = 2(x – v0t)/t2. Substituting x =
24.0 m, v0 = 56.0 km/h = 15.55 m/s and t = 2.00 s, we find
a=
2 ( 24.0m − ( 15.55m/s )
( 2.00s )
( 2.00s ) )
2
= − 3.56m/s 2 ,
or | a |= 3.56 m/s 2 . The negative sign indicates that the acceleration is opposite to the direction of motion of the
car. The car is slowing down.
(b) We evaluate v = v0 + at as follows:
c
v = 1555
. m / s − 356
. m / s2
hb2.00 sg= 8.43 m / s
which can also be converted to 30.3 km/h.
35. (a) From the figure, we see that x0 = –2.0 m. From Table 2-1, we can apply x – x0 = v0t +
and then again with t = 2.0 s. This yields two equations for the two unknowns, v0 and a:
1
2
at2 with t = 1.0 s,
1
2
0.0 − ( −2.0 m ) = v0 (1.0 s ) + a (1.0 s )
2
1
2
6.0 m − ( −2.0 m ) = v0 ( 2.0 s ) + a ( 2.0 s ) .
2
Solving these simultaneous equations yields the results v0 = 0 and a = 4.0 m/s2.
(b) The fact that the answer is positive tells us that the acceleration vector points in the +x direction.
38. We take the direction of motion as +x, so a = –5.18 m/s2, and we use SI units, so v0 = 55(1000/3600) =
15.28 m/s.
(a) The velocity is constant during the reaction time T, so the distance traveled during it is
dr = v0T – (15.28 m/s) (0.75 s) = 11.46 m.
We use Eq. 2-16 (with v = 0) to find the distance db traveled during braking:
2
2
0
v = v + 2ad b
(15.28 m/s)2
⇒ db = −
2 ( −5.18 m/s 2 )
which yields db = 22.53 m. Thus, the total distance is dr + db = 34.0 m, which means that the driver is able to
stop in time. And if the driver were to continue at v0, the car would enter the intersection in t = (40 m)/(15.28
m/s) = 2.6 s which is (barely) enough time to enter the intersection before the light turns, which many people
29
would consider an acceptable situation.
(b) In this case, the total distance to stop (found in part (a) to be 34 m) is greater than the distance to the
intersection, so the driver cannot stop without the front end of the car being a couple of meters into the
intersection. And the time to reach it at constant speed is 32/15.28 = 2.1 s, which is too long (the light turns in
1.8 s). The driver is caught between a rock and a hard place.
39. The displacement (∆x) for each train is the “area” in the graph (since the displacement is the integral of the
velocity). Each area is triangular, and the area of a triangle is 1/2( base) × (height). Thus, the (absolute value of
the) displacement for one train (1/2)(40 m/s)(5 s) = 100 m, and that of the other train is (1/2)(30 m/s)(4 s) = 60
m. The initial “gap” between the trains was 200 m, and according to our displacement computations, the gap
has narrowed by 160 m. Thus, the answer is 200 – 160 = 40 m.
42. In this solution we elect to wait until the last step to convert to SI units. Constant acceleration is indicated,
so use of Table 2-1 is permitted. We start with Eq. 2-17 and denote the train’s initial velocity as vt and the
locomotive’s velocity as vl (which is also the final velocity of the train, if the rear-end collision is barely
avoided). We note that the distance ∆x consists of the original gap between them D as well as the forward
distance traveled during this time by the locomotive vlt . Therefore,
vt + vl ∆x D + vlt D
=
=
= + vl .
2
t
t
t
We now use Eq. 2-11 to eliminate time from the equation. Thus,
vt + vl
D
=
+ vl
2
vl − vt / a
b g
v −v I
1
Fv + v − v IJF
a=G
H 2 KG
HD JK= − 2 D bv − v g.
which leads to
t
l
l
2
t
l
l
t
Hence,
a=−
F
G
H
1
km
km
29
− 161
2(0.676 km)
h
h
IJ = −12888 km / h
K
2
2
which we convert as follows:
1h I
F1000 mIJF
a=c
−12888 km / h hG
G
H1 km KH3600 sJK= −0.994 m / s
2
2
2
so that its magnitude is |a| = 0.994 m/s2. A graph is
the case where a collision is just avoided (x along
in meters and t along the horizontal axis is in
(straight) line shows the motion of the locomotive
curve shows the motion of the passenger train.
shown below for
the vertical axis is
seconds). The top
and the bottom
The other case (where the collision is not quite
similar except that the slope of the bottom curve
than that of the top line at the point where they
avoided) would be
would be greater
meet.
30
44. Neglect of air resistance justifies setting a = –g = –9.8 m/s2 (where down is our –y direction) for the
duration of the fall. This is constant acceleration motion, and we may use Table 2-1 (with ∆y replacing ∆x).
(a) Using Eq. 2-16 and taking the negative root (since the final velocity is downward), we have
v = − v02 − 2 g ∆y = − 0 − 2(9.8 m/s 2 )(−1700 m) = −183 m/s .
Its magnitude is therefore 183 m/s.
(b) No, but it is hard to make a convincing case without more analysis. We estimate the mass of a raindrop to be
about a gram or less, so that its mass and speed (from part (a)) would be less than that of a typical bullet, which
is good news. But the fact that one is dealing with many raindrops leads us to suspect that this scenario poses an
unhealthy situation. If we factor in air resistance, the final speed is smaller, of course, and we return to the
relatively healthy situation with which we are familiar.
45. We neglect air resistance, which justifies setting a = –g = –9.8 m/s2 (taking down as the –y direction) for the
duration of the fall. This is constant acceleration motion, which justifies the use of Table 2-1 (with ∆y replacing
∆x).
(a) Starting the clock at the moment the wrench is dropped (v0 = 0), then v 2 = vo2 − 2 g ∆y leads to
(−24 m/s)2
∆y = −
= −29.4 m
2(9.8 m/s 2 )
so that it fell through a height of 29.4 m.
(b) Solving v = v0 – gt for time, we find:
t=
v0 − v 0 − (−24 m/s)
=
= 2.45 s.
g
9.8 m/s 2
(c) SI units are used in the graphs, and the initial position is taken as the coordinate origin. In the interest of
saving space, we do not show the acceleration graph, which is a horizontal line at –9.8 m/s2.
47. We neglect air resistance for the duration of the motion (between “launching” and “landing”), so a = –g = –
9.8 m/s2 (we take downward to be the –y direction). We use the equations in Table 2-1 (with ∆y replacing ∆x)
because this is a = constant motion.
(a) At the highest point the velocity of the ball vanishes. Taking y0 = 0, we set v = 0 in v 2 = v02 − 2 gy , and solve
31
for the initial velocity: v0 = 2 gy . Since y = 50 m we find v0 = 31 m/s.
(b) It will be in the air from the time it leaves the ground until the time it returns to the ground (y = 0). Applying
Eq. 2-15 to the entire motion (the rise and the fall, of total time t > 0) we have
1
2v
y = v0t − gt 2 ⇒ t = 0
2
g
which (using our result from part (a)) produces t = 6.4 s. It is possible to obtain this without using part (a)’s
result; one can find the time just for the rise (from ground to highest point) from Eq. 2-16 and then double it.
(c) SI units are understood in the x and v graphs shown. In the interest of saving space, we do not show the
graph of a, which is a horizontal line at –9.8 m/s2.
49. We neglect air resistance, which justifies setting a = –g = –9.8 m/s2 (taking down as the –y direction) for the
duration of the motion. We are allowed to use Table 2-1 (with ∆y replacing ∆x) because this is constant
acceleration motion. We are placing the coordinate origin on the ground. We note that the initial velocity of the
package is the same as the velocity of the balloon, v0 = +12 m/s and that its initial coordinate is y0 = +80 m.
(a) We solve y = y0 + v0t − 12 gt 2 for time, with y = 0, using the quadratic formula (choosing the positive root to
yield a positive value for t).
t=
b gbg
2
v0 + v02 + 2 gy0 12 + 12 + 2 9.8 80
=
= 5.4 s
g
9.8
(b) If we wish to avoid using the result from part (a), we could use Eq. 2-16, but if that is not a concern, then a
variety of formulas from Table 2-1 can be used. For instance, Eq. 2-11 leads to
v = v0 – gt = 12 – (9.8)(5.4) = – 41 m/s.
Its final speed is 41 m/s.
53. (a) With upward chosen as the +y direction, we use Eq. 2-11 to find the initial velocity of the package:
v = vo + at ⇒ 0 = vo – (9.8 m/s2)(2.0 s)
which leads to vo = 19.6 m/s. Now we use Eq. 2-15:
∆y = (19.6 m/s)(2.0 s) + Erro!(–9.8 m/s2)(2.0 s)2 ≈ 20 m .
32
We note that the “2.0 s” in this second computation refers to the time interval 2 < t < 4 in the graph (whereas
the “2.0 s” in the first computation referred to the 0 < t < 2 time interval shown in the graph).
(b) In our computation for part (b), the time interval (“6.0 s”) refers to the 2 < t < 8 portion of the graph:
∆y = (19.6 m/s)(6.0 s) + Erro!(–9.8 m/s2)(6.0 s)2 ≈ –59 m ,
or | ∆y |= 59 m .
59. We neglect air resistance, which justifies setting a = –g = –9.8 m/s2 (taking down as the –y direction) for the
duration of the motion. We are allowed to use Table 2-1 (with ∆y replacing ∆x) because this is constant
acceleration motion. The ground level is taken to correspond to the origin of the y axis.
(a) The time drop 1 leaves the nozzle is taken as t = 0 and its time of landing on the floor t1 can be computed
from Eq. 2-15, with v0 = 0 and y1 = –2.00 m.
1
−2 y
−2(−2.00 m)
y1 = − gt12 ⇒ t1 =
=
= 0.639 s .
2
g
9.8 m/s 2
At that moment, the fourth drop begins to fall, and from the regularity of the dripping we conclude that drop 2
leaves the nozzle at t = 0.639/3 = 0.213 s and drop 3 leaves the nozzle at t = 2(0.213 s) = 0.426 s. Therefore,
the time in free fall (up to the moment drop 1 lands) for drop 2 is t2 = t1 – 0.213 s = 0.426 s. Its position at the
moment drop 1 strikes the floor is
1
1
y2 = − gt22 = − (9.8 m/s 2 )(0.426 s) 2 = −0.889 m,
2
2
or 89 cm below the nozzle.
(b) The time in free fall (up to the moment drop 1 lands) for drop 3 is t3 = t1 –0.426 s = 0.213 s. Its position at
the moment drop 1 strikes the floor is
1
1
y3 = − gt32 = − (9.8 m/s 2 )(0.213 s)2 = −0.222 m,
2
2
or 22 cm below the nozzle.
z
67. Since v = dx / dt (Eq. 2-4), then ∆x = v dt , which corresponds to the area under the v vs t graph. Dividing
the total area A into rectangular (base × height) and triangular 12 base × height areas, we have
A
b
= A0 <t < 2 + A2 < t <10 + A10 <t <12 + A12 <t <16
=
F
G
H
g
IJ
K
1
1
(2)(8) + (8)(8) + (2)(4) + (2)(4) + (4)(4)
2
2
with SI units understood. In this way, we obtain ∆x = 100 m.
69. The problem is solved using Eq. 2-26:
33
 area between the acceleration curve 
v1 − v0 = 

 and the time axis, from t0 to t1

To compute the speed of the unhelmeted, bare head at t1=7.0 ms, we divide the area under the a vs. t graph into
4 regions: From 0 to 2 ms, region A has the shape of a triangle with area
1
area A = (0.0020 s) (120 m/s 2 ) = 0.12 m/s.
2
From 2 ms to 4 ms, region B has the shape of a trapezoid with area
area B =
1
(0.0020 s) (120 + 140) m/s 2 = 0.26 m/s.
2
From 4 to 6 ms, region C has the shape of a trapezoid with area
area C =
1
(0.0020 s) (140 + 200) m/s 2 = 0.34 m/s.
2
From 6 to 7 ms, region D has the shape of a triangle with area
area D =
1
(0.0010 s) (200 m/s 2 )= 0.10 m/s.
2
Substituting these values into Eq. 2-26, with v0=0 then gives
vunhelmeted = 0.12 m/s + 0.26 m/s + 0.34 m/s + 0.10 m/s = 0.82 m/s.
Carrying out similar calculations for the helmeted head, we have the following results: From 0 to 3 ms, region
A has the shape of a triangle with area
1
(0.0030 s) (40 m/s 2 ) = 0.060 m/s.
2
From 3 ms to 4 ms, region B has the shape of a rectangle with area
area A =
area B = (0.0010 s) (40 m/s 2 )= 0.040 m/s.
From 4 to 6 ms, region C has the shape of a trapezoid with area
area C =
1
(0.0020 s) (40 + 80) m/s 2 = 0.12 m/s.
2
From 6 to 7 ms, region D has the shape of a triangle with area
area D =
1
(0.0010 s) (80 m/s 2 )= 0.040 m/s.
2
34
Substituting these values into Eq. 2-26, with v0=0 then gives
vhelmeted = 0.060 m/s + 0.040 m/s + 0.12 m/s + 0.040 m/s = 0.26 m/s.
Thus, the difference in the speed is
∆v = vunhelmeted − vhelmeted = 0.82 m/s − 0.26 m/s = 0.56 m/s.
35
Chapter 4
21. (a) From Eq. 4-22 (with θ0 = 0), the time of flight is
t=
2h
2(45.0 m)
=
= 3.03 s.
g
9.80 m/s 2
(b) The horizontal distance traveled is given by Eq. 4-21:
∆x = v0t = (250 m/s)(3.03 s) = 758 m.
(c) And from Eq. 4-23, we find
v y = gt = (9.80 m/s 2 )(3.03 s) = 29.7 m/s.
23. Using Eq. (4-26), the take-off speed of the jumper is
v0 =
gR
(9.80 m/s 2 )(77.0 m)
=
= 43.1 m/s
sin 2θ 0
sin 2(12.0°)
25. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are directly
applicable. The initial velocity is horizontal so that v 0 y = 0 and v0 x = v0 = 10 m s.
(a) With the origin at the initial point (where the dart leaves the thrower’s hand), the y coordinate of the dart is
given by y = − 21 gt 2 , so that with y = –PQ we have PQ =
1
2
( 9.8 m/s ) ( 0.19 s )
2
2
= 0.18 m.
(b) From x = v0t we obtain x = (10 m/s)(0.19 s) = 1.9 m.
53. We denote h as the height of a step and w as the width. To hit step n, the ball must fall a distance nh and
travel horizontally a distance between (n – 1)w and nw. We take the origin of a coordinate system to be at the
point where the ball leaves the top of the stairway, and we choose the y axis to be positive in the upward
direction. The coordinates of the ball at time t are given by x = v0xt and y = − 12 gt 2 (since v0y = 0). We equate y
36
to
–nh and solve for the time to reach the level of step n:
t=
2nh
.
g
The x coordinate then is
x = v0 x
2nh
2n(0.203 m)
= (1.52 m/s)
= (0.309 m)
g
9.8 m/s 2
n.
The method is to try values of n until we find one for which x/w is less than n but greater than n – 1. For n = 1, x
= 0.309 m and x/w = 1.52, which is greater than n. For n = 2, x = 0.437 m and x/w = 2.15, which is also greater
than n. For n = 3, x = 0.535 m and x/w = 2.64. Now, this is less than n and greater than n – 1, so the ball hits the
third step.
37
54. We apply Eq. 4-21, Eq. 4-22 and Eq. 4-23.
(a) From ∆x = v0 x t , we find v0 x = 40 m / 2 s = 20 m/s.
(b) From ∆y = v0 y t − 21 gt 2 , we find v0 y = ( 53 m + 12 (9.8 m/s 2 )(2 s) 2 ) / 2 = 36 m/s.
(c)
From
v y = v0 y − gt ′
with
=
vy
0
as
the
condition
for
maximum
height,
we
obtain
t′ = (36 m/s) /(9.8 m/s 2 ) = 3.7 s. During that time the x-motion is constant, so x′ − x0 = (20 m/s)(3.7 s) = 74 m.
55. Let y0 = h0 = 1.00 m at x0 = 0 when the ball is hit. Let y1 = h (the height of the wall) and x1 describe the point
where it first rises above the wall one second after being hit; similarly, y2 = h and x2 describe the point where it
passes back down behind the wall four seconds later. And yf = 1.00 m at xf = R is where it is caught. Lengths are
in meters and time is in seconds.
(a) Keeping in mind that vx is constant, we have x2 – x1 = 50.0 m = v1x (4.00 s), which leads to v1x = 12.5 m/s.
Thus, applied to the full six seconds of motion:
xf – x0 = R = vx(6.00 s) = 75.0 m.
(b) We apply y − y0 = v0 y t − 12 gt 2 to the motion above the wall,
y2 − y1 = 0 = v1 y ( 4.00 s ) −
1
2
g ( 4.00 s )
2
and obtain v1y = 19.6 m/s. One second earlier, using v1y = v0y – g(1.00 s), we find v0 y = 29.4 m/s . Therefore, the
velocity of the ball just after being hit is
r
v = v0 x ˆi + v0 y ˆj = (12.5 m/s) iˆ + (29.4 m/s) ˆj
r
Its magnitude is | v |= (12.5 m/s)2 +(29.4 m/s)2 = 31.9 m/s.
(c) The angle is
38
 vy
 vx
θ = tan −1 

− 1  29.4 m/s 
 = tan 
 = 67.0 ° .
 12.5 m/s 

We interpret this result as a velocity of magnitude 31.9 m/s, with angle (up from rightward) of 67.0°.
(d) During the first 1.00 s of motion, y = y0 + v0 y t − 21 gt 2 yields
h = 1.0 m + ( 29.4 m/s )(1.00 s ) −
1
2
( 9.8 m/s ) (1.00 s )
2
2
= 25.5 m.
56. (a) During constant-speed circular motion, the velocity vector is perpendicular to the acceleration vector at
every instant. Thus, v,
→
· a,
→
= 0.
(b) The acceleration in this vector, at every instant, points towards the center of the circle, whereas the position
vector points from the center of the circle to the object in motion. Thus, the angle between r,
→
→
180º at every instant, so r, × a, = 0.
→
and a,
→
is
57. (a) Since the wheel completes 5 turns each minute, its period is one-fifth of a minute, or 12 s.
(b) The magnitude of the centripetal acceleration is given by a = v2/R, where R is the radius of the wheel, and v
is the speed of the passenger. Since the passenger goes a distance 2πR for each revolution, his speed is
v=
b g
2π 15 m
= 7.85 m / s
12 s
b7.85 m / sg= 4.1 m / s .
and his centripetal acceleration is a =
2
2
15 m
(c) When the passenger is at the highest point, his centripetal acceleration is downward, toward the center of the
orbit.
(d) At the lowest point, the centripetal acceleration is a = 4.1 m/s 2 , same as part (b).
(e) The direction is up, toward the center of the orbit.
58. The magnitude of the acceleration is
39
b
g= 4.0 m / s .
10 m / s
v2
a=
=
r
25 m
2
2
59. We apply Eq. 4-35 to solve for speed v and Eq. 4-34 to find centripetal acceleration a.
(a) v = 2πr/T = 2π(20 km)/1.0 s = 126 km/s = 1.3 × 105 m/s.
(b) The magnitude of the acceleration is
b
g= 7.9 × 10 m / s .
126 km / s
v2
a=
=
r
20 km
2
5
2
(c) Clearly, both v and a will increase if T is reduced.
60. We apply Eq. 4-35 to solve for speed v and Eq. 4-34 to find acceleration a.
(a) Since the radius of Earth is 6.37 × 106 m, the radius of the satellite orbit is
r = (6.37 × 106 + 640 × 103 ) m = 7.01 × 106 m.
Therefore, the speed of the satellite is
c
h
g
2π 7.01 × 106 m
2πr
v=
=
= 7.49 × 103 m / s.
T
98.0 min 60 s / min
b
gb
(b) The magnitude of the acceleration is
c
h= 8.00 m / s .
7.49 × 103 m / s
v2
a=
=
r
7.01 × 106 m
2
2
61. The magnitude of centripetal acceleration (a = v2/r) and its direction (towards the center of the circle) form
the basis of this problem.
r
(a) If a passenger at this location experiences a = 183
. m / s2 east, then the center of the circle is east of this
location. The distance is r = v2/a = (3.66 m/s)2/(1.83 m/s2) = 7.32 m.
40
(b) Thus, relative to the center, the passenger at that moment is located 7.32 m toward the west.
r
(c) If the direction of a experienced by the passenger is now south—indicating that the center of the merry-goround is south of him, then relative to the center, the passenger at that moment is located 7.32 m toward the
north.
62. (a) The circumference is c = 2πr = 2π(0.15 m) = 0.94 m.
(b) With T = (60 s)/1200 = 0.050 s, the speed is v = c/T = (0.94 m)/(0.050 s) = 19 m/s. This is equivalent to
using Eq. 4-35.
(c) The magnitude of the acceleration is a = v2/r = (19 m/s)2/(0.15 m) = 2.4 × 103 m/s2.
(d) The period of revolution is (1200 rev/min)–1 = 8.3 × 10–4 min which becomes, in SI units, T = 0.050 s = 50
ms.
63. Since the period of a uniform circular motion is T = 2π r / v , where r is the radius and v is the speed, the
centripetal acceleration can be written as
2
v 2 1  2π r 
4π 2 r
=
a= = 
.

r r T 
T2
Based on this expression, we compare the (magnitudes) of the wallet and purse accelerations, and find their
ratio is the ratio of r values. Therefore, awallet = 1.50 apurse . Thus, the wallet acceleration vector is
ˆ
a = 1.50[(2.00 m/s 2 )iˆ +(4.00 m/s 2 )j]=(3.00
m/s 2 )iˆ +(6.00 m/s 2 )jˆ .
64. The fact that the velocity is in the +y direction, and the acceleration is in the +x direction at t1= 4.00 s
implies that the motion is clockwise. The position corresponds to the “9:00 position.” On the other hand, the
position at t2=10.0 s is in the “6:00 position” since the velocity points in the -x direction and the acceleration is
in the +y direction. The time interval ∆ t = 10.0 s − 4.00 s = 6.00 s is equal to 3/4 of a period:
3
6.00 s = T ⇒ T = 8.00 s.
4
Eq. 4-35 then yields
r=
vT (3.00 m/s)(8.00 s)
=
= 3.82 m.
2π
2π
41
(a) The x coordinate of the center of the circular path is x = 5.00 m + 3.82 m = 8.82 m.
(b) The y coordinate of the center of the circular path is y = 6.00 m.
In other words, the center of the circle is at (x,y) = (8.82 m, 6.00 m).
→
65. We first note thata1,
(the acceleration at t1 = 2.00 s) is perpendicular to a2,
s), by taking their scalar (dot) product.:
→
(the acceleration at t2=5.00
r r
ˆ
ˆ ⋅ [(4.00 m/s 2 )i+(
ˆ − 6.00 m/s 2 )j]=0.
ˆ
a1 ⋅ a2 = [(6.00 m/s 2 )i+(4.00
m/s 2 )j]
Since the acceleration vectors are in the (negative) radial directions, then the two positions (at t1 and t2) are a
quarter-circle apart (or three-quarters of a circle, depending on whether one measures clockwise or
counterclockwise). A quick sketch leads to the conclusion that if the particle is moving counterclockwise (as
the problem states) then it travels three-quarters of a circumference in moving from the position at time t1 to the
position at time t2 . Letting T stand for the period, then t2 – t1 = 3.00 s = 3T/4. This gives T = 4.00 s. The
magnitude of the acceleration is
a = ax2 + a y2 = (6.00 m/s 2 ) 2 + (4.00 m/s)2 = 7.21 m/s 2 .
Using Eq. 4-34 and 4-35, we have a = 4π 2 r / T 2 , which yields
aT 2 (7.21 m/s 2 )(4.00 s)2
=
= 2.92 m.
r=
4π 2
4π 2
66. When traveling in circular motion with constant speed, the instantaneous acceleration vector necessarily
points towards the center. Thus, the center is “straight up” from the cited point.
(a) Since the center is “straight up” from (4.00 m, 4.00 m), the x coordinate of the center is 4.00 m.
(b) To find out “how far up” we need to know the radius. Using Eq. 4-34 we find
2
v 2 ( 5.00 m/s )
r= =
= 2.00 m.
a
12.5 m/s 2
Thus, the y coordinate of the center is 2.00 m + 4.00 m = 6.00 m. Thus, the center may be written as (x, y) =
42
(4.00 m, 6.00 m).
67. To calculate the centripetal acceleration of the stone, we need to know its speed during its circular motion
(this is also its initial speed when it flies off). We use the kinematic equations of projectile motion (discussed in
§4-6) to find that speed. Taking the +y direction to be upward and placing the origin at the point where the stone
leaves its circular orbit, then the coordinates of the stone during its motion as a projectile are given by x = v0t
and y = − 21 gt 2 (since v0y = 0). It hits the ground at x = 10 m and y = –2.0 m. Formally solving the second
equation for the time, we obtain t = −2 y / g , which we substitute into the first equation:
9.8 m / s2
−
= 15.7 m / s.
2 −2.0 m
b g b
g
v0 = x −
= 10 m
2y
g
Therefore, the magnitude of the centripetal acceleration is
b
g= 160 m / s .
15.7 m / s
v2
a=
=
r
15
. m
2
2
68. We note that after three seconds have elapsed (t2 – t1 = 3.00 s) the velocity (for this object in circular motion
of period T ) is reversed; we infer that it takes three seconds to reach the opposite side of the circle. Thus, T =
2(3.00 s) = 6.00 s.
(a) Using Eq. 4-35, r = vT/2π, where v = (3.00 m/s) 2 + (4.00 m/s)2 = 5.00 m/s , we obtain r = 4.77 m . The
magnitude of the object’s centripetal acceleration is therefore a = v2/r = 5.24 m/s2.
(b) The average acceleration is given by Eq. 4-15:
r r
r
v2 − v1 (− 3.00iˆ − 4.00ˆj) m/s − (3.00iˆ + 4.00ˆj) m/s
ˆ − 2.67 m/s 2 )ˆj
aavg =
=
= (− 2.00 m/s 2 )i+(
t2 − t1
5.00 s − 2.00 s
r
which implies | aavg |= (− 2.00 m/s 2 )2 + (− 2.67 m/s 2 )2 = 3.33 m/s 2 .
69. We use Eq. 4-15 first using velocities relative to the truck (subscript t) and then using velocities relative to
the ground (subscript g). We work with SI units, so 20 km / h → 5.6 m / s , 30 km / h → 8.3 m / s , and
45 km / h → 12.5 m / s . We choose east as the + $i direction.
43
(a) The velocity of the cheetah (subscript c) at the end of the 2.0 s interval is (from Eq.
4-44)
r
r
r
vc t = vc g − vt g = (12.5 m/s) ˆi − (−5.6 m/s) iˆ = (18.1 m/s) iˆ
relative to the truck. Since the velocity of the cheetah relative to the truck at the beginning of the 2.0 s interval
is (−8.3 m/s)iˆ , the (average) acceleration vector relative to the cameraman (in the truck) is
r
(18.1 m/s)iˆ − ( −8.3 m/s)iˆ
ˆ
aavg =
= (13 m/s 2 )i,
2.0 s
r
or | aavg |= 13 m/s 2 .
r
(b) The direction of aavg is +iˆ , or eastward.
(c) The velocity of the cheetah at the start of the 2.0 s interval is (from Eq. 4-44)
r
r
r
v0cg = v0ct + v0tg = ( −8.3 m/s)iˆ + ( − 5.6 m/s)iˆ = ( −13.9 m/s)iˆ
relative to the ground. The (average) acceleration vector relative to the crew member (on the ground) is
r
(12.5 m/s)iˆ − ( −13.9 m/s)iˆ
ˆ |ar |= 13 m/s 2
aavg =
= (13 m/s 2 )i,
avg
2.0 s
identical to the result of part (a).
r
(d) The direction of aavg is +iˆ , or eastward.
70. We use Eq. 4-44, noting that the upstream corresponds to the +iˆ direction.
(a) The subscript b is for the boat, w is for the water, and g is for the ground.
r
r
r
ˆ
vbg = vbw + vwg = (14 km/h) ˆi + (−9 km/h) ˆi = (5 km/h) i.
44
r
Thus, the magnitude is | vbg |= 5 km/h.
r
(b) The direction of vbg is +x, or upstream.
(c) We use the subscript c for the child, and obtain
r
r
r
vc g = vc b + vb g = ( −6 km / h) $i + (5 km / h) $i = ( −1 km / h) $i .
r
The magnitude is | vcg |= 1 km/h.
r
(d) The direction of vcg is −x, or downstream.
71. While moving in the same direction as the sidewalk’s motion (covering a distance d relative to the ground in
time t1 = 2.50 s), Eq. 4-44 leads to
vsidewalk + vman running = Erro! .
While he runs back (taking time t2 = 10.0 s) we have
vsidewalk − vman running = − Erro! .
Dividing these equations and solving for the desired ratio, we get Erro! = Erro! = 1.67.
r
72. We denote the velocity of the player with vPF and
relative velocity between the player and the ball be
r
the velocity vBF of the ball relative to the field is given
r
r
r
vBF = vPF + vBP . The smallest angle θmin corresponds to
r
r
when vBF ⊥ vPF . Hence,
θ min = 180° − cos
−1
r
 | vPF | 
−1  4.0 m/s 
 r  = 180° − cos 
 = 130°.
 6.0 m/s 
 | vBP | 
the
r
vBP . Then
by
the
case
45
73. The velocity vectors (relative to the shore) for ships A and B are given by
r
v A = − (v A cos 45°) ˆi + (v A sin 45°) ˆj
r
vB = − (vB sin 40°) ˆi − (vB cos 40°) ˆj,
with vA = 24 knots and vB = 28 knots. We take east as + $i and north as $j .
(a) Their relative velocity is
r
r r
v A B = v A − vB = (vB sin 40° − v A cos 45°) ˆi + (vB cos 40° + v A sin 45°) ˆj
r
the magnitude of which is | v A B |= (1.03 knots) 2 + (38.4 knots)2 ≈ 38 knots.
r
(b) The angle θ which v A B makes with north is given by
 v AB , x
 v AB , y

θ = tan −1 

−1  1.03 knots 
 = tan 
 = 1.5°
38.4
knots



r
which is to say that v A B points 1.5° east of north.
(c) Since they started at the same time, their relative velocity describes at what rate the distance between them
is increasing. Because the rate is steady, we have
| ∆r | 160
t = r AB =
= 4.2 h.
| v AB | 38.4
r
(d) The velocity v A B does not change with time in this problem, and
they started at the same time. Reversing the points of view, we have
r
r
rA B is in the same direction as v A B since
r
r
r
r
v A B = − v B A so that rA B = − rB A (i.e., they
are 180° opposite to each other). Hence, we conclude that B stays at a bearing of 1.5° west of south relative to A
during the journey (neglecting the curvature of Earth).
→
74. The destination is D,
^
= 800 km j, where we orient axes so that +y points north and +x points east. This
takes two hours, so the (constant) velocity of the plane (relative to the ground) is vpg,
→
^
= (400 km/h) j, .
46
This must be the vector sum of the plane’s velocity with respect to the air which has (x,y) components
(500cos70º, 500sin70º) and the velocity of the air (wind) relative to the ground vag,
^
^
^
→
. Thus,
(400 km/h) j, = (500 km/h) cos70º i, + (500 km/h) sin70º j, + vag,
→
which yields
vag,
→
^
^
=( –171 km/h)i, –( 70.0 km/h)j, .
r
r
(a) The magnitude of vag is | vag |= (− 171 km/h) 2 + (− 70.0 km/h)2 = 185 km/h.
r
(b) The direction of vag is
 − 70.0 km/h 
 = 22.3 ° (south of west).
 − 171 km/h 
θ = tan −1 
75. Relative to the car the velocity of the snowflakes has a vertical component of 8.0 m/s and a horizontal
component of 50 km/h = 13.9 m/s. The angle θ from the vertical is found from
v 13.9 m/s
tan θ = h =
= 1.74
vv
8.0 m/s
which yields θ = 60°.
76. Velocities are taken to be constant; thus, the velocity of the plane relative to the ground is
r
vPG = (55 km)/(1/4 hour) ˆj= (220 km/h)jˆ . In addition,
r
ˆ
v AG = (42 km/h)(cos 20 °ˆi − sin 20 °ˆj) = (39 km/h)iˆ − (14 km/h)j.
r
r
r
Using vPG = vPA + v AG , we have
r
r
r
ˆ
vPA = vPG − vAG = −(39 km/h)iˆ + (234 km/h)j.
r
which implies | vPA |= 237 km/h , or 240 km/h (to two significant figures.)
77. Since the raindrops fall vertically relative to the train, the horizontal component of the velocity of a raindrop
is vh = 30 m/s, the same as the speed of the train. If vv is the vertical component of the velocity and θ is the
47
angle between the direction of motion and the vertical, then tan θ = vh/vv. Thus vv = vh/tan θ = (30 m/s)/tan
70° = 10.9 m/s. The speed of a raindrop is
v = vh2 + vv2 = ( 30 m / s) 2 + (10.9 m / s) 2 = 32 m / s .
78. This is a classic problem involving two-dimensional relative motion. We align our coordinates so that east
r
r
r
corresponds to +x and north corresponds to +y. We write the vector addition equation as v BG = v BW + vWG . We
r
r
have vWG = (2.0∠0° ) in the magnitude-angle notation (with the unit m/s understood), or vWG = 2.0$i in unitr
vector notation. We also have v BW = (8.0∠120° ) where we have been careful to phrase the angle in the
r
ˆ
ˆ m/s.
‘standard’ way (measured counterclockwise from the +x axis), or vBW = ( −4.0i+6.9j)
r
(a) We can solve the vector addition equation for v BG :
r
r
ˆ
ˆ m/s = ( − 2.0 m/s)iˆ + (6.9 m/s) ˆj.
vBG = vBW + vWG = (2.0 m/s) ˆi + ( − 4.0i+6.9j)
r
Thus, we find | v BG | = 7.2 m/s.
r
(b) The direction of vBG is θ = tan −1[(6.9 m/s) /(− 2.0 m/s)] = 106°(measured counterclockwise from the +x
axis), or 16° west of north.
(c) The velocity is constant, and we apply y – y0 = vyt in a reference frame. Thus, in the ground reference frame,
we have (200 m) = (7.2 m/s)sin(106°)t → t = 29 s. Note: if a student obtains “28 s”, then the student has
probably neglected to take the y component properly (a common mistake).
79. We denote the police and the motorist with subscripts p and m, respectively. The coordinate system is
indicated in Fig. 4-49.
(a) The velocity of the motorist with respect to the police car is
r
r r
vm p = vm − v p = (− 60 km/h)ˆj − (− 80 km/h)iˆ = (80 km/h) ˆi − (60 km/h)ˆj.
r
(b) vm p does happen to be along the line of sight. Referring to Fig. 4-49, we find the vector pointing from one
48
r
r
car to another is r = (800 m)iˆ − (600 m) ˆj (from M to P). Since the ratio of components in r is the same as in
r
vm p , they must point the same direction.
(c) No, they remain unchanged.
80. We make use of Eq. 4-44 and Eq. 4-45.
The velocity of Jeep P relative to A at the instant is
r
vPA = (40.0 m/s)(cos 60°ˆi + sin 60°ˆj) = (20.0 m/s)iˆ + (34.6 m/s)ˆj.
Similarly, the velocity of Jeep B relative to A at the instant is
r
vBA = (20.0 m/s)(cos 30°ˆi + sin 30°ˆj) = (17.3 m/s)iˆ + (10.0 m/s)ˆj.
Thus, the velocity of P relative to B is
r
r
r
ˆ
vPB = vPA − vBA = (20.0iˆ + 34.6ˆj) m/s − (17.3iˆ + 10.0ˆj) m/s = (2.68 m/s)iˆ + (24.6 m/s)j.
r
r
(a) The magnitude of vPB is | vPB |= (2.68 m/s)2 + (24.6 m/s)2 = 24.8 m/s.
r
(b) The direction of vPB is θ = tan −1[(24.6 m/s) /(2.68 m/s)] = 83.8° north of east (or 6.2º east of north).
(c) The acceleration of P is
r
aPA = (0.400 m/s 2 )(cos 60.0°ˆi + sin 60.0°ˆj) = (0.200 m/s 2 )iˆ + (0.346 m/s 2 ) ˆj,
r
r
r
and aPA = aPB . Thus, we have | aPB |= 0.400 m/s 2 .
(d) The direction is 60.0° north of east (or 30.0° east of north).
49
81. Here, the subscript W refers to the water. Our coordinates are chosen with +x being east and +y being
north. In these terms, the angle specifying east would be 0° and the angle specifying south would be –90° or
270°. Where the length unit is not displayed, km is to be understood.
r
r
r
(a) We have v A W = v A B + v B W , so that
r
v A B = (22 ∠ – 90°) – (40 ∠ 37°) = (56 ∠ – 125°)
in the magnitude-angle notation (conveniently done with a vector-capable calculator in polar mode).
Converting to rectangular components, we obtain
r
v A B = ( −32km/h) ˆi − (46 km/h) ˆj .
Of course, this could have been done in unit-vector notation from the outset.
(b) Since the velocity-components are constant, integrating them to obtain the position is straightforward
r r
r
(r − r0 = v dt )
z
r
r = (2.5 − 32t ) ˆi + (4.0 − 46t ) ˆj
with lengths in kilometers and time in hours.
r
(c) The magnitude of this r is r = ( 2.5 − 32t ) 2 + ( 4.0 − 46t ) 2 . We minimize this by taking a derivative and
requiring it to equal zero — which leaves us with an equation for t
dr 1
6286t − 528
=0
=
dt 2 (2.5 − 32t ) 2 + (4.0 − 46t ) 2
which yields t = 0.084 h.
(d) Plugging this value of t back into the expression for the distance between the ships (r), we obtain r = 0.2
km. Of course, the calculator offers more digits (r = 0.225…), but they are not significant; in fact, the
uncertainties implicit in the given data, here, should make the ship
captains worry.
82. We construct a right triangle starting from the clearing on the
drawing a line (200 m long) due north (upward in our sketch) across
a line due west (upstream, leftward in our sketch) along the north
south
bank,
the river, and then
bank for a distance
50
(82 m) + (1.1 m/s)t , where the t-dependent contribution is the distance that the river will carry the boat
downstream during time t.
The hypotenuse of this right triangle (the arrow in our sketch) also depends on t and on the boat’s speed
(relative to the water), and we set it equal to the Pythagorean “sum” of the triangle’s sides:
b4.0gt =
b
2002 + 82 + 11
.t
g
2
which leads to a quadratic equation for t
46724 + 180.4t − 14.8t 2 = 0.
We solve this and find a positive value: t = 62.6 s.
The angle between the northward (200 m) leg of the triangle and the hypotenuse (which is measured “west of
north”) is then given by
θ = tan −1
82 + 11
. tI
F
F151 I
= tan G J= 37° .
G
J
H200 K H200 K
−1
83. Using displacement = velocity × time (for each constant-velocity part of the trip), along with the fact that 1
hour = 60 minutes, we have the following vector addition exercise (using notation appropriate to many vector
capable calculators):
(1667 m ∠ 0º) + (1333 m ∠ −90º) + (333 m ∠ 180º) + (833 m ∠ −90º) + (667 m ∠ 180º) + (417 m ∠ −90º) =
(2668 m ∠ −76º).
(a) Thus, the magnitude of the net displacement is 2.7 km.
(b) Its direction is 76° clockwise (relative to the initial direction of motion).
84. We compute the coordinate pairs (x, y) from x = (v0 cosθ )t and y = v0 sin θ t − 12 gt 2 for t = 20 s and the
speeds and angles given in the problem.
(a) We obtain
( xA , y A ) = (10.1 km, 0.56 km )
( xC , yC ) = (14.3 km, 2.68 km )
( xB , yB ) = (12.1 km, 1.51 km )
( xD , yD ) = (16.4 km, 3.99 km )
51
and (xE, yE) = (18.5 km, 5.53 km) which we plot in the next part.
(b) The vertical (y) and horizontal (x) axes are in kilometers. The graph does not start at the origin. The curve
to “fit” the data is not shown, but is easily imagined (forming the “curtain of death”).
85. Let vo = 2π(0.200 m)/(0.00500 s) ≈ 251 m/s (using Eq. 4-35) be the speed it had in circular motion and θo =
(1 hr)(360º/12 hr [for full rotation]) = 30.0º. Then Eq. 4-25 leads to
y = (2.50 m) tan 30.0° −
(9.8 m/s 2 )(2.50 m) 2
≈ 1.44 m
2(251 m/s)2 (cos 30.0°)2
which means its height above the floor is 1.44 m + 1.20 m = 2.64 m.
r
86. For circular motion, we must have v with direction perpendicular to r,
→
and (since the speed is constant)
magnitude v = 2π r / T where r = (2.00 m)2 + (−3.00 m)2 and T = 7.00 s . The r,
→
(given in the problem
statement) specifies a point in the fourth quadrant, and since the motion is clockwise then the velocity must
have both components negative. Our result, satisfying these three conditions, (using unit-vector notation which
r r
r
makes it easy to double-check that r ⋅ v = 0 ) for v = (–2.69 m/s)i,^ + (–1.80 m/s)j,^.
87. Using Eq. 2-16, we obtain v 2 = v02 − 2 gh , or h = (v02 − v 2 ) / 2 g .
(a)
Since
v = 0 at
the
maximum
height
of
an
upward
motion,
with
v0 = 7.00 m/s ,
we
have h = (7.00 m/s)2 / 2(9.80 m/s2 ) = 2.50 m.
(b) The relative speed is vr = v0 − vc = 7.00 m/s − 3.00 m/s = 4.00 m/s with respect to the floor. Using the above
equation we obtain h = (4.00 m/s)2 / 2(9.80 m/s 2 ) = 0.82 m.
(c) The acceleration, or the rate of change of speed of the ball with respect to the ground is 9.80 m/s2
(downward).
52
(d) Since the elevator cab moves at constant velocity, the rate of change of speed of the ball with respect to
the cab floor is also 9.80 m/s2 (downward).
→
^
^
88. Relative to the sled, the launch velocity is vo rel,
= vox i, + voy j, . Since the sled’s motion is in the
negative direction with speed vs (note that we are treating vs as a positive number, so the sled’s velocity is
^
actually –vs i, ), then the launch velocity relative to the ground is vo,
horizontal and vertical displacement (relative to the ground) are therefore
→
= (vox – vs) i,
^
^
+ voy j, . The
xland – xlaunch = ∆xbg = (vox – vs) tflight
yland – ylaunch = 0 = voy tflight + Erro!(−g)(tflight)2 .
Combining these equations leads to
∆xbg = Erro! – Erro!vs.
The first term corresponds to the “y intercept” on the graph, and the second term (in parentheses) corresponds
to the magnitude of the “slope.” From Figure 4-54, we have
∆ xbg = 40 − 4vs .
This implies voy = (4.0 s)(9.8 m/s2)/2 = 19.6 m/s, and that furnishes enough information to determine vox.
(a) vox = 40g/2voy = (40 m)(9.8 m/s2)/(39.2 m/s) = 10 m/s.
(b) As noted above, voy = 19.6 m/s.
(c) Relative to the sled, the displacement ∆xbs does not depend on the sled’s speed, so ∆xbs = vox tflight = 40 m.
(d) As in (c), relative to the sled, the displacement ∆xbs does not depend on the sled’s speed, and ∆xbs = vox tflight
= 40 m.
89. We establish coordinates with $i pointing to the far side of the river (perpendicular to the current) and $j
pointing in the direction of the current. We are told that the magnitude (presumed constant) of the velocity of
r
the boat relative to the water is | vbw | = 6.4 km/h. Its angle, relative to the x axis is θ. With km and h as the
r
ˆ
understood units, the velocity of the water (relative to the ground) is vwg = (3.2 km/h)j.
53
r
(a) To reach a point “directly opposite” means that the velocity of her boat relative to ground must be vbg = vbg ˆi
r
r
r
where vbg > 0 is unknown. Thus, all $j components must cancel in the vector sum vbw + vwg = vbg , which
means the vrbw sin θ = (–3.2 km/h) $j , so
θ = sin–1 [(–3.2 km/h)/(6.4 km/h)] = –30°.
(b) Using the result from part (a), we find vbg = vbw cosθ = 5.5 km/h. Thus, traveling a distance of l = 6.4 km
requires a time of (6.4 km)/(5.5 km/h) = 1.15 h or 69 min.
(c) If her motion is completely along the y axis (as the problem implies) then with vwg = 3.2 km/h (the water
speed) we have
ttotal =
vbw
D
D
+
= 1.33 h
+ vwg
vbw − vwg
where D = 3.2 km. This is equivalent to 80 min.
(d) Since
D
D
D
D
+
=
+
vbw +vwg vbw − vwg
vbw − vwg vbw + vwg
the answer is the same as in the previous part, i.e., ttotal = 80 min .
(e) The shortest-time path should have θ = 0°. This can also be shown by noting that the case of general θ leads
to
r
r
r
vbg = vbw + vwg = vbw cosθ ˆi + (vbwsin θ + vwg ) ˆj
r
where the x component of vbg must equal l/t. Thus,
t =
which can be minimized using dt/dθ = 0.
l
vbw cos θ
54
(f) The above expression leads to t = (6.4 km)/(6.4 km/h) = 1.0 h, or 60 min.
90. We use a coordinate system with +x eastward and +y upward.
(a) We note that 123° is the angle between the initial position and later position vectors, so that the angle from
+x to the later position vector is 40° + 123° = 163°. In unit-vector notation, the position vectors are
r
r1 = (360 m)cos(40°) ˆi + (360 m)sin(40°) ˆj = (276 m)iˆ + (231 m) ˆj
r
r2 = (790 m) cos(163°) ˆi + (790 m) sin(163°) ˆj = ( − 755 m )iˆ + (231 m )jˆ
respectively. Consequently, we plug into Eq. 4-3
r
∆r = [( − 755 m) − (276 m)]iˆ + (231 m − 231 m) ˆj = −(1031 m) ˆi.
r
r
The magnitude of the displacement ∆ r is | ∆ r | = 1031 m.
r
(b) The direction of ∆ r is −î , or westward.
91. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are directly
applicable.
(a) With the origin at the firing point, the y coordinate of the bullet is given by y = − 21 gt 2 . If t is the time of
flight and y = – 0.019 m indicates where the bullet hits the target, then
t=
2 ( 0.019 m )
= 6.2 × 10−2 s.
9.8 m/s2
(b) The muzzle velocity is the initial (horizontal) velocity of the bullet. Since x = 30 m is the horizontal position
of the target, we have x = v0t. Thus,
v0 =
x
30 m
=
= 4.8 × 102 m/s.
−2
t 6.3 × 10 s
92. Eq. 4-34 describes an inverse proportionality between r and a, so that a large acceleration results from a
small radius. Thus, an upper limit for a corresponds to a lower limit for r.
55
(a) The minimum turning radius of the train is given by
b
g
b gc h
2
rmin
216 km / h
v2
=
=
= 7.3 × 103 m.
2
a max
0.050 9.8 m / s
(b) The speed of the train must be reduced to no more than
v = amax r = 0.050 ( 9.8 m/s 2 )(1.00 × 103 m ) = 22 m/s
which is roughly 80 km/h.
93. (a) With r = 0.15 m and a = 3.0 × 1014 m/s2, Eq. 4-34 gives
v = ra = 6.7 × 106 m / s.
(b) The period is given by Eq. 4-35:
T=
2πr
= 1.4 × 10 −7 s.
v
94. We use Eq. 4-2 and Eq. 4-3.
r
r
(a) With the initial position vector as r1 and the later vector as r2 , Eq. 4-3 yields
∆r = [( − 2.0 m) − 5.0 m]iˆ + [(6.0 m) − ( − 6.0 m)]jˆ + (2.0 m − 2.0 m) kˆ = (−7.0 m) ˆi + (12 m) ˆj
for the displacement vector in unit-vector notation.
(b) Since there is no z component (that is, the coefficient of k̂ is zero), the displacement vector is in the xy
plane.
b g
95. We write our magnitude-angle results in the form R ∠ θ with SI units for the magnitude understood (m
for distances, m/s for speeds, m/s2 for accelerations). All angles θ are measured counterclockwise from +x, but
we will occasionally refer to angles φ which are measured counterclockwise from the vertical line between the
circle-center and the coordinate origin and the line drawn from the circle-center to the particle location (see r in
the figure). We note that the speed of the particle is v = 2πr/T where r = 3.00 m and T = 20.0 s; thus, v = 0.942
m/s. The particle is moving counterclockwise in Fig. 4-56.
56
(a) At t = 5.0 s, the particle has traveled a fraction of
t 5.00 s 1
=
=
T 20.0 s 4
of a full revolution around the circle (starting at the origin). Thus, relative to the circle-center, the particle is at
1
4
φ = (360° ) = 90°
measured from vertical (as explained above). Referring to Fig. 4-56, we see that this position (which is the “3
o’clock” position on the circle) corresponds to x = 3.0 m and y = 3.0 m relative to the coordinate origin. In our
magnitude-angle notation, this is expressed as ( R ∠θ ) = ( 4.2 ∠ 45° ) . Although this position is easy to analyze
without resorting to trigonometric relations, it is useful (for the computations below) to note that these values of
x and y relative to coordinate origin can be gotten from the angle φ from the relations
x = r sin φ ,
y = r − r cos φ .
Of course, R = x 2 + y 2 and θ comes from choosing the appropriate possibility from tan–1 (y/x) (or by using
particular functions of vector-capable calculators).
(b) At t = 7.5 s, the particle has traveled a fraction of 7.5/20 = 3/8 of a revolution around the circle (starting at
the origin). Relative to the circle-center, the particle is therefore at φ = 3/8 (360°) = 135° measured from vertical
in the manner discussed above. Referring to Fig. 4-56, we compute that this position corresponds to
x = (3.00 m)sin 135° = 2.1 m
y = (3.0 m) – (3.0 m)cos 135° = 5.1 m
relative to the coordinate origin. In our magnitude-angle notation, this is expressed as (R ∠ θ ) = (5.5 ∠ 68°).
(c) At t = 10.0 s, the particle has traveled a fraction of 10/20 = 1/2 of a revolution around the circle. Relative to
the circle-center, the particle is at φ = 180° measured from vertical (see explanation, above). Referring to Fig. 456, we see that this position corresponds to x = 0 and y = 6.0 m relative to the coordinate origin. In our
magnitude-angle notation, this is expressed as ( R ∠ θ ) = ( 6.0 ∠ 90° ) .
(d) We subtract the position vector in part (a) from the position vector in part (c):
57
( 6.0 ∠ 90° ) − ( 4.2 ∠ 45° ) = ( 4.2 ∠135° )
using magnitude-angle notation (convenient when using vector-capable calculators). If we wish instead to use
unit-vector notation, we write
r
∆R = (0 − 3.0 m) ˆi + (6.0 m − 3.0 m) ˆj = (−3.0 m)iˆ + (3.0 m)ˆj
r
which leads to | ∆R |= 4.2 m and θ = 135°.
r
r
(e) From Eq. 4-8, we have vavg = ∆R / ∆t . With ∆t = 5.0 s , we have
r
vavg = (−0.60 m/s) ˆi + (0.60 m/s) ˆj
in unit-vector notation or (0.85 ∠ 135°) in magnitude-angle notation.
(f) The speed has already been noted (v = 0.94 m/s), but its direction is best seen by referring again to Fig. 4-56.
r
The velocity vector is tangent to the circle at its “3 o’clock position” (see part (a)), which means v is vertical.
Thus, our result is ( 0.94 ∠ 90° ) .
(g) Again, the speed has been noted above (v = 0.94 m/s), but its direction is best seen by referring to Fig. 4-56.
r
The velocity vector is tangent to the circle at its “12 o’clock position” (see part (c)), which means v is
horizontal. Thus, our result is ( 0.94 ∠ 180° ) .
(h) The acceleration has magnitude a = v2/r = 0.30 m/s2, and at this instant (see part (a)) it is horizontal (towards
the center of the circle). Thus, our result is ( 0.30 ∠ 180° ) .
(i) Again, a = v2/r = 0.30 m/s2, but at this instant (see part (c)) it is vertical (towards the center of the circle).
Thus, our result is ( 0.30 ∠ 270° ) .
r
96. Noting that v2 = 0 , then, using Eq. 4-15, the average acceleration is
58
(
)
r
ˆ
ˆ
r
∆v 0 − 6.30 i − 8.42 j m/s
aavg =
=
= −2.1iˆ + 2.8 ˆj m/s 2
∆t
3s
(
)
r
97. (a) The magnitude of the displacement vector ∆r is given by
r
| ∆r | = (21.5 km) 2 + (9.7 km) 2 + (2.88 km) 2 = 23.8 km.
Thus,
r
r
| ∆r | 23.8 km
| vavg | =
=
= 6.79 km/h.
∆t
3.50 h
(b) The angle θ in question is given by


 = 6.96°.
 (21.5 km)2 + (9.7 km)2 


2.88 km
θ = tan −1 
98. The initial velocity has magnitude v0 and because it is horizontal, it is equal to vx the horizontal component
of velocity at impact. Thus, the speed at impact is
v02 + v y2 = 3v0
where vy = 2 gh and we have used Eq. 2-16 with ∆x replaced with h = 20 m. Squaring both sides of the first
equality and substituting from the second, we find
bg
v02 + 2 gh = 3v0
2
which leads to gh = 4v02 and therefore to v0 = (9.8 m/s2 )(20 m) / 2 = 7.0 m/s.
r
99. We choose horizontal x and vertical y axes such that both components of v0 are positive. Positive angles are
counterclockwise from +x and negative angles are clockwise from it. In unit-vector notation, the velocity at
each instant during the projectile motion is
r
v = v0 cos θ 0 ˆi + ( v0 sin θ 0 − gt ) ˆj.
59
r
r
(a) With v0 = 30 m/s and θ0 = 60°, we obtain v = (15 ˆi +6.4 ˆj) m/s , for t = 2.0 s. The magnitude of v is
r
| v |= (15 m/s)2 + (6.4 m/s)2 = 16 m/s.
r
(b) The direction of v is
θ = tan − 1[(6.4 m/s) /(15 m/s)] = 23 °,
measured counterclockwise from +x.
(c) Since the angle is positive, it is above the horizontal.
r
(d) With t = 5.0 s, we find v = (15 iˆ − 23 ˆj) m/s , which yields
r
| v |= (15 m/s)2 + (− 23 m/s)2 = 27 m/s.
r
(e) The direction of v is θ = tan −1[(− 23 m/s) /(15 m/s)] = − 57 ° , or 57± measured clockwise from +x.
(f) Since the angle is negative, it is below the horizontal.
100. The velocity of Larry is v1 and that of Curly is v2. Also, we denote the length of the corridor by L. Now,
Larry’s time of passage is t1 = 150 s (which must equal L/v1), and Curly’s time of passage is t2 = 70 s (which
must equal L/v2). The time Moe takes is therefore
L
1
1
t=
=
= 1
= 48s.
v1 + v2 v1 / L + v2 / L 150 s + 701 s
101. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are directly
applicable. The coordinate origin is at the initial position for the football as it begins projectile motion in the
sense of §4-5), and we let θ0 be the angle of its initial velocity measured from the +x axis.
(a) x = 46 m and y = –1.5 m are the coordinates for the landing point; it lands at time t = 4.5 s. Since x = v0xt,
v0 x =
Since y = v0 y t − 21 gt 2 ,
x 46 m
=
= 10.2 m / s.
t 4.5 s
60
y+
v0 y =
1 2
1
gt
( −15
. m) + (9.8 m / s2 )(4.5 s) 2
2
2
=
= 217
. m / s.
t
4.5 s
The magnitude of the initial velocity is
v0 = v02 x + v02 y = (10.2 m / s) 2 + (21.7 m / s) 2 = 24 m / s.
(b) The initial angle satisfies tan θ0 = v0y/v0x. Thus, θ0 = tan–1 [(21.7 m/s)/(10.2 m/s) ]= 65°.
102. We assume the ball’s initial velocity is perpendicular to the plane of the net. We choose coordinates so that
(x0, y0) = (0, 3.0) m, and vx > 0 (note that v0y = 0).
(a) To (barely) clear the net, we have
y − y0 = v0 y t −
1 2
1
gt ⇒ 2.24 m − 3.0 m = 0 − ( 9.8 m/s 2 ) t 2
2
2
which gives t = 0.39 s for the time it is passing over the net. This is plugged into the x-equation to yield the
(minimum) initial velocity vx = (8.0 m)/(0.39 s) = 20.3 m/s.
(b) We require y = 0 and find t from y − y0 = v0 y t − 12 gt 2 . This value (t = 2 ( 3.0 m ) /(9.8 m/s2 ) = 0.78 s) is
plugged into the x-equation to yield the (maximum) initial velocity vx = (17.0 m)/(0.78 s) = 21.7 m/s.
103. (a) With ∆x = 8.0 m, t = ∆t1, a = ax , and vox = 0, Eq. 2-15 gives
8.0 m = Erro!ax(∆t1)2 ,
and the corresponding expression for motion along the y axis leads to
∆y = 12 m = Erro!ay(∆t1)2 .
Dividing the second expression by the first leads to a y / ax = 3 / 2 = 1.5.
(b) Letting t = 2∆t1, then Eq. 2-15 leads to ∆x = (8.0 m)(2)2 = 32 m, which implies that its x coordinate is now
(4.0 + 32) m = 36 m. Similarly, ∆y = (12 m)(2)2 = 48 m, which means its y coordinate has become (6.0 + 48) m
= 54 m.
61
104. We apply Eq. 4-34 to solve for speed v and Eq. 4-35 to find the period T.
(a) We obtain
v = ra =
b5.0 mgb7.0gc9.8 m / s h= 19 m / s.
2
(b) The time to go around once (the period) is T = 2πr/v = 1.7 s. Therefore, in one minute (t = 60 s), the
astronaut executes
t 60 s
=
= 35
T 1.7 s
revolutions. Thus, 35 rev/min is needed to produce a centripetal acceleration of 7g when the radius is 5.0 m.
(c) As noted above, T = 1.7 s.
105. The radius of Earth may be found in Appendix C.
(a) The speed of an object at Earth’s equator is v = 2πR/T, where R is the radius of Earth (6.37 × 106 m) and T is
the length of a day (8.64 × 104 s):
v = 2π(6.37 × 106 m)/(8.64 × 104 s) = 463 m/s.
The magnitude of the acceleration is given by
b
g
2
463 m / s
v2
a=
=
= 0.034 m / s2 .
6
R 6.37 × 10 m
(b) If T is the period, then v = 2πR/T is the speed and the magnitude of the acceleration is
a=
v 2 (2π R / T )2 4π 2 R
=
=
.
R
R
T2
Thus,
R
T = 2π
= 2π
a
6.37 × 106 m
= 51
. × 103 s = 84 min.
2
9.8 m / s
62
106. When the escalator is stalled the speed of the person is v p = l t , where l is the length of the escalator
and t is the time the person takes to walk up it. This is vp = (15 m)/(90 s) = 0.167 m/s. The escalator moves at ve
= (15 m)/(60 s) = 0.250 m/s. The speed of the person walking up the moving escalator is
v = vp + ve = 0.167 m/s + 0.250 m/s = 0.417 m/s
and the time taken to move the length of the escalator is
t = l / v = (15 m) / (0.417 m / s) = 36 s.
If the various times given are independent of the escalator length, then the answer does not depend on that
length either. In terms of l (in meters) the speed (in meters per second) of the person walking on the stalled
escalator is l 90 , the speed of the moving escalator is l 60 , and the speed of the person walking on the moving
escalator is v = ( l 90 ) + ( l 60 ) = 0.0278l . The time taken is t = l v = l 0.0278l = 36 s and is independent of l .
107. (a) Eq. 2-15 can be applied to the vertical (y axis) motion related to reaching the maximum height (when t
= 3.0 s and vy = 0):
ymax – y0 = vyt – Erro!gt2 .
With ground level chosen so y0 = 0, this equation gives the result ymax = Erro! g(3.0 s)2 = 44 m.
(b) After the moment it reached maximum height, it is falling; at t = 2.5 s, it will have fallen an amount given
by Eq. 2-18:
yfence – ymax = (0)(2.5 s) – Erro! g(2.5 s)2
which leads to yfence = 13 m.
(c) Either the range formula, Eq. 4-26, can be used or one can note that after passing the fence, it will strike the
ground in 0.5 s (so that the total "fall-time" equals the "rise-time"). Since the horizontal component of velocity
in a projectile-motion problem is constant (neglecting air friction), we find the original x-component from 97.5
m = v0x(5.5 s) and then apply it to that final 0.5 s. Thus, we find v0x = 17.7 m/s and that after the fence
∆x = (17.7 m/s)(0.5 s) = 8.9 m.
108. With gB = 9.8128 m/s2 and gM = 9.7999 m/s2, we apply Eq. 4-26:
63
R M − RB =
F
IJ
G
H K
v02 sin 2θ 0 v02 sin 2θ 0 v02 sin 2θ 0 g B
−
=
−1
gM
gB
gB
gM
which becomes
 9.8128 m/s 2

RM − RB = RB 
− 1
2
 9.7999 m/s

and yields (upon substituting RB = 8.09 m) RM – RB = 0.01 m = 1 cm.
109. We make use of Eq. 4-25.
(a) By rearranging Eq. 4-25, we obtain the initial speed:
v0 =
x
cosθ 0
g
2( x tan θ 0 − y )
which yields v0 = 255.5 ≈ 2.6 × 102 m/s for x = 9400 m, y = –3300 m, and θ0 = 35°.
(b) From Eq. 4-21, we obtain the time of flight:
t=
x
9400 m
=
= 45 s.
v0 cos θ 0 (255.5 m/s) cos 35°
(c) We expect the air to provide resistance but no appreciable lift to the rock, so we would need a greater
launching speed to reach the same target.
110. When moving in the same direction as the jet stream (of speed vs), the time is
t1 = Erro!,
where d = 4000 km is the distance and vja is the speed of the jet relative to the air (1000 km/h). When moving
against the jet stream, the time is
t2 = Erro! ,
where t2 – t1 = Erro!h . Combining these equations and using the quadratic formula to solve gives vs = 143
km/h.
111. Since the x and y components of the acceleration are constants, we can use Table 2-1 for the motion along
64
both axes. This can be handled individually (for ∆x and ∆y) or together with the unit-vector notation (for ∆r).
Where units are not shown, SI units are to be understood.
r
(a) Since r0 = 0 , the position vector of the particle is (adapting Eq. 2-15)
r r
1r
1
r = v0t + at 2 = 8.0 ˆj t + 4.0 ˆi + 2.0 ˆj t 2 = 2.0t 2 ˆi + 8.0t + 1.0t 2 ˆj.
2
2
( )
(
)
(
) (
)
Therefore, we find when x = 29 m, by solving 2.0t2 = 29, which leads to t = 3.8 s. The y coordinate at that time
is y = (8.0 m/s)(3.8 s) + (1.0 m/s2)(3.8 s)2 = 45 m.
(b) Adapting Eq. 2-11, the velocity of the particle is given by
r r r
v = v0 + at .
Thus, at t = 3.8 s, the velocity is
r
v = (8.0 m/s) ˆj + (4.0 m/s 2 ) ˆi + (2.0 m/s 2 ) ˆj ( 3.8 s ) = (15.2 m/s) ˆi + ( 15.6 m/s) ˆj
(
)
which has a magnitude of
v = vx2 + v y2 = (15.2 m/s) 2 + (15.6 m/s)2 = 22 m/s.
112. We make use of Eq. 4-34 and Eq. 4-35.
(a) The track radius is given by
r=
v 2 (9.2 m/s)2
=
= 22 m .
a
3.8 m/s2
(b) The period of the circular motion is T = 2π(22 m)/(9.2 m/s) = 15 s.
113. Since this problem involves constant downward acceleration of magnitude a, similar to the projectile
motion situation, we use the equations of §4-6 as long as we substitute a for g. We adopt the positive direction
choices used in the textbook so that equations such as Eq. 4-22 are directly applicable. The initial velocity is
horizontal so that v 0 y = 0 and
65
v0 x = v0 = 1.00 × 109 cm/s.
(a) If l is the length of a plate and t is the time an electron is between the plates, then l = v0 t , where v0 is the
initial speed. Thus
t=
l
2.00 cm
=
= 2.00 × 10− 9 s.
9
v0
1.00 × 10 cm/s
(b) The vertical displacement of the electron is
2
1
1
y = − at 2 = − (1.00 × 1017 cm/s 2 )( 2.00 × 10 −9 s ) = − 0.20 cm = − 2.00 mm,
2
2
or | y | = 2.00 mm.
(c) The x component of velocity does not change: vx = v0 = 1.00 × 109 cm/s = 1.00 × 107 m/s.
(d) The y component of the velocity is
v y = a y t = (1.00 × 1017 cm/s2 )( 2.00 × 10− 9 s ) = 2.00 × 108 cm/s = 2.00 × 106 m/s.
114. We neglect air resistance, which justifies setting a = –g = –9.8 m/s2 (taking down as the –y direction) for
the duration of the motion of the shot ball. We are allowed to use Table 2-1 (with ∆y replacing ∆x) because the
ball has constant acceleration motion. We use primed variables (except t) with the constant-velocity elevator (so
v ' = 10 m/s ), and unprimed variables with the ball (with initial velocity v0 = v′ + 20 = 30 m/s , relative to the
ground). SI units are used throughout.
(a) Taking the time to be zero at the instant the ball is shot, we compute its maximum height y (relative to the
ground) with v 2 = v02 − 2 g ( y − y0 ) , where the highest point is characterized by v = 0. Thus,
y = yo +
v02
= 76 m
2g
where yo = yo′ + 2 = 30 m (where yo′ = 28 m is given in the problem) and v0 = 30 m/s relative to the ground as
noted above.
(b) There are a variety of approaches to this question. One is to continue working in the frame of reference
adopted in part (a) (which treats the ground as motionless and “fixes” the coordinate origin to it); in this case,
66
one describes the elevator motion with y ′ = yo′ + v′t and the ball motion with Eq. 2-15, and solves them for
the case where they reach the same point at the same time. Another is to work in the frame of reference of the
elevator (the boy in the elevator might be oblivious to the fact the elevator is moving since it isn’t accelerating),
which is what we show here in detail:
1
∆ye = v0e t − gt 2
2
2
⇒
t=
v0e + v0 e − 2 g∆ye
g
where v0e = 20 m/s is the initial velocity of the ball relative to the elevator and ∆ye =
–2.0 m is the ball’s displacement relative to the floor of the elevator. The positive root is chosen to yield a
positive value for t; the result is t = 4.2 s.
115. (a) With v = c /10 = 3 ×107 m/s and a = 20 g = 196 m/s2 , Eq. 4-34 gives
r = v 2 / a = 4.6 ×1012 m.
(b) The period is given by Eq. 4-35: T = 2π r / v = 9.6 ×105 s. Thus, the time to make a quarter-turn is T/4 = 2.4 ×
105 s or about 2.8 days.
116. Using the same coordinate system assumed in Eq. 4-25, we rearrange that equation to solve for the initial
speed:
v0 =
x
cos θ 0
g
2 ( x tan θ 0 − y )
which yields v0 = 23 ft/s for g = 32 ft/s2, x = 13 ft, y = 3 ft and θ0 = 55°.
r
117. The (box)car has velocity vc g = v1 $i relative to the ground, and the bullet has velocity
r
v0b g = v2 cosθ $i + v2 sin θ $j
relative to the ground before entering the car (we are neglecting the effects of gravity on the bullet). While in
r
the car, its velocity relative to the outside ground is vbg = 0.8v2 cosθ $i + 0.8v2 sin θ $j (due to the 20% reduction
mentioned in the problem). The problem indicates that the velocity of the bullet in the car relative to the car is
r
(with v3 unspecified) vb c = v3 $j . Now, Eq. 4-44 provides the condition
r
r
r
vb g = vb c + vc g
0.8v2 cosθ ˆi + 0.8v2 sin θ ˆj = v3 ˆj + v1 ˆi
67
so that equating x components allows us to find θ. If one wished to find v3 one could also equate the y
components, and from this, if the car width were given, one could find the time spent by the bullet in the car,
but this information is not asked for (which is why the width is irrelevant). Therefore, examining the x
components in SI units leads to
1000 m/km
 v1 
−1  85 km/h ( 3600 s/h ) 
=
cos



 0.8v2 
 0.8 (650 m/s) 
θ = cos−1 
r
which yields 87° for the direction of vb g (measured from $i , which is the direction of motion of the car). The
problem asks, “from what direction was it fired?” — which means the answer is not 87° but rather its
supplement 93° (measured from the direction of motion). Stating this more carefully, in the coordinate system
we have adopted in our solution, the bullet velocity vector is in the first quadrant, at 87° measured
counterclockwise from the +x direction (the direction of train motion), which means that the direction from
which the bullet came (where the sniper is) is in the third quadrant, at
–93° (that is, 93° measured clockwise from +x).
2
118. Since v y2 = v0 y − 2 g ∆y , and vy=0 at the target, we obtain
(
)
v0 y = 2 9.80 m/s 2 ( 5.00 m ) = 9.90 m/s
(a) Since v0 sin θ0 = v0y, with v0 = 12.0 m/s, we find θ0 = 55.6°.
(b) Now, vy = v0y – gt gives t = (9.90 m/s)/(9.80 m/s2) = 1.01 s. Thus, ∆x = (v0 cos θ0)t = 6.85 m.
(c) The velocity at the target has only the vx component, which is equal to v0x = v0 cos θ0 = 6.78 m/s.
119. From the figure, the three displacements can be written as
68
r
d1 = d1 (cos θ1ˆi + sin θ1ˆj) = (5.00 m)(cos 30°iˆ + sin 30°ˆj) = (4.33 m)iˆ + (2.50 m)ˆj
r
ˆ = (8.00 m)(cos160°ˆi + sin160°ˆj)
d 2 = d 2 [cos(180° + θ1 − θ 2 )iˆ + sin(180° + θ1 − θ 2 )j]
= (−7.52 m)iˆ + (2.74 m)ˆj
r
ˆ = (12.0 m)(cos 260°ˆi + sin 260°ˆj)
d 3 = d3 [cos(360° − θ3 − θ 2 + θ1 )iˆ + sin(360° − θ3 − θ 2 + θ1 )j]
= (−2.08 m)iˆ − (11.8 m)ˆj
where the angles are measured from the +x axis. The net displacement is
r r r r
d = d1 + d 2 + d 3 = (− 5.27 m)iˆ − (6.58 m)ˆj.
(a) The magnitude of the net displacement is
r
| d |= (− 5.27 m) 2 + (− 6.58 m) 2 = 8.43 m.
r
(b) The direction of d is
 dy 
− 1  − 6.58 m 
 = tan 
 = 51.3 ° or 231° .
 − 5.27 m 
 dx 
θ = tan −1 
We choose 231° (measured counterclockwise from +x) since the desired angle is in the third quadrant. An
equivalent answer is − 129 ° (measured clockwise from +x).
120. With v0 = 30.0 m/s and R = 20.0 m, Eq. 4-26 gives
sin 2θ 0 =
gR
= 0.218.
v02
Because sin φ = sin (180° – φ), there are two roots of the above equation:
2θ 0 = sin −1 (0.218) = 12.58° and 167.4°.
which correspond to the two possible launch angles that will hit the target (in the absence of air friction and
related effects).
(a) The smallest angle is θ0 = 6.29°.
69
(b) The greatest angle is and θ0 = 83.7°.
An alternative approach to this problem in terms of Eq. 4-25 (with y = 0 and 1/cos2 = 1 + tan2) is possible —
and leads to a quadratic equation for tanθ0 with the roots providing these two possible θ0 values.
121. On the one hand, we could perform the vector addition of the displacements with a vector-capable
calculator in polar mode ((75 ∠ 37°) + (65 ∠ − 90°) = (63 ∠ − 18°)), but in keeping with Eq. 3-5 and Eq. 3-6
we will show the details in unit-vector notation. We use a ‘standard’ coordinate system with +x East and +y
North. Lengths are in kilometers and times are in hours.
(a) We perform the vector addition of individual displacements to find the net displacement of the camel.
r
∆r1 = (75 km)cos(37°) ˆi + (75 km) sin(37°) ˆj
r
∆r2 =( − 65 km) ˆj
r
r
r
∆r = ∆r1 + ∆r2 = (60 km) ˆi − (20 km )ˆj .
If it is desired to express this in magnitude-angle notation, then this is equivalent to a vector of length
r
| ∆r |= (60 km) 2 +( − 20 km)2 = 63 km .
r
(b) The direction of ∆ r is θ = tan −1[(− 20 km) /(60 km)] = − 18 ° , or 18° south of east.
(c) We use the result from part (a) in Eq. 4-8 along with the fact that ∆t = 90 h. In unit vector notation, we
obtain
r
(60 iˆ − 20 ˆj) km
vavg =
= (0.67 ˆi − 0.22 ˆj) km/h.
90 h
r
This leads to | vavg | = 0.70 km/h.
r
(d) The direction of vavg is θ = tan −1[(− 0.22 km/h) /(0.67 km/h)] = − 18 ° , or 18° south of east.
(e) The average speed is distinguished from the magnitude of average velocity in that it depends on the total
distance as opposed to the net displacement. Since the camel travels 140 km, we obtain (140 km)/(90 h) = 1.56
km/h ≈ 1.6 km/h .
70
(f) The net displacement is required to be the 90 km East from A to B. The displacement from the resting
r
place to B is denoted ∆r3 . Thus, we must have
r
r
r
∆r1 + ∆r2 + ∆r3 = (90 km) ˆi
r
which produces ∆r3 = (30 km)iˆ + (20 km)jˆ in unit-vector notation, or (36 ∠ 33° ) in magnitude-angle notation.
Therefore, using Eq. 4-8 we obtain
r
| vavg | =
36 km
= 1.2 km/h.
(120 − 90) h
r
r
(g) The direction of vavg is the same as r3 (that is, 33° north of east).
122. We make use of Eq. 4-21 and Eq.4-22.
(a) With vo = 16 m/s, we square Eq. 4-21 and Eq. 4-22 and add them, then (using Pythagoras’ theorem) take the
square root to obtain r:
r = ( x − x0 ) 2 + ( y − y0 )2 = (v0 cos θ 0t ) 2 + (v0 sin θ 0t − gt 2 / 2)2
= t v02 − v0 g sin θ 0t + g 2 t 2 / 4
Below we plot r as a function of time for θo = 40.0º:
(b) For this next graph for r versus t we set θo = 80.0º.
71
(c) Differentiating r with respect to t, we obtain
dr v02 − 3v0 gt sin θ 0 / 2 + g 2t 2 / 2
=
dt
v02 − v0 g sin θ 0 t + g 2t 2 / 4
Setting dr / dt = 0 , with v0 = 16.0 m/s and θ 0 = 40.0 ° , we have 256 − 151t + 48t 2 = 0 . The equation has no
real solution. This means that the maximum is reached at the end of the flight, with
ttotal = 2v0 sin θ 0 / g = 2(16.0 m/s) sin(40.0°) /(9.80 m/s 2 ) = 2.10 s.
(d) The value of r is given by
r = (2.10) (16.0)2 − (16.0)(9.80)sin 40.0°(2.10) + (9.80)2 (2.10)2 / 4 = 25.7 m.
(e) The horizontal distance is rx = v0 cos θ0 t = (16.0 m/s) cos 40.0°(2.10 s) = 25.7 m.
(f) The vertical distance is ry = 0 .
(g) For the θ0 = 80º launch, the condition for maximum r is 256 − 232t + 48t 2 = 0 , or t = 1.71 s (the other
solution, t = 3.13 s, corresponds to a minimum.)
(h) The distance traveled is
r = (1.71) (16.0)2 − (16.0)(9.80) sin 80.0°(1.71) + (9.80) 2 (1.71)2 / 4 = 13.5 m.
72
(i) The horizontal distance is
rx = v0 cos θ 0 t = (16.0 m/s) cos80.0°(1.71 s) = 4.75 m.
(j) The vertical distance is
ry = v0 sin θ 0t −
gt 2
(9.80 m/s 2 )(1.71s)2
= (16.0 m/s) sin 80°(1.71 s) −
= 12.6 m.
2
2
123. Using the same coordinate system assumed in Eq. 4-25, we find x for the elevated cannon from
y = x tan θ 0 −
b
gx 2
2 v0 cos θ 0
g
2
where y = −30 m.
Using the quadratic formula (choosing the positive root), we find
Fv sin θ
x = v cosθ G
G
H
0
0
0
bv sin θ g− 2 gy IJ
JK
g
2
+
0
0
0
which yields x = 715 m for v0 = 82 m/s and θ0 = 45°. This is 29 m longer than the 686 m found in that Sample
Problem. Since the “9” in 29 m is not reliable, due to the low level of precision in the given data, we write the
answer as 3 × 101 m.
124. (a) Using the same coordinate system assumed in Eq. 4-25, we find
y = x tan θ 0 −
b
gx 2
2 v0 cosθ 0
=−
g
2
gx 2
2v02
if θ 0 = 0.
Thus, with v0 = 3.0 × 106 m/s and x = 1.0 m, we obtain y = –5.4 × 10–13 m which is not practical to measure
(and suggests why gravitational processes play such a small role in the fields of atomic and subatomic physics).
(b) It is clear from the above expression that |y| decreases as v0 is increased.
125. At maximum height, the y-component of a projectile’s velocity vanishes, so the given 10 m/s is the
(constant) x-component of velocity.
(a) Using v0y to denote the y-velocity 1.0 s before reaching the maximum height, then (with vy = 0) the equation
vy = v0y – gt leads to v0y = 9.8 m/s. The magnitude of the velocity vector (or speed) at that moment is therefore
73
vx2 + v0 y 2 = (10 m/s) 2 + (9.8 m/s)2 = 14 m/s.
(b) It is clear from the symmetry of the problem that the speed is the same 1.0 s after reaching the top, as it was
1.0 s before (14 m/s again). This may be verified by using vy = v0y – gt again but now “starting the clock” at the
highest point so that v0y = 0 (and t = 1.0 s ). This leads to vy = –9.8 m/s and
2
(10 m/s)2 + ( −9.8 m/s ) = 14 m/s .
(c) The x0 value may be obtained from x = 0 = x0 + (10 m/s)(1.0s), which yields x0 = −10 m.
(d) With v0y = 9.8 m/s denoting the y-component of velocity one second before the top of the trajectory, then we
have y = 0 = y0 + v0 y t −
1
2
gt 2 where t = 1.0 s. This yields y0 = −4.9 m.
(e) By using x – x0 = (10 m/s)(1.0 s) where x0 = 0, we obtain x = 10 m.
(f) Let t = 0 at the top with y0 = v0 y = 0 . From y − y0 = v0 y t − 12 gt 2 , we have, for t = 1.0 s,
y = −(9.8 m/s 2 )(1.0 s)2 / 2 = −4.9 m.
126. With no acceleration in the x direction yet a constant acceleration of 1.4 m/s2 in the y direction, the
position (in meters) as a function of time (in seconds) must be
F
G
H
IJ
K
r
1
r = (6.0t ) $i +
(14
. )t 2 $j
2
r
and v is its derivative with respect to t.
r
(a) At t = 3.0 s, therefore, v = (6.0iˆ + 4.2ˆj) m/s.
r
ˆ m.
(b) At t = 3.0 s, the position is r = (18iˆ + 6.3j)
127. We note that
r
r
r
v PG = v PA + v AG
r
r
describes a right triangle, with one leg being v PG (east), another leg being v AG (magnitude = 20, direction =
74
r
south), and the hypotenuse being v PA (magnitude = 70). Lengths are in kilometers and time is in hours. Using
the Pythagorean theorem, we have
r
r
r
r
vPA = |vPG |2 + |v AG |2 ⇒ 70 km/h = | vPG |2 + (20 km/h) 2
r
which is easily solved for the ground speed: | v PG | = 67 km / h.
128. The figure offers many interesting points to analyze, and others are easily inferred (such as the point of
maximum height). The focus here, to begin with, will be the final point shown (1.25 s after the ball is released)
which is when the ball returns to its original height. In English units, g = 32 ft/s2.
(a) Using x – x0 = vxt we obtain vx = (40 ft)/(1.25 s) = 32 ft/s. And y − y0 = 0 = v0 y t − 12 gt 2 yields
v0 y =
1
2
( 32 ft/s ) (1.25 s ) = 20 ft/s. Thus, the initial speed is
2
r
v0 = |v0 |= (32 ft/s)2 + (20 ft/s)2 = 38 ft/s.
(b) Since vy = 0 at the maximum height and the horizontal velocity stays constant, then the speed at the top is
the same as vx = 32 ft/s.
(c) We can infer from the figure (or compute from v y = 0 = v0 y − gt ) that the time to reach the top is 0.625 s.
With this, we can use y − y0 = v0 y t − 21 gt 2 to obtain 9.3 ft (where y0 =
3 ft has been used). An alternative approach is to use v 2y = v02 y − 2 g ( y − y0 ) .
r
129. We denote v PG as the velocity of the plane relative to the
r
the velocity of the air relative to the ground, and v PA as the
ground,
r
v AG
as
velocity
of
the
plane relative to the air.
r
r
r
(a) The vector diagram is shown on the right: v PG = v PA + v AG .
Since
the
magnitudes vPG and vPA are equal the triangle is isosceles, with
length.
two sides of equal
Consider either of the right triangles formed when the bisector
r
dashed line). It bisects v AG , so
of θ is drawn (the
75
sin (θ / 2 ) =
vAG
70.0 mi/h
=
2vPG 2 (135 mi/h )
r
which leads to θ = 30.1°. Now v AG makes the same angle with the E-W line as the dashed line does with the
N-S line. The wind is blowing in the direction 15.0° north of west. Thus, it is blowing from 75.0° east of south.
r
(b) The plane is headed along v PA , in the direction 30.0° east of north. There is another solution, with the plane
headed 30.0° west of north and the wind blowing 15° north of east (that is, from 75° west of south).
r
130. Taking derivatives of r = 2t ˆi + 2sin(π t / 4)ˆj (with lengths in meters, time in seconds and angles in radians)
provides expressions for velocity and acceleration:
r
π
r dr
πt 
v=
= 2iˆ + cos   ˆj
dt
2
 4
r
2
r dv
π
 πt 
a=
= − sin   ˆj.
dt
8
 4
Thus, we obtain:
time t
→
(a)
r,
position
(b)
v,
velocity
(c)
a,
acceleration
→
→
0.0
1.0
2.0
3.0
4.0
x
0.0
2.0
4.0
6.0
8.0
y
0.0
1.4
2.0
1.4
0.0
vx
2.0
2.0
2.0
vy
1.1
0.0
−1.1
ax
0.0
0.0
0.0
ay
−0.87
−1.2
−0.87
And the path of the particle in the xy plane is shown in the following graph. The arrows indicating the
velocities are not shown here, but they would appear as tangent-lines, as expected.
131. We make use of Eq. 4-24 and Eq. 4-25.
(a) With x = 180 m, θo = 30º, and vo = 43 m/s, we obtain
y = tan(30°)(180 m) −
(9.8 m/s 2 )(180 m)2
= −11 m
2(43 m/s)2 (cos 30°) 2
76
or | y | = 11 m . This implies the rise is roughly eleven meters above the fairway.
(b) The horizontal component (in the absence of air friction) is unchanged, but the vertical component increases
(see Eq. 4-24). The Pythagorean theorem then gives the magnitude of final velocity (right before striking the
ground): 45 m/s.
132. We let gp denote the magnitude of the gravitational acceleration on the planet. A number of the points on
the graph (including some “inferred” points — such as the max height point at x = 12.5 m and t = 1.25 s) can be
analyzed profitably; for future reference, we label (with subscripts) the first ((x0, y0) = (0, 2) at t0 = 0) and last
(“final”) points ((xf, yf) = (25, 2) at tf = 2.5), with lengths in meters and time in seconds.
(a) The x-component of the initial velocity is found from xf – x0 = v0x tf. Therefore, v0 x = 25 / 2.5 = 10 m/s. And
we try to obtain the y-component from y f − y0 = 0 = v0 y t f − 12 g pt 2f . This gives us v0y = 1.25gp, and we see we
need another equation (by analyzing another point, say, the next-to-last one) y − y0 = v0 y t − 12 g pt 2 with y = 6
and t = 2; this produces our second equation v0y = 2 + gp. Simultaneous solution of these two equations
produces results for v0y and gp (relevant to part (b)). Thus, our complete answer for the initial velocity is
r
v = (10 m/s)iˆ + (10 m/s)jˆ .
(b) As a by-product of the part (a) computations, we have gp = 8.0 m/s2.
(c) Solving for tg (the time to reach the ground) in y g = 0 = y0 + v0 y t g −
1
2
g p t g2 leads to a positive answer: tg =
2.7 s.
(d) With g = 9.8 m/s2, the method employed in part (c) would produce the quadratic equation
−4.9t g2 + 10t g + 2 = 0 and then the positive result tg = 2.2 s.