Name:_________________________________________
22. The velocity-time graph for an object
moving along a straight path is shown in
Figure P2.22. (a) Find the average
accelerations of this object during the time
intervals
0 to 5.0 s
5.0 s to 15 s
0 to 20 s.
IB 1D motion review
32. A jet plane lands with a speed of 100
m/s and can accelerate at a maximum rate
of –5.00 m/s2 as it comes to rest. (a) From
the instant the plane touches the runway,
what is the minimum time needed before it
can come to rest?
(b) Find the instantaneous accelerations at
2.0 s________ 10 s________18 s________
(b) Can this plane land on a small tropical
island airport where the runway is 0.800 km
long?
48. A ball thrown vertically upward is
caught by the thrower after 2.00 s. Find
25. Jules Verne in 1865 proposed sending
men to the Moon by firing a space capsule
from a 220-m-long cannon with final speed
of 10.97 km/s. What would have been the
unrealistically large acceleration
experienced by the space travelers during
launch? (A human can stand an acceleration
of 15g for a short time.)
(a) the initial velocity of the ball and
(b) the maximum height it reaches.
26. A truck covers 40.0 m in 8.50 s while
smoothly slowing down to final speed
2.80 m/s. Find its original speed
49. A model rocket is launched straight
upward with an initial speed of 50.0 m/s. It
accelerates with a constant upward
acceleration of 2.00 m/s2 until its engines
stop at an altitude of 150 m.
(a) What is the maximum height reached by
the rocket?
63. Using a rocket pack with full throttle, a
lunar astronaut accelerates upward from
the Moon’s surface with a constant
acceleration of 2.00 m/s2. At a height of
5.00 m, a bolt comes loose. (The free-fall
acceleration on the Moon’s surface is about
1.67 m/s2.)
(a) How fast is the astronaut moving at that
time?
(b) How long after lift-off does the rocket
reach its maximum height?
(b) When will the bolt hit the Moon’s
surface?
(c) How long is the rocket in the air?
(c) How fast will it be moving then?
51. A student throws a set of keys vertically
upward to her sorority sister, who is in a
window 4.00 m above. The keys are caught
1.50 s later by the sister’s outstretched
hand.
(a) With what initial velocity were the keys
thrown?
(b) What was the velocity of the keys just
before they were caught?
2.22
(a) From t  0 to t  5.0 s ,
a
v f  vi
t f  ti

2.32
00
 0 .
5.0 s  0
(b) The minimum distance needed to
stop the plane is
From t  5.0 s to t  15 s ,
a
8.0 m s   8.0 m s 
15 s  5.0 s

 v f  vi 
 0  100 m s 
x  vt  
t 
  20.0 s   1000


2
 2 
1.00 km .
1.6 m s 2 ,
and from t  0 to t  20 s ,
a
8.0 m s   8.0 m s 
20 s  0

Thus, the plane cannot stop in 0.8
km.
0.80 m s 2 .
(b) At t  2.0 s , the slope of the
2.48
tangent line to the curve is 0 .
At t  10 s , the slope of the
2
tangent line is 1.6 m s , and
at t  18 s , the slope of the
tangent line is 0 .
2.25
3

m s 2  0  2 a 220 m  so
that a  2.74  105 m s 2 which is
2.79  104 times g!
2.26
(a)
 v f  vi 
x  v  t   
t
 2 
becomes
 2.80m s  v i 
40.0 m  
  8.50 s  ,

2
which yields v i  6.61 m s .
(a) Consider the relation
1
y  v i t  at 2 with a   g .
2
When the ball is at the throwers
hand, the displacement y is
1
zero, or 0  v i t  gt 2 . This
2
equation has two solutions,
t  0 which corresponds to
when the ball was thrown, and
t  2v i g corresponding to
when the ball is caught.
Therefore, if the ball is caught at
t  2.00 s , the initial velocity
must have been
From v 2f  v i2  2 a x  , we have
 10.97  10
(a) The time required to stop is
v f  v i 0  100 m s
t


a
 5.00 m s 2
20.0 s .


9.80 m s 2  2.00 s 
gt


2
2
9.80 m s .
vi 
(b)
From v 2f  v i2  2a y  , with v f  0 at
the maximum height,
(b)
v f  vi
2.80 m s  6.61 m s
a


t
8.50 s
 0.448 m s 2
 y max 
4.90 m .
v 2f  v i2
2a

0   9.80 m s 

2
2  9.80 m s 2


2.49
(a) When it reaches a height of 150
m, the speed of the rocket is
(b) The total time of the upward
motion of the rocket is the sum
of two intervals. The first is the
time for the rocket to go from
2
v  50.0 m s at the ground to a
v f  v i2  2a y    50.0 m s   2  2.00 m s 2  150 mi   55.7 m s
velocity of v f  55.7 m s at an
.
altitude of 150 m. This time is
given by
After the engines stop, the
rocket continues moving
upward with an initial velocity
of v i  55.7 m s and
acceleration a   g  9.80 m s
. When the rocket reaches
maximum height, v f  0 . The
2
displacement of the rocket
above the point where the
engines stopped (i.e., above the
150 m level) is
y 
v 2f  v i2
2a

0   55.7 m s 

2  9.80 m s 2
t1 

.
The maximum height above
ground that the rocket reaches
is then given by
hmax  150 m  158 m= 308 m .
v1

2 150 m 
 2.84 s
 55.7  50.0 m s
.
The second interval is the time
to rise 158 m starting with
v i  55.7 m s and ending with
v f  0 . This time is
2
 158 m
 y 1
t1 
 y 2
v2

2 158 m 
 5.67 s .
0+55.7 m s
The total time of the upward
flight is then
t up  t1  t 2   2.84  5.67  s 
8.51 s .
2.51
(a) The keys have acceleration
a   g  -9.80 m s 2 from the
release point until they are
caught 1.50 s later. Thus,
1
y  v i t  at 2 gives
2
2.63
(a) At 5.00 m above the surface, the
velocity of the astronaut is
given by


v f  vi2  2 a y   0  2 2.00 m s 2  5.00 m  
 4.47 m s .


2
 2 The bolt begins free-fall with an
y  at 2 2   4.00 m    9.80 m s 1.50 s (b)
vi 

  10.0 m s
initial velocity of
t
1.50 s
v f  4.47 m s at 5.00 m above
,
2
the surface. When it reaches the
1
surface, y  v i t  at 2 gives
2
(b) The velocity of the keys just
1
5.00 m   4.47 m s  t  1.67 m s 2 t 2
before the catch was
2
which has a positive solution of
t  6.31 s . Thus, the bolt hits
v f  vi  at  10.0 m s   9.80 m s 2 1.50 s    4.68 m
s s after it is released .
6.31
,
(c) When it reaches the surface, the
velocity of the bolt is
or 4.68 m s downward .
or v i  10.0 m s upward .






v f  vi  at   4.47 m s  1.67 m s 2 6.31 s  
 6.06 m s .