PROBLEM 11.34
A motorist is traveling at 54 km/h when she
observes that a traffic light 240 m ahead of her
turns red. The traffic light is timed to stay red for
24 s. If the motorist wishes to pass the light
without stopping just as it turns green again,
determine (a) the required uniform deceleration of
the car, (b) the speed of the car as it passes the
light.
SOLUTION
Uniformly accelerated motion:
x0  0
(a)
x  x0  v0t 
v0  54 km/h  15 m/s
1 2
at
2
when t  24s, x  240 m:
240 m  0  (15 m/s)(24 s) 
1
a (24 s) 2
2
a  0.4167 m/s 2
(b)
a  0.417 m/s 2 
v  v0  a t
when t  24s:
v  (15 m/s)  ( 0.4167 m/s)(24 s)
v  5.00 m/s
v  18.00 km/h
v  18.00 km/h 
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PROBLEM 11.38
A sprinter in a 100-m race accelerates uniformly for the first 35 m and then runs with
constant velocity. If the sprinter’s time for the first 35 m is 5.4 s, determine (a) his
acceleration, (b) his final velocity, (c) his time for the race.
SOLUTION
0  x  35 m, a  constant
Given:
35 m  x  100 m, v  constant
At t  0, v  0 when
x  35 m, t  5.4 s
Find:
(a)
a
(b)
v when x  100 m
(c)
t when x  100 m
(a)
We have
At t  5.4 s:
or
x  0  0t 
35 m 
1 2
at
2
for
0  x  35 m
1
a(5.4 s)2
2
a  2.4005 m/s 2
a  2.40 m/s 2 
(b)
First note that v  vmax for 35 m  x  100 m.
Now
(c)
v 2  0  2a( x  0)
for
0  x  35 m
When x  35 m:
2
vmax
 2(2.4005 m/s 2 )(35 m)
or
vmax  12.9628 m/s
We have
When x  100 m:
or
x  x1  v0 (t  t1 )
vmax  12.96 m/s 
for
35 m  x  100 m
100 m  35 m  (12.9628 m/s)(t 2  5.4) s
t 2  10.41 s 
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PROBLEM 11.59
The system shown starts from rest, and each component moves with a constant
acceleration. If the relative acceleration of block C with respect to collar B is 60
mm/s 2 upward and the relative acceleration of block D with respect to block A is
110 mm/s 2 downward, determine (a) the velocity of block C after 3 s, (b) the
change in position of block D after 5 s.
SOLUTION
From the diagram
2 y A  2 y B  yC  constant
Cable 1:
Then
2v A  2vB  vC  0
(1)
and
2 a A  2 a B  aC  0
(2)
( y D  y A )  ( y D  y B )  constant
Cable 2:
Then
and
(3)
 a A  aB  2aD  0
(4)
At t  0, v  0; all accelerations constant;
aC/B  60 mm/s2 , aD /A  110 mm/s 2
Given:
(a)
 v A  vB  2vD  0
We have
aC/B  aC  a B  60
or
a B  aC  60
and
a D/A  a D  a A  110
or
a A  a D  110
Substituting into Eqs. (2) and (4)
Eq. (2):
or
Eq. (4):
or
2( a D  110)  2( aC  60)  aC  0
3aC  2 a D  100
(5)
 ( a D  110)  ( aC  60)  2 a D  0
 aC  a D  50
(6)
PROBLEM 11.59 (Continued)
Solving Eqs. (5) and (6) for aC and a D
aC  40 mm/s2
aD  10 mm/s 2
Now
vC  0  aC t
At t  3 s:
vC  (40 mm/s 2 )(3 s)
v C  120.0 mm/s 
or
(b)
We have
At t  5 s:
or
yD  ( yD )0  (0)t 
yD  ( yD )0 
1
aD t 2
2
1
(10 mm/s 2 )(5 s)2
2
y D  ( y D ) 0  125.0 mm 
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PROBLEM 11.97
An airplane used to drop water on brushfires is flying
horizontally in a straight line at 180 mi/h at an altitude of
300 ft. Determine the distance d at which the pilot should
release the water so that it will hit the fire at B.
SOLUTION
First note
v0  180 km/h  264 ft/s
Place origin of coordinates at Point A.
Vertical motion. (Uniformly accelerated motion)
y  0  (0)t 
1 2
gt
2
1
300 ft   (32.2 ft/s 2 )t 2
2
At B:
t B  4.31666 s
or
Horizontal motion. (Uniform)
x  0  (vx )0 t
At B:
or
d  (264 ft/s)(4.31666 s)
d  1140 ft 
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PROBLEM 11.99
A baseball pitching machine
“throws”
baseballs
with
a
horizontal velocity v0. Knowing
that height h varies between 788
mm and 1068 mm, determine
(a) the range of values of v0, (b) the
values of  corresponding to
h  788 mm and h  1068 mm.
SOLUTION
(a)
y0  1.5 m, (v y )0  0
Vertical motion:
or
t 
2( y0  y)
g
y h
or
tB 
2( y0  h)
g
When h  788 mm  0.788 m,
tB 
(2)(1.5  0.788)
 0.3810 s
9.81
When h  1068 mm  1.068 m,
tB 
(2)(1.5  1.068)
 0.2968 s
9.81
y  y0  (v y )0 t 
At Point B,
Horizontal motion:
1 2
gt
2
x0  0, (vx )0  v0 ,
x  v0t
or
v0 
x
x
 B
t
tB
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PROBLEM 11.99 (Continued)
v0 
12.2
 32.02 m/s
0.3810
and
v0 
12.2
 41.11 m/s
0.2968
32.02 m/s  v0  41.11 m/s
or
Vertical motion:
v y  (v y )0  gt   gt
Horizontal motion:
vx  v0
With xB  12.2 m,
(b)
we get
tan   
115.3 km/h  v0  148.0 km/h 
(v y ) B
dy
gt

 B
dx
(vx ) B
v0
For h  0.788 m,
tan  
(9.81)(0.3810)
 0.11673,
32.02
  6.66 
For h  1.068 m,
tan  
(9.81)(0.2968)
 0.07082,
41.11
  4.05 
PROBLEM 11.142
At a given instant in an airplane race, airplane A is flying
horizontally in a straight line, and its speed is being increased
at the rate of 8 m/s 2 . Airplane B is flying at the same altitude
as airplane A and, as it rounds a pylon, is following a circular
path of 300-m radius. Knowing that at the given instant the
speed of B is being decreased at the rate of 3 m/s2 , determine,
for the positions shown, (a) the velocity of B relative to A,
(b) the acceleration of B relative to A.
SOLUTION
First note
v A  450 km/h vB  540 km/h  150 m/s
(a)
v B  v A  v B/A
We have
The graphical representation of this equation is then as shown.
We have
vB2/A  4502  5402  2(450)(540) cos 60°
vB/A  501.10 km/h
and
540 501.10

sin  sin 60
  68.9
v B/A  501 km/h
(b)
First note
Now
a A  8 m/s2
(aB ) n 
v 2B
B

(a B )t  3 m/s 2
Then
60
(150 m/s) 2
300 m
(a B )n  75 m/s2

68.9 
30 
a B  (a B )t + (a B ) n
 3( cos 60 i  sin 60 j) + 75(cos 30 i  sin 30 j)
= (66.452 m/s 2 )i  (34.902 m/s 2 ) j
Finally
a B  a A  a B/ A
a B/A  (66.452i  34.902 j)  (8i )
 (74.452 m/s 2 )i  (34.902 m/s 2 ) j
a B/A  82.2 m/s2
25.1 
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PROBLEM 11.145
A golfer hits a golf ball from Point A with an initial velocity of 50 m/s at an angle
of 25° with the horizontal. Determine the radius of curvature of the trajectory
described by the ball (a) at Point A, (b) at the highest point of the trajectory.
SOLUTION
(a)
We have
or
(a A ) n 
A 
v A2
A
(50 m/s) 2
(9.81 m/s 2 ) cos 25
 A  281 m 
or
(b)
We have
( aB ) n 
vB2
B
where Point B is the highest point of the trajectory, so that
vB  (v A ) x  v A cos 25
Then
or
B 
[(50 m/s) cos 25°]2
9.81 m/s 2
 B  209 m 
PROBLEM 11.148
From measurements of a photograph, it has been found that as the
stream of water shown left the nozzle at A, it had a radius of
curvature of 25 m. Determine (a) the initial velocity vA of the
stream, (b) the radius of curvature of the stream as it reaches its
maximum height at B.
SOLUTION
(a)
We have
(a A ) n 
v A2
A
or
4

v A2   (9.81 m/s 2 )  (25 m)
5

or
v A  14.0071 m/s
vA  14.01 m/s
(b)
We have
Where
Then
or
( aB ) n 
vB2
B
vB  (v A ) x 
B
36.9° 
4
vA
5
 4  14.0071 m/s 
 5
2
9.81 m/s2
 B  12.80 m 
PROBLEM 11.161
The oscillation of rod OA about O is defined by the relation   3/ sin  t 
where and t are expressed in radians and seconds, respectively. Collar B slides
along the rod so that its distance from O is r  6(1  e2t ) where r and t
are expressed in inches and seconds, respectively. When t  1 s, determine (a) the
velocity of the collar, (b) the acceleration of the collar, (c) the acceleration of the
collar relative to the rod.
SOLUTION
Calculate the derivatives with respect to time.
3
r  6  6e 2t in.

r  12e2t in/s
  3cos t rad/s

r  24e 2t in/s 2
  3 sin  t rad/s 2

sin  t rad
At t  1 s,
(a)
3
r  6  6e 2  5.1880 in.

r  12e 2  1.6240 in/s
  3cos   3 rad/s

r  24e2  3.2480 in/s 2
  3 sin   0

sin   0
Velocity of the collar.
v  rer  re  1.6240 er  (5.1880)(3)e
v  (1.624 in/s)er  (15.56 in/s)e 
(b)
Acceleration of the collar.
a  (
r  r2 )er  (r  2r)e
 [3.2480  (5.1880)(3)2 ]er  (5.1880)(0)  (2)(1.6240)(3)]e
(49.9 in/s2 )er  (9.74 in/s 2 )e 
(c)
Acceleration of the collar relative to the rod.
a B /OA  
re r
a B /OA  (3.25 in/s2 )er 
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