EGR 232 Dynamics: Homework Set 02
Fall 2014
Problem 11.31
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, determine a) the required uniform deceleration of the car b) the
speed of the car as it passes the light.
__________________________________________________________________________
Solution:
a=?
v0 = 54 km/h
x0 = 0 km
x = 240 m
v0 = 54 km/h =15 m/s
v=?
t = 24 s
This is a constant acceleration
problem.
Principle:
v  v0  at
x  x0  v0t 
1 2
at
2
x  x0 
Acceleration can be found using:
1

x  x0  v0t  at 2
2
a
1
(v  v0 )t
2
a
v 2  v02  2a(x  x0 )
2(x  x0  v0t )
t2
2(x  x0  v0t ) 2(240m  0  (15m / s)(24s))

 0.4167m / s2
t2
(24s)2
Final Velocity is found using:
v  v0  at
v  15m / s  (0.4167m / s2 )(24s)  5m / s  18km / h
EGR 232 Dynamics: Homework Set 02
Fall 2014
Problem 11.36
Automobile A starts from O and accelerates at the constant rate of 0.75 m/s2. A short time later it is
passed by bus B which is traveling in the opposite direction at a constant speed of 6 m/s. Knowing
that bus B passes point O 20 s after automobile A started from there, determine when and where
the vehicles passed each other.
__________________________________________________________________________
Solution:
Car: Uniform accel.
xA0 = 0
xA = 0
vA0 = 0
vA =?
aA = 0.75 m/s2
tA = ?
Total time of travel:
t A  tB  20 s
For Car:
v  v0  at
Bus: uniform motion
xB0 = x
xB = 0
vB0 = -6 m/s
aB = 0
tB = ?
x  x0  v0t 
1 2
at
2
x  x0 
1
(v  v0 )t
2
v 2  v02  2a(x  x0 )
so:
x  x0  v0t 
tA 
1 2
at
2
2(x A  x A0  v A0t A )

aA
For Bus:
so
2(x A  x A0  v A0t A )
aA
2(x  0  0)
 1.634 x
0.75
x  x0  vt
x  x0  vt
tB 
tA 


tB 
xB  xB 0
vB
xB  xB0 0  x

 0.1667x
vB
6
Combining in the total time equation:
t A  tB  20 s
1.634 x  0.1667x  20
1.634 x  20  0.1667x

1.634 x

2
 20  0.1667x 
2
2.67x  400  6.667x  0.02779x2
0.02779x2  9.333x  400  0
solving with the quadratic equation:
b  b2  4ac (9.333)  (9.333)2  4(0.02779)(400)
x

2a
2(0.02779)
9.333  6.526

 285 or 50.5
0.05558
to figure out which one makes better sense:
using x = 28.42 then t A  1.634 285  27.6 s and tB  0.1667(285)  47.5 s
This one doesn’t make any sense….check the second answer.
using x = 50.5 then t A  1.634 50.5  11.6 s and tB  0.1667(285)  8.4 s
Yes, this one makes sense.
therefore the correct answer is x = 50.5 m and they passed at 11.6 seconds from the start.
EGR 232 Dynamics: Homework Set 02
Fall 2014
Problem 11.40
The elevator shown in the figure moves downward with a constant velocity of 5 m/s. Determine
a) the velocity of the cable C,
b) the velocity of the counterweight W,
c) the relative velocity of the cable C with respect to the elevator,
d) the relative velocity of the counterweight W with respect to the elevator.
__________________________________________________________________________
Solution: (assume down is + direction)
vE = 5 m/s
Constrained motion and uniform velocity:
For Rope L1:
L1  xC  xE  xE  L1  xC  2xE
For Rope L2:
L2  xE  xW

xC
xE
L1
L2
L2  xW  xE
Taking the derivative of each of the constrained
motion equations:
dL1 dxC
dx
dL2 dxW dxE

2 E


dt
dt
dt
dt
dt
dt
0  vW  vE
0  vC  2vE
vC  2vE
vW  vE
Since
vE  5m / s (downward)
then
vC  2vE  2(5)  10m / s  10m / s upward
and
vW  vE  (5)  5m / s  5m / s upward
Relative velocity of C with respect to the elevator:
vC / E  vC  vE  10m / s  5m / s  15m / s  15m / s downward
Relative velocity of W with respect to the elevator:
vW / E  vW  vE  5m / s  5m / s  10m / s  10m / s downward
xW
EGR 232 Dynamics: Homework Set 02
Fall 2014
Problem 11.48
Collar A starts from rest at t = 0 and moves upward with a constant acceleration of 3.6 in/s2,
Knowing that collar B moves downward with a constant velocity of 16 in/s, determine
a) the time at which the velocity of block C is zero.
b) the corresponding position of block C.
__________________________________________________________________________
Solution:
Block A: constant acceleration:
t=0
vA0=0
aA=3.6in/s2
therefore:
1
x A  x A0  v A0t  aAt 2
2
1
xA
--> xA  1.8t 2
x A  0  0t  (3.6)t 2
2
and
xB
v A  v A0t  aAt
v A  0t  3.6t
 v A  3.6t
Block B: constant velocity
vB=-16 in/s
xB  xB0  vBt
xB  0  (16)t
--> xB  16t
and
vB  16
Constrained motion equations: (four rope sections)
L  (x A  xB )  (x A  xC )  (xB  xC )  (xB  xC )
L  2x A  xB  3xC
therefore:
0  2v A  vB  3vC
0  2aA  aB  3aC
or
2
1
2
1
vC  v A  vB  (3.6t )  (16)  2.4t  5.333
3
3
3
3
2
1
2
1
aC  aA  aB  (3.6)  (0)  2.4
3
3
3
3
xC
The velocity of C is zero at
5.333
 2.22 s
2.4
the position of C has changed by: vc0 = vC (t=0) = -5.333
1
1
xC  xC 0  vC 0t  aC t 2  (5.333)(2.22)  (3.6)(2.22)2  2.97 in
2
2
0  2.4t  5.333
 t 