Question

Two cars start from rest at a red stop light. When the light turns green, both...

Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates uniformly at a rate of 5.3 m/s2 for 4.7 seconds. It then continues at a constant speed for 14.3 seconds, before applying the brakes such that the car

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Homework Answers

Answer #1

1 )

Data for blue car ::

initial velocity of blue car = Vbi = 0 m/s

acceleration of blue car = ab = 5.3 m/s2

time of travel = t = 1.9 sec

using the kinematics equation ::

Vf = Vi + at

Vf = 0 + (5.3 m/s2 ) (1.9 sec)

Vf = 10.07 m/s

2)

initial velocity of blue car = Vbi = 0 m/s

acceleration of blue car = ab = 5.3 m/s2

time of travel = t = 8.7 sec

using the kinematics equation ::

Vbf = Vbi + abt

Vf = 0 + (5.3 m/s2 ) (8.7 sec)

Vbf = 46.11 m/s

3) Velocity of blue car after 4.7 sec can be given using the equation

Vbf = Vbi + abt

Vbf = 0 + (5.3 m/s2 ) (4.7 sec)

Vbf = 24.91 m/s

Distance travelled in 4.7 sec can be calculated using ::

d = Vbi t + (0.5) ab t2

d = 0 x 4.7 + (0.5) (5.3 m/s2 ) (4.7 sec)2

d = 58.54 m

distance travelled in 14.3 sec at constant velocity of 24.91 m/s :::

d' = (24.91 m/s) (14.3 sec) = 356.2 m

d' = 356.2 m

Total distance travelled by blue car before applying the brakes

d + d' = 58.54 m + 356.2 = 414.74 m

4) total distance covered by the blue car = 444.02 m

total distance covered by the blue car before applying brakes = 414.74 m

distance covered by blue car after applying brakes = D = 444.02 - 414.74 = 29.28 m

D = 29.28 m

initial velocity before applying brakes = Vi = 24.91 m/s

final velocity = Vf =0                  (since the car stops)

Using the kinematics equation ::

Vf2 = Vi2 + 2 a D

02 = 24.912 + 2 a (29.28)

a = -10.6 m/s2

5) time for acceleration = 4.7 sec

time for constant velocity = 14.3 sec

time for retardation can be calculated using the equation ::

Vf = Vi + at

0 = 4.91 m/s + (-10.6 m/s2 ) t

t = 0.463 sec

So total time of travel = 4.7 sec + 14.3 sec + 0.436 sec = 19.436 sec

6 ) distance travelled by yellow car = d = 444.02 m

initial velocity = Vi =0

time of travel= 19.436 sec

using the equation ::

d = Vi t + 1/2 a t2

444.02 = 0 t + (0.5) a (19.436)2

a = 2.35 m/s2

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