Question

a car sits in an entrance ramp to a freeway, waiting for a break in traffic. the driver sees a small gap between two vehicle and accelerate with constant acceleration along the ramp onto the freeway. the car starts from rest, moves in a straight line ,and has speed of 20 m/s when it reaches the end of the 120m ramp. what is the acceleration of the car?

a) how much time does it take the car to reach the end of the ramp?
b) the traffic on the freeway is moving at a constant speed of 20m/s. what distance does traffic travel while the car is moving the length of the ramp?​

Answer 1

Answer:

Use the kinematic relation

v22 - v12 = 2ad

v2 = 20.0 m/s

v1 = 0 [starts from rest]

d = 125 m

a = v22/(2d)

Plug in v2 and d and get a in m/s2.

(2)

d = v1t + at2/2

v1 = 0

d = at2/2

t = √(2d/a)

Plug in d and the result for a from part (1) and get t in seconds.

(3)

u = 20.0 m/s

It moves a distance ut, where t is the result from part (2).