A car sits in an entrance ramp to a freeway, waiting for a break in the traffic.

Question

A car sits in an entrance ramp to a freeway, waiting for a break in the traffic. The driver sees a small gap between a van and an 18-wheel truck and accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 22.0 m/s when it reaches the end of the ramp, which has length 125 m. What is the acceleration of the car? How much time does it take the car to travel the length of the ramp? The traffic on the freeway is moving at a constant speed of 22.0 m/s. What distance does the traffic travel while the car is moving the length of the ramp?

Explanation / Answer

Using v2 = u2 + 2as we get,

222 = 02 + 2*a*125

This gives, acceleration, a = 1.936 m/s2

Using, v = u + at we get,

22 = 0 + 1.936*t

This gives, time t = 11.36 s

Using distance = Speed*Time we get, Distance = 22*11.36 = 250 m

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