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 19.0 m/s when it reaches the end of the ramp, which has length 126 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 23.0 . What distance does the traffic travel while the car is moving the length of the ramp?

Answer 1

Part a)

`a = 1.43 m/s^2`

Part b)

`t = 13.3 s`

Part c)

`d = 305 m`

Step-by-step explanation:

Part a)

Car start from rest and reached to final speed of 19 m/s when it will cover a distance of 126 m

So we will have

`v_f^2 - v_i^2 = 2 a d`

now we will have

`19^2 - 0 = 2(a)(126)`

`a = 1.43 m/s^2`

Part b)

in order to find the time taken by the car we can use another kinematics equation

`x = ((v_f + v_i)/(2)) t`

`126 = ((19 + 0)/(2))t`

`t = 13.3 s`

Part c)

If the traffic on free way is moving with speed 23 m/s

so we can say that traffic will move by distance

`d = v t`

`d = (23)(13.3)`

`d = 305 m`