Question

A resting car starts accelerating uniformly at `a_1` rate a then moves at a constant velocity =4m/s then decelerates at a constant rate `a_2`. the velocity of the car is greater than 2m/s for 10sec. What is the distance traversed by the car?

Answer 1

The total distance was `40 m`.

Explanation:

The car accelerates uniformly from rest to `4 m/s` over some time `t_1`. Then it decelerates uniformly for another time, `t_2`. So the greatest speed was `4 m/s`. We do not really know if the story continues until the car comes to a stop.We are asked to find total distance, but over how much of the story? Until the deceleration brings the car to a stop?

I will assume we need to find distance from start until the deceleration brings the car to a stop.

We are also told that the velocity was greater than `2 m/s` for 10 s. It would have reached `2 m/s` in a time of `t_1/2`. And during the deceleration, it would have decreased to `2 m/s` in a time of `t_2/2`. Therefore

`t_1/2 + t_2/2 = 10 s and t_1 + t_2 = 20 s`

Because both `a_1 and a_2` are constant, and the peak velocity was `4 m/s`, you can verify that the average velocity was `2 m/s` using the formula

#v_"ave" = (u + v)/2

Therefore the total distance was

`2 m/s * (t_1 + t_2) = 2 m/s * 20 s = 40 m`

I hope this helps,
Steve