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?

1 Answer
Jul 21, 2018

Answer:

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