Cars A and B move in the same direction in adjacent lanes. The position x of car

Question

Cars A and B move in the same direction in adjacent lanes. The position x of car A is given in the figure, from time t = 0 to t = 7.0 s. At t = 0, car B is at x = 0, with a velocity of 4.0 m/s and a negative constant acceleration aB.

(a) What mustaBbe such that the cars are side by side (momentarily at the same value ofx) att=20.s?
m/s2

(b) For that value ofaB, how many times are the cars side by side?


(c) How many times will the cars be side by side if the magnitude of accelerationaBis more than the answer to part (a)?


(d) How many times will the cars be side by side if the magnitude of accelerationaBis less than the answer to part (a)?

Explanation / Answer

Xa at t= 2 sec = 24 m

4*t + 0.5 *aB*t*t = 24

aB = 8 m/s2 (positive)

the Velocity of B keeps on increasing while Va = 2 m/s

So it is once side by side of the car A

If aB is increased then also the car B will be once side by side of the car A

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