Question

Cars A and B move in the same direction in adjacent lanes. The position of car A is given in the figure. At t=0, car B is at x=0, with a velocity of 10 m/s and a negative acceleration. What must be that acceleration such that the cars arc side by side when t=6 sec? For that value of the acceleration, how many times will the cars be side by side?

Answer 1

First of all the velocity of car A is 20/5=4m/s

now if both car moves side by side at t=6sec that means they have same velocity at t=6sec

therefore for Va=Ua+f*6

or Va=4+6f

and for car B

Vb=Ub-f*6

or Vb=10-6f [for B accleration is negative

now Va=Vb

therefore

or 4+6f= 10-6f

or f=.5m/s^2

b) for the value of same f Car B's final velocity=10-6*.5= 7m/s

then for negative accleration the velocity will be zero after coving s distance

therefore using V^2-U^2=2fs we get 0-7^2=2*.5*s

or s=4.9m

therefore the cars will be side by side for two times once when the velocity is 7m/s and secondly when B comes to rest at 4.9m