Question

particles A and B are moving along horizontal straight lines which lie in the west east direction. At a certain time, A passes a point P moving to the east with speed 5ms^-1 and constant acceleration 1ms^-2. Simultaneously, B passes a point Q east of P with speed 4ms^-1 towards the west but with constant acceleration 2ms^-2 towards the east. Given that the distance PQ is 28m, Calculate
a) The distance of A from P when B reverse direction
b) The time taken after the start for A to overtake
c) The further time taken for B to overtake A

Answer 1

B reverses direction when velocity hits 0
2t -4 = 0 ---> t = 2

plug t=2 into A position function
x(2) = 12

A is 12m away from point P

Answer 2

for other parts, simply equate the two position functions and solve for t

you will get 2 solutions, the time A overtakes B and then the later time when B overtakes A

--> .5t^2 +5t = t^2 -4t +28

--> .5t^2 -9t +28 = 0

--> t^2 -18t + 56 = 0

---> (t-4)(t-14) = 0