Question

A car B travellin at constant velocity of 25m/s overtakes a stationary car B.Two seconds later,carB sets off in pursuit accelerating at a uniform 6m/s^2.How far does B travel before catching up with carA?

Answer 1

It would be easier to set references.
In this case, I would set t=0 when B starts pursuit, at which time A would be ahead by 25 m/s *2 s = 50 m.

For distance reference, I would set B's initial position as zero.

This means that
A's position as a function of time, A(t)
= 25t+50

B's position as a function of time, B(t)
= vi*t + (1/2)at²
= 0 + (1/2)*6*t²
=3t²

Equate car positions when B catches up:
A(t)=B(t)
25t+50=3t²
Solve for t:
(3t+5)(t-10)=0
Reject negative root.

Answer 2

Can I have it a bit more detailed

Answer 3

can it be detailed please

Answer 4

car A travelling at a constant velocity of 25m/s overtakes a stationary car B. Two seconds later car B sets off in pursuit accelerating at a uniform 6m/s. How far does B travel before catching up with A