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?

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.

Can I have it a bit more detailed

can it be detailed please

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

To find the distance Car B travels before catching up with Car A, we can follow these steps:

Step 1: Find the time it takes for Car B to catch up with Car A.
In this scenario, Car B starts moving 2 seconds after Car A. Since Car B is accelerating, we need to calculate the time it takes for Car B to match the velocity of Car A. To do this, we can use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

Given:
u (initial velocity of Car B) = 0 (since it is stationary)
v (final velocity of Car A and Car B) = 25 m/s
a (acceleration of Car B) = 6 m/s^2
t (time for Car B to catch up) = ?

Using the formula v = u + at and rearranging the terms, we get:
t = (v - u) / a

t = (25 - 0) / 6
t = 25 / 6
t ≈ 4.167 seconds

So it will take approximately 4.167 seconds for Car B to catch up with Car A.

Step 2: Find the distance Car B travels during this time.
To calculate the distance, we can use the formula s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.

Given:
u (initial velocity of Car B) = 0 (since it is stationary)
a (acceleration of Car B) = 6 m/s^2
t (time for Car B to catch up) ≈ 4.167 seconds

Using the formula s = ut + (1/2)at^2, we get:
s = 0 + (1/2) * 6 * (4.167)^2
s = (1/2) * 6 * 17.36
s = 3 * 17.36
s ≈ 52.08 meters

Therefore, Car B will travel approximately 52.08 meters before catching up with Car A.