Question: A car is travelling at a constant velocity with Vc. At the instant the car passes a motor traffic officer, the motor officer accelerates from rest with Am.

Let d be the distance the motorcycle travels before catching up with the car. In terms of d, how far has the motorcycle travelled when their velocities are equal?

Solution: Since the acceleration is constant u = A*t and distance traveled given by s = (1/2)At^2. We will use the latter first.

s = (1/2)At^2 ........ motorcycle starts from rest
d = (1/2)At^2 ........ when the motorcycle reaches the car
t = SQRT[2*d/A] ... solving this for the time t

So the motorcycle reaches the car at time SQRT[2*d/A]

Now use the first equation u = A*t
If T is the time it takes the motorcycle to reach the car then u = A*T is the speed of the motorcycle when it reaches the car.
We want to find when the speed of the motorcycle is V/2 so when the speed is A*T/2
At any time we have u = A*t = A*T/2 and t = T/2
So the motorcycle reaches the same speed as the car when the time is 1/2 the time it takes the motorcycle to reach the car.

Now use the distance equation again to find how far the motorcycle has traveled in this time T/2
s = (1/2)A*[SQRT(2*d/A)/2]^2 = (A/2)*[(2*d/A)/4] = d/4

So the motorcycle has travelled 1/4 the distance d when it has a speed equal to that of the car

Note. Another way to realize the time to reach a speed equal to that of the car is to think about the average speed. The average speed is 1/2 the final speed (since the motorcylce starts from 0). Since velocty is linear with respect to time (u = A*t) this will occur when the time is half the time it takes to reach the car.

Note: This is not solved by me.
NOTE: PLOT A GRAPH AND U CAN GET THE ANSWER =DD