To solve this problem, we can use the equations of motion for uniformly accelerated motion. Let's assume that car A surpasses car B at time t, and let's denote the distance covered by car A at that time as S.
For car A: Initial velocity (u) = 0 m/s (since it starts from rest) Acceleration (a) = 2 m/s^2 Time (t) = t
Using the equation of motion S = ut + (1/2)at^2, we can calculate the distance covered by car A in terms of time:
S = 0 * t + (1/2) * 2 * t^2 S = t^2
For car B: Initial velocity (u) = 20 m/s Velocity (v) = 20 m/s (constant velocity) Time (t) = t
The distance covered by car B is given by the formula S = ut, where u is the initial velocity and t is the time:
S = 20 * t S = 20t
Now, we want to find the time at which car A surpasses car B, so we can set S (distance covered by car A) equal to S (distance covered by car B):
t^2 = 20t
Rearranging the equation:
t^2 - 20t = 0
Factoring out t:
t(t - 20) = 0
So, either t = 0 or (t - 20) = 0.
Since t cannot be zero (as it represents the time when car A surpasses car B), we have:
t - 20 = 0
Solving for t:
t = 20 seconds
Now we can find the distance covered by car A at time t = 20 seconds:
S = t^2 = (20)^2 = 400 meters
Therefore, car A surpasses car B at a distance of 400 meters.