two cars A & B at rest at same point initially if A starts with uniform velocity of 40m/s and B starts in the same direction with constant acceleration of 4m/s^2, then B will catch A after how much time
20 second
To find out after how much time car B will catch car A, we need to determine the time it takes for car B to cover the distance between them.
First, let's calculate the distance car A will cover before car B catches up. We can use the formula:
distance = initial velocity * time
For car A:
initial velocity of A = 40 m/s
time taken by A = t (Let's assume this to be the time when car B catches up)
So, the distance covered by car A = 40 * t = 40t
Now, let's determine the distance covered by car B at time t. We can use the second equation of motion:
distance = initial velocity * time + (1/2) * acceleration * time^2
For car B:
initial velocity of B = 0 m/s (as it starts from rest)
acceleration of B = 4 m/s^2
time taken by B = t
So, the distance covered by car B = 0 * t + (1/2) * 4 * t^2 = 2t^2
Now, we know that car B catches up with car A when the distances they cover are equal. Therefore, we can set up the equation:
40t = 2t^2
Simplifying the equation:
20t = t^2
t^2 - 20t = 0
t(t - 20) = 0
From the equation, we find two solutions for t: t = 0 and t = 20.
Since t = 0 means that car B catches up with car A at the starting point (which is not possible since car A has a non-zero velocity), we discard t = 0.
Therefore, the time it takes for car B to catch car A is 20 seconds.