Question

Two cars A and B are 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/s2,then B will catch A after how much time?​

Answer 1

Explanation:

Let distance covered by A be S1

Let distance covered by B be S2

B will catch A when S1 = S2

S = ut + 1/2 * a * t^2

Therefore, 

40t + (1/2 * 0 * t^2) = 0t + 1/2 * 4 * t^2

40t = 2t^2

20t = t^2

Therefore, t = 20

hope this helps ✌️✌️✌️

Answer 2

Let,

distance covered by A be S1

Let,

distance covered by B be S2

B will catch A when S1 = S2

S = ut + 1/2 * a * t^2

Therefore, 

40t + (1/2 * 0 * t^2) = 0t + 1/2 * 4 * t^2

40t = 2t^2

20t = t^2

Therefore, t = 20