Question

Two bikes $A$ and $B$ start from a point. A move uniform speed $40m/s$ $B$ starts from rest with uniform acceleration $2m/s^2$. If $B$ starts at $t=0$ and $A$ starts from the same point at $t=10s$, then the time interval during the journey in which $A$ was ahead of $B$ is

• A. $20s$
• B. $8s$
• C. $10s$
• D. $A$ is never ahead of $B$

Answer 1

The correct option is A $20s$

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