Question

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

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

Answer 1

Answer: (D)

A will be ahead of B when $x_{A} \ge\,x_{B}$
$40(t-10) \ge (0) t +\frac{1}{2} (2) t^{2}$
as $A$ is $10$ sec late than $B$
$\Rightarrow t^{2}-40t+400 \le\,0$
$\Rightarrow (t-20)^{2} \le 0$
which is not possible, so A will never be ahead at B