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/s2 . 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

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/s2 . 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

Tardigrade Admin · Accepted Answer

A will be ahead of B when xA ge xB 40(t-10) ge (0) t +(1/2) (2) t2 as A is 10 sec late than B ⇒ t2-40t+400 le 0 ⇒ (t-20)2 le 0 which is not possible, so A will never be ahead at B

Q. 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

$20 s$

$8 s$

$10 s$

$A$ is never ahead of $B$

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

Q. Two bikes A and B start from a point. A moves with uniform speed 40m/s and B starts from rest with uniform acceleration 2m/s2 . If B starts at t=10 and A starts from the same point at t=10s, then the time during the journey in which A was ahead of B is

20s

8s

10s

A is never ahead of B