First slide

Rectilinear Motion

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 . lf 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

Moderate

A

20 s
B

8 s
C

10 s
D

A is never ahead of B

Solution

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

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