Question

A cyclist starts from rest and moves with a constant acceleration of $1{\text{m/s}}^2$. A boy who is $48\text{m}$ behind the cyclist starts moving with a constant velocity of $10\text{m/s}$. After how much time the boy meets the cyclist? Choose the appropriate?

• A. $8\text{s}$
• B. $12\text{s}$
• C. $10\text{s}$
• D. both (1) and (2)

Answer 1

The correct option is D both (1) and (2)


Let the boy meet the cyclist at $P$.
Boy (uniform motion): $10t=48+d\left(i\right)$
Cyclist accelerated motion : $d=\frac{1}{2}.1.t^2\left(ii\right)$
$10t=48+\frac{t^2}{2}\Rightarrow t^2-20t+96=0$
$(t-8)(t-12)=0$
The boy meets the cyclist at $t=8\text{s},12\text{s}$
At $t=8\text{s}$, velocity of cyclist, $v_c=1\times 8=8\text{m/s}$
$v_b>v_c$, boy crosses the cyclist
At $=12\text{s}$, velocity of cyclist, $v_c=1\times 12=12\text{m/s}$
$v_c>v_b$, cyclist crosses the boy and will be always ahead.

Note:
Apply relative velocity approach
$x_{BC}=u_{BC}t+\frac{1}{2}{a}_{BC}{t}^2$
$\Rightarrow +48=[10-0]t+\frac{1}{2}[0-1]$
$t^2-20t+96=0$