Question

The figure shows velocity $(v)$ versus time $(t)$ graphs for two cyclists moving along the same straight segment of a highway from the same point. The second cyclist starts moving at $t=3 \min$. At what time (in minute) do the two cyclists meet?

Answer 1

Let the common velocity at $t=4 \min$ be $v_{0}$.
Then, $a_{A}=\frac{v_{0}}{4}$ and $a_{B}=\frac{v_{0}}{1}$
Let at time $t$ both the cyclisțs meet.
Then, $s_{A}$ [in time interval $(t-0)$ ]
$=s_{B} [$ in time interval $(t-3)]$
$\Rightarrow \frac{1}{2} a_{A} t^{2}=\frac{1}{2} a_{B}(t-3)^{2}$
$\Rightarrow \frac{1}{2} \frac{v_{0}}{4} t^{2}=\frac{1}{2} \frac{v_{0}}{1}(t-3)^{2}$
$\Rightarrow \frac{t}{2}=t-3$
$\Rightarrow t=6 \min$