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

Answer: 6

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 $(t-3)]$
$\Rightarrow \frac{1}{2}{a}_A{t}^2=\frac{1}{2}{a}_B(t-3{)}^2$
$\Rightarrow \frac{1}{2}\frac{v_0}{4}{t}^2=\frac{1}{2}\frac{v_0}{1}(t-3{)}^2$
$\Rightarrow \frac{t}{2}=t-3$
$\Rightarrow t=6\min$