Question

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

Answer 1

Let the common velocity at $t=4min$ be $v_0$
Then acceleration of cyclist

$a_A=\frac{v_0}{4}$

and

$a_B=\frac{v_0}{1}$

Let at time $t$ both the cyclists meet. then,

$S_A$ [in time interval $(t-0)$]

$=S_B$ [in time interval $(t-3)$].

$\frac{1}{2}{a}_A{t}^2$ = $\frac{1}{2}{a}_B(t-3{)}^2$

$\frac{1}{2}\frac{V_0}{4}{t}^2=\frac{1}{2}\frac{V_0}{1}(t-3{)}^2$

$\frac{t}{2}=t-3$

$t=6\text{min}$
Final Answer: $t=6min$