A bus starts moving with acceleration $2 \mathrm{~m} / \mathrm{s}^{2}$. A cyclist $96 \mathrm{~m}$ behind…
A bus starts moving with acceleration $2 \mathrm{~m} / \mathrm{s}^{2}$. A cyclist $96 \mathrm{~m}$ behind the bus starts simultaneously towards the bus at $20 \mathrm{~m} / \mathrm{s}$. After what time will he be able to overtake the bus?
$4 \mathrm{sec}$
$8 \mathrm{sec}$
$18 \mathrm{sec}$
$16 \mathrm{sec}$
Solution
Velocity of bus $\mathrm{V}_{\mathrm{b}}=0$, Velocity of cyclist $\mathrm{V}_{c}=20 \mathrm{~m} / \mathrm{s}$,
Acceleration of bus $\mathrm{a}_{\mathrm{b}}=2 \mathrm{~m} / \mathrm{s}^{2}$ Acceleration of cyclist $\mathrm{a}_{\mathrm{c}}=0$
Relative velocity $\mathrm{V}_{\mathrm{cb}}=\mathrm{V}_{\mathrm{c}}-\mathrm{V}_{\mathrm{b}}=20 \mathrm{~m} / \mathrm{s}$
Relative acceleration $\mathrm{a}_{\mathrm{c} b}=\mathrm{a}_{\mathrm{c}}-\mathrm{a}_{\mathrm{b}}=-2 \mathrm{~m} / \mathrm{s}^{2}$
Relative separation $=96 \mathrm{~m}$
Using $s=u t+\frac{1}{2} a t^{2}$ $\Rightarrow 96=20 t-\frac{1}{2} 2 t^{2} \Rightarrow t^{2}-20 t+96=0$
Solving the equation, we get $t=8$ s and $t=12$ s
Hence, after 8 s, cyclist will overtake the bus.
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Asked in: JEE Mains - Motion In One Dimension - Test 2