A bus is beginning to move with an acceleration of $1 {m/s}^{2}$ . A boy who is $48 m$ behind the bus starts running with constant speed of $10 m/s$ . The earliest time when the boy can catch the bus is

8 sec

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10 sec

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12 sec

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14 sec

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Solution

The correct option is A 8 sec

Initial velocity of boy with respect to bus $= 10 {ms}^{- 1}$
Acceleration of boy with respect to bus $= - 1 {ms}^{- 1}$
Solving with respect to bus,
$s = u t + \frac{1}{2} a t^{2}$
$48 = 10 t - \frac{1}{2} t^{2}$
$t^{2} - 20 t + 96 = 0$
$t^{2} - 12 t - 8 t + 96 = 0$
$(t - 8) (t - 12) = 0$
$t = 8 s$ and $12 s$
$∴$ (a)

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A bus is beginning to move with an acceleration of 1 m/s2. A boy who is 48 m behind the bus starts running with constant speed of 10 m/s. The earliest time when the boy can catch the bus is

The correct option is A 8 sec Initial velocity of boy with respect to bus =10 ms−1 Acceleration of boy with respect to bus =−1 ms−1 Solving with respect to bus, s=ut+12at2 48=10t−12t2 t2−20t+96=0 t2−12t−8t+96=0 (t−8)(t−12)=0 t=8 s and 12 s ∴ (a)

A bus is beginning to move with an acceleration of $1 {m/s}^{2}$ . A boy who is $48 m$ behind the bus starts running with constant speed of $10 m/s$ . The earliest time when the boy can catch the bus is