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A student is standing at a distance of 50 metres from a bus. As soon as the bus begins its motion (starts moving away from student) with an acceleration of 1 ms^-2, the student starts running towards the bus with a uniform velocity u. Assuming the motion to be along a straight road. The minimum value of u, so that the student is able to catch the bus is:

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5 $m s^{- 1}$

8 $m s^{- 1}$

10 $m s^{- 1}$

12 $m s^{- 1}$

answer is C.

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Detailed Solution

50 +x = ut …(1) and x = $\frac{t^{2}}{2} ... (2)$

From (1) and (2)

$50 + \frac{t^{2}}{2} = u t \Rightarrow t^{2} - 2 u t + 100 = 0$

For real roots, discriminant = 0

For minimum velocity, discriminant = 0

$\Rightarrow 4 u^{2} - 400 = 0 \Rightarrow u = 10 m / s$

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