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A student is standing at a distance of 50 meters from the bus. As soon as the bus starts its motion with an acceleration of $1 m s^{- 2}$ , 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 (in m/s), so that the students is able to catch the bus is:

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answer is 10.

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Detailed Solution

Let student will catch the bus after t sec.

so, it will cover distance ut.

Similarly, distance travelled by the bus will be $\frac{1}{2} a t^{2}$ for the given condition

$\begin{array}{l} u t = 50 + \frac{1}{2} a t^{2} = 50 + \frac{t^{2}}{2} \\ \Rightarrow u = \frac{50}{t} + \frac{t}{2} (As a = 1 m / s^{2}) \end{array}$

To find the minimum value of $u, \frac{d u}{d t} = 0$

So we get t = 10 sec

Then u = 10 m/s.

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