A student is standing at a distance of 50 , m | vedclass

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(C) Let the student catch the bus after time $t \, s$. The distance covered by the student is $s_s = ut$.The distance covered by the bus starting from

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$10$

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(C) Let the student catch the bus after time $t \, s$. The distance covered by the student is $s_s = ut$.The distance covered by the bus starting from rest is $s_b = \frac{1}{2}at^2 = \frac{1}{2}(1)t^2 = \frac{t^2}{2}$.For the student to catch the bus,the total distance covered by the student must equal the initial distance plus the distance covered by the bus:$ut = 50 + \frac{t^2}{2}$.Rearranging for $u$:$u = \frac{50}{t} + \frac{t}{2}$.To find the minimum velocity $u$,we differentiate $u$ with respect to $t$ and set it to zero:$\frac{du}{dt} = -\frac{50}{t^2} + \frac{1}{2} = 0$.Solving for $t$:$\frac{50}{t^2} = \frac{1}{2} \implies t^2 = 100 \implies t = 10 \, s$.Substituting $t = 10 \, s$ back into the equation for $u$:$u = \frac{50}{10} + \frac{10}{2} = 5 + 5 = 10 \, m/s$.

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