A student is standing at a distance of $50\,metres$ from the bus. As soon as the bus begins its motion 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.........$ms^{-1}$
$5$
$8$
$10$
$12 $
A student is standing at a distance of $50\,metres$ from the bus. As soon as the bus begins its motion 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.........$ms^{-1}$
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
$ut = 50 + \frac{1}{2}a{t^2} = 50 + \frac{{{t^2}}}{2}$ [$a = 1\;m/{s^2}$]
$⇒$ $ u = \frac{{50}}{t} + \frac{t}{2}$
To find the minimum value of $u$
$\frac{{du}}{{dt}} = 0$, so we get $t = 10\sec $, then $u = 10\;m/s$
Similarly distance travelled by the bus will be $\frac{1}{2}a{t^2}$ for the given condition
$ut = 50 + \frac{1}{2}a{t^2} = 50 + \frac{{{t^2}}}{2}$ [$a = 1\;m/{s^2}$]
$⇒$ $ u = \frac{{50}}{t} + \frac{t}{2}$
To find the minimum value of $u$
$\frac{{du}}{{dt}} = 0$, so we get $t = 10\sec $, then $u = 10\;m/s$
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