A man is 45 m behind the bus when the bus start accelerating from rest with acceleration 2.5 m/s 2 . With what minimum velocity should the man start running to catch the bus ?
Text Solution
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Let man will catch the bus after ' t ' sec . So he will cover distance ut.
Similarly distance travelled by the bus will be \(\frac{1}{2at^{2}}\) . For the given condition
\(u t = 45 + \frac{1}{2} a t^{2}\) = 45 + 1.25 t^{2} \(\left[ \text{As } a = 2.5 \ \mathrm{m/s^{2}} \right]\)
⇒ ⇒ \(u = \frac{45}{t} + 1.25 t\)
To find the minimum value of u
\(\frac{du}{dt} = 0\) so we get \(t = 6 \text{ sec}\) then,
\(u = \frac{45}{6} + 1.25 \times 6 = 7.5 + 7.5 = 15 m/s\)
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