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 \(\frac{1}{2} \alpha^{2}\) . For the given condition
\(vt - 45 + \frac{1}{2}gt^{2}\) = 45 + 1.25 t^{2} [As a = 2.5 m/s^{2}]
⇒ ⇒ \(\nu - \frac{45}{\tau} + 1.25 t\)
To find the minimum value of u
\(\frac{d u}{d t} = 0\) so we get \(l = 6 \sec x\) then,
\(v = \frac{45}{6} + 1.25 \times 6 - 7.5 + 7.5 - 15 \text{ m/s}\)
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