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A particle starting from rest, travels a distance S with uniform acceleration and immediately after travels a distance of 3S with uniform velocity followed by a distance 5S with uniform deceleration, and comes to rest. Then the ratio of average speed to the maximum speed of the particle is

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$\frac{3}{5}$

$\frac{2}{5}$

$\frac{1}{5}$

$\frac{4}{5}$

answer is C.

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Detailed Solution

$v_{o} \to maximum speed$

$s = \frac{v_{o} + 0}{2} t_{1} \Rightarrow t_{1} = \frac{2 s}{v_{o}}$

$t_{2} = \frac{3 s}{v_{o}}$

5s = $\frac{v_{o} + 0}{2} t_{3} \Rightarrow t_{3} = \frac{10 s}{v_{o}}$

$v_{av} = \frac{s + 3 s + 5 s}{t_{1} + t_{2} + t_{3}}$

$v_{av} = \frac{9 s}{\frac{2 s}{v_{o}} + \frac{3 s}{v_{o}} + \frac{10 s}{v_{o}}} \Rightarrow \frac{v_{av}}{v_{o}} = \frac{3}{5}$

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