Question

A particle starts from rest and moves along a straight line covers distance 2s with constant acc then covers distance 2s with constant speed and 3s with uniform retardation and finally comes at rest. Find ratio of avg velocity and maximum velocity

Answer 1

Given:

• u=0 m/s

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solution :

we know,

$s = ut + \dfrac{1}{2} a {t}^{2}$

Using equation of motion we get:

$S= \dfrac{1}{2} \times T1 \times \: Vmax \\ \implies \: T1 = \dfrac{2S}{Vmax}---i$

$2S = T2 \times \: Vmax \\ \implies \: T2 = \dfrac{2S}{Vmax}----ii$

$3S = \dfrac{1}{2} \times T3 \times \: Vmax \\ \implies \: T3 = \dfrac{6S}{Vmax} ----iii$

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from I, ii, and iii total time:-

T1+T2+T3

=$\dfrac{2s+2s+6s}{Vmax}$

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$avg \: V = \dfrac{total \: displacement}{time} \\ \implies \: avg \: V = \dfrac{6S}{ \dfrac{2S + 2S + 6S}{Vmax} }$

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$\dfrac{avg \: V}{vmax} = \dfrac{ \dfrac{3}{5}Vmax }{Vmax} =$ $\red{\dfrac{3}{5}}$