Question

An particle starts from rest and moves along a straight line covers a distance of S. with constant speed then 3S with uniform retardation and finally comes at rest. ▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪ Find the ratio if average velocity and maximum velocity.

Answer 1

Given:

u=0 m/s

We know the area of a v-t graph gives the displacement.

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solution :

here,

Lets consider the 1st triangle first,

then,

$area = \dfrac{1}{2} \times T1 \times Vmax$

$\dfrac{1}{2} \times T1 \times Vmax = S$

or,

${\pink{T1 = \dfrac{2S}{Vmax}}}$

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consider the rectangle,

$area = \: T2 \times Vmax = 2S$

${\pink{T2 = \dfrac{2S}{Vmax}}}$

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Consider the last triangle,

$t3 \times \dfrac{1}{2} \times Vmax = 3S$

${\pink{=> T3 = \dfrac{6S}{Vmax}}}$

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$average \: velocity \: = \dfrac{total \: displacement}{total \: time}$

$= > V \: avg = { \dfrac{6S} {\dfrac{2S + 2S + 6S}{Vmax}}}$

$= > \dfrac{6S}{10S} \times Vmax$

${\blue{ = > \dfrac{3}{5} Vmax}}$

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$\dfrac{Vavg}{Vmax} = \dfrac{ \dfrac{3}{5} \times Vmax }{Vmax}$

$\bold{\red{= \dfrac{3}{5}}}$