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Question

(c)
Velocity of particle at the end of $\frac{2}{3}$ distance is $v_1$.
Now, by equation of motion,
$\Rightarrow v^2-u^2=2 a s, 	ext { we have }$

Vedclass · Accepted Answer

$\sqrt{3 d / f}$

Vedclass · Answer

(c)
Velocity of particle at the end of $\frac{2}{3}$ distance is $v_1$.
Now, by equation of motion,
$\Rightarrow v^2-u^2=2 a s, 	ext { we have }$
$\Rightarrow v_1^2-0^2=2 f 	imes \frac{2}{3} d$
$\Rightarrow v_1^2=\frac{4}{3} f d \Rightarrow v_1=\frac{2}{\sqrt{3}} \cdot \sqrt{f d}$
As, final velocity is zero, so for next part of journey is
$v^2-u^2=2 a s$
$	ext { Gives, } 0-v_1^2 =2 a\left(\frac{1}{3} dight)$
$\Rightarrow -\frac{4}{3} f d=\frac{2}{3} a d$
or deacceleration, $a=-2 f$
Now, using $v=u+a t$, for first part of journey,
$v=u+a t$
$\Rightarrow \quad v_1=0+f t_1$
$\Rightarrow \frac{2}{\sqrt{3}} \sqrt{f d}=f t_1 \Rightarrow t_1=\frac{2}{\sqrt{3}} \cdot \sqrt{\frac{d}{f}}$
For second part of journey,
$v=u+a t$
$\Rightarrow 0=v_1-2 f t_2$
$\Rightarrow \frac{2}{\sqrt{3}} \sqrt{f d}=2 f t_2$
$\Rightarrow t_2=\frac{1}{\sqrt{3}} \sqrt{\frac{d}{f}}$
So, total time is
$t=t_1+t_2$
$=\frac{2}{\sqrt{3}} \sqrt{\frac{d}{f}}+\frac{1}{\sqrt{3}} \sqrt{\frac{d}{f}}$
$=\sqrt{3} \sqrt{\frac{d}{f}}=\sqrt{\frac{3 d}{f}}$

A particle starts moving along a line from zero initial velocity and comes to rest after moving distance $d$. During its motion, it had a constant acceleration $f$ over $2 / 3$ of the distance and covered the rest of the distance with constant retardation. The time taken to cover the distance is

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