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Q.

A particle initially (i.e, at t = 0) moving with a velocity u is subjected to a retarding force, as a result of which it decelerates at a rate a = - $k \sqrt{v}$ where v is the instantaneous velocity and k is a positive constant. The time T taken by the particle to come to rest is given by

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a

T = $\frac{2 u^{\frac{3}{2}}}{k}$

b

T = $\frac{2 \sqrt{u}}{k}$

c

T = $\frac{2 u}{k} v$

d

T = $\frac{2 u^{2}}{k}$

answer is A.

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Detailed Solution

Given a = - $k \sqrt{v}$ or $\frac{d v}{d t}$ = $k \sqrt{v}$ . Thus $v^{\frac{- 1}{2}}$ $d v = - k d t$ Integrating, we get $2 v^{\frac{1}{2}} = - k t + c$ . Using the given initial condition $(v = u a t t = 0),$ we get $c = 2 \sqrt{u}$ . Thus, we have

2 $(v^{\frac{1}{2}} - u^{\frac{1}{2}})$ = $- k t$

Now, use $t = T$ and $v = 0$ .

we get,

$T = \frac{2 \sqrt{u}}{k}$

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