Question

A radius vector of a particle varies with time $t$ as $r=at(1-\alpha t)$, where $a$ is a constant vector and $\alpha$ is a positive constant. Find the distance $s$ covered before the particle comes to rest

• A. $s=\frac{a}{\alpha }$
• B. $s=\frac{a}{3\alpha }$
• C. $s=\frac{a}{2\alpha }$
• D. $s=\frac{3a}{2\alpha }$

Answer 1

The correct option is C $s=\frac{a}{2\alpha }$

We have the radius vector as $r=at(1-\alpha t)$
This radius vector becomes zero at $t=0$ and $t=\frac{1}{\alpha }$
Now the rate of change of radius vector is given as $\frac{dr}{dt}=a(1-2\alpha t)$.
This becomes 0 for $t=\frac{1}{2\alpha }$.
Thus we get the equation for $\frac{dr}{dt}$ as
$a(1-2\alpha t)$ for $t<\frac{1}{2\alpha }$ and
$a(2\alpha t-1)$ for $t>\frac{1}{2\alpha }$
Thus we get
$s={\int }_0^{\frac{1}{2\alpha }}a(1-2\alpha t)dt+{\int }_{\frac{1}{2\alpha }}^{\frac{1}{\alpha }}a(2\alpha t-1)dt$
$=at-\frac{\text{/}2a\alpha t^2}{\text{/}2}{|}_0^{\frac{1}{2\alpha }}+\frac{\text{/}2a\alpha t^2}{\text{/}2}-at{|}_{\frac{1}{2\alpha }}^{\frac{1}{\alpha }}$
Solving this we get
$s=\frac{a}{2\alpha }$