Question

Two particles start simultaneously from the same point and move along two straight lines, one with uniform velocity v and other with a uniform acceleration a. If $\alpha$ is the angle between the lines of motion of two particles then the least value of magnitude of relative velocity will be at time given by

• A. $\frac{v}{a}sin\alpha$
• B. $\frac{v}{a}cos\alpha$
• C. $\frac{v}{a}tan\alpha$
• D. $\frac{v}{a}cot\alpha$

Answer 1

The correct option is B $\frac{v}{a}cos\alpha$



Consider the particles going in paths inclined at $\alpha$ as shown

AT time 't' velocity of particle 1,

${\overrightarrow{V}}_1=Vcos\alpha \hat{i}+Vsin\alpha \hat{j}$
& at time 't' velocity of particle 2,

${\overrightarrow{V}}_2=at\hat{i}$

Relative velocity =${\overrightarrow{V}}_1-{\overrightarrow{V}}_2=(Vcos\alpha -at)\hat{i}+\left(vsin\alpha \right)\hat{j}$

Magnitude of relative velocity = $\left[\right(Vcos\alpha -at{)}^2+(Vsin\alpha {)}^2{]}^{\frac{1}{2}}$

For minimum magnitude its derivative should be zero.

$\Rightarrow \frac{d}{dt}\left[\right(Vcos\alpha -at{)}^2+(Vsin\alpha {)}^2{]}^{\frac{1}{2}}=0$

$\Rightarrow \frac{1}{2}\left[\right(Vcos\alpha -at{)}^2+\left(Vsin\alpha {)}^2{]}^{-\frac{1}{2}}\right[2(Vcos\alpha -at)(-a)+0]=0$

$1^{st}$ square bracket expression cannot be zero, so second term is zero

i.e $Vcos\alpha -at=0$

$\Rightarrow [t=\frac{Vcos\alpha }{a}]$