Question

Two particles instantaneously at $A$ and $B$ are $5m$, apart and they are moving with uniform velocities, the former towards $B$ at $4m/s$ and the latter perpendicular to $AB$ at $3m/s$. They are nearest at the instant (in seconds)

• A. $\frac{4}{5}$
• B. $\frac{2}{5}$
• C. $1$
• D. $2$

Answer 1

The correct option is A $\frac{4}{5}$

Velocity of A, $v_A=4㎧$ towards B

Velocity of B, $v_B=3㎧$ perpendicular to AB

Distance between them, $l=5m$

Let they are closest after time t seconds

And A and B reach to $A^{{}^{\prime }}$ & $B^{{}^{\prime }}$respectively.

The distance traveled by A

$=4\times t$

$\therefore A^{{}^{\prime }}B=(5-4t)$

And distance traveled by $B=3\times t$

$\therefore B{B}^{{}^{\prime }}=3\times t$

AAt this distance the distance between A and B is $A^{{}^{\prime }}{B}^{{}^{\prime }}$

In $△A^{{}^{\prime }}B{B}^{{}^{\prime }}:{\left(A^{{}^{\prime }}{B}^{{}^{\prime }}\right)}^2={\left(A^{{}^{\prime }}B\right)}^2+{\left(B{B}^{{}^{\prime }}\right)}^2$

$={(5-4t)}^2+{\left(3t\right)}^2$

$=25-40t+16t^2+9t^2$

$=25t^2-40t+25$

$A^{{}^{\prime }}{B}^{{}^{\prime }}=\sqrt{{(5t-4)}^2+9}$

Now $A^{{}^{\prime }}{B}^{{}^{\prime }}$ is minimum when

${(5t-4)}^2=0$

$\Rightarrow 5t=4$

$t=\frac{4}{5}$

At this time particles are nearest to each other