Question

Two particles P and Q are launched simultaneously as shown in figure. Find the minimum distance between particles in meters.

Answer 1

The minimum distance will be found when Q moves with $10m/s$ velocity which can be found by,

$v=u+at$

$10=0+2t$

$t=5sec$.

Distance traveled by P in 5 seconds is $5\times 10=50m$.

And distance traveled by Q in 5 seconds is$=0.5\times 2\times 5^2=25m$.

So minimum distance between P and Q will be $40+25-50=15m$