Question

Two particles A and B are initially 20 m apart. A is behind B. Particle A starts moving with a uniform velocity of 5 m/s towards B. Particle B starting from rest has an acceleration of 1 $m/s^2$ in the direction of velocity of A. The minimum distance between the particles is 1.5n(metres). Then n is ___

Answer 1

Take the origin at the initial position of A and positive x-axis towards B. At a time t, $x_A$ and $x_B$ are the Co-ordinates of the particles. Gap = G = $x_B-x_A$
$G=(20+\frac{1}{2}\times 1\times t^2)-\left(5t\right)$
$\therefore$ The gap is minimum when $\frac{dG}{dt}=0$
$\Rightarrow$ t=5 seconds
Minimum gap $=20+\frac{1}{2}\times (5{)}^2-5(5)=7.5m=1.5\times 5(m)$