Question

Two cars A and B are moving west to east and south to north respectively along crossroads. A moves with a speed of $18km{h}^{-1}$ and is $125m$ away from the point of intersection of crossroads and B moves with a speed of $13.5km{h}^{-1}$ and is $100m$ away from the point of intersection of crossroads. Find the shortest distance between them (in $m$)

Answer 1

Converting velocities in SI units,
Velocity of A $=18\times \frac{5}{18}=5m/s$
Velocity of B $=13.5\times \frac{5}{18}=3.75m/s$


After time $t$, let us plot the components of velocity of A and B in the direction along A'B'. (Where ${\alpha }_f$ is the angle made by the line A'B' with the x-axis)

When the distance between the two is minimum, the relative velocity of approach is zero.

$5\cos {\alpha }_f+(-3.75\sin {\alpha }_f)=0$
$5\cos {\alpha }_f=-3.75\sin {\alpha }_f$
$\therefore \tan {\alpha }_f=-\frac{5}{3.75}=-\frac{4}{3}$
Distance of particle B from the origin after time $t=100-3.75t$
Distance of particle A from the origin after time $t=125-5t$
$\therefore \tan {\alpha }_f=\frac{100-3.75t}{125-5t}=-\frac{4}{3}$
$\therefore t=\frac{128}{5}s$
So, OB'$=4m$ and OA'$=-3m$
A'B'$=\sqrt{4^2+(-3{)}^2}=5m$