Question

Two cars $A$ and $B$ are moving west to east and south to north respectively along crossroads. $A$ moves with a speed of $72{\text{kmh}}^{-1}$ and is $500\text{m}$ away from point of intersection of cross roads and $B$ moves with a speed of $54{\text{kmh}}^{-1}$ and is 400 m away from point of intersection of cross roads. Find the shortest distance between them in metres?

Answer 1

Method-I (Using the concept of relative velocity)
In this method we watch the velocity of $A$ w.r.t. $B$. To do this we plot the resultant velocity $V_r$. Since the accelerations of both the bodies is zero, so the relative acceleration between them is also zero. Hence the relative velocity will remain constant. So, the path of $A$ with respect to $B$ will be straight line and along the direction of relative velocity of $A$ with respect to $B$. The shortest distance between $A$ & $B$ is when $A$ is at point F (i.e. when we drop a perpendicular from $B$ on the line of motion of $A$ with respect to $B$).

$\tan \theta =\frac{V_B}{V_A}=\frac{54}{72}=\frac{3}{4}\dots \dots \left(i\right)$
This $\theta$ is the angle made by the resultant velocity vector with the $x-$ axis.
From the figure, we can say
$\tan \theta =\frac{OE}{OA}=\frac{OE}{500}\dots \dots \left(ii\right)$
From equation $\left(i\right)$ and $\left(ii\right)$ we get
$OE=375\text{m}$
$\therefore BE=OB-OE=400-375=25\text{m}$
But the shortest distance is BF.
From the magnified figure we see that $BF=BE\cos \theta =25\times \frac{4}{5}$
$\therefore BF=20\text{m}$
Method II (using the concept of relative velocity of approach)
After time $t$ let us plot the components of velocity of $A$ and $B$ in the direction along $A^{{}^{\prime }}{B}^{{}^{\prime }}$.

When the distance between the two is minimum, the relative velocity of approach is zero.
$\therefore V_A\cos {\alpha }_f+V_B\sin {\alpha }_f=0$
(Where ${\alpha }_f$ is the angle made by the line $A^{{}^{\prime }}{B}^{{}^{\prime }}$ with the $x-$ axis)
$20\cos {\alpha }_f=-15\sin {\alpha }_f$
$\therefore \tan {\alpha }_f=-\frac{20}{15}=-\frac{4}{3}$
(Here do not confuse this angle with the angle $\theta$ in method (l) because that $\theta$ is the angle made by the net relative velocity with $x-$ axis, but ${\alpha }_f$ is the angle made by the line joining the two particles with $x-$ axis when velocity of approach is zero.)
Distance of particle $B$ from the origin after time $t=400-15t$
Distance of particle $A$ from the origin after time $t=500-20t$
$\therefore \tan {\alpha }_f=\frac{400-15t}{500-20t}=-\frac{4}{3}$
$\therefore t=\frac{128}{5}$
So, $O{B}^{{}^{\prime }}=16\text{m}$ and $O{A}^{{}^{\prime }}=-12\text{m}$
$A^{{}^{\prime }}{B}^{{}^{\prime }}=\sqrt{{16}^2+(-12{)}^2}=20\text{m}$