Question

Two cars $A$ and $B$ are moving west to east and south to north, respectively, along crossroads. $A$ moves with a speed of $20m{s}^{-1}$ and is $500m$ away from the point of intersection of cross roads and $B$ moves with a speed of $15m{s}^{-1}$ and is $400m$ away from the point of intersection of cross roads. Find the shortest distance between them.

Answer 1

Using the concept of relative velocity

Let us assume the velocity of A w.r.t. B. To do this, we plot the resultant velocity, ${\overrightarrow{V}}_{AB}$

${\overrightarrow{V}}_{AB}={\overrightarrow{V}}_A-{\overrightarrow{V}}_B={\overrightarrow{V}}_A+(-{\overrightarrow{V}}_B)$

As the accelerations of both the cars 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 $A$ and $B$ is a perpendicular from $B$ on the line of motion of $A$ with respect to $B$.

From the figure

$tan\theta =\frac{V_B}{V_A}=\frac{15}{20}=\frac{3}{4}$ ....(i)

This $\theta$ is the angle made by the resultant velocity vector $|{\overrightarrow{V}}_{AB}|$ with the $x$-axis.

Also we know that from Fig.

$OC=\frac{x}{500}=\frac{3}{4}$

From equation (i) and (ii), we get $x$ = 375 m.

$\therefore AB=OB-OC=400-375=25m$

But the shortest distance is BP.

From diagram, it is clear that $BP=BCcos\theta =25\times \frac{4}{5}$

$\therefore BP=20m$