Question

A car is moving towards south with a speed of $20m/s$. A motorcyclist is moving towards east with a speed of $15m/s$. At a certain instant ($t=0$) the motorcyclist is due south of the car and is at a distance of $50m$ from the car. The shortest distance between the motorcyclist and the car is and the time after which they are nearest to each other after $t=0$ :

• A. $10m,1.6sec$.
• B. $20m,1sec$.
• C. $30m,1.6sec$.
• D. $40m,1sec$.

Answer 1

The correct option is C $30m,1.6sec$.

If we will observe the situation according to the car then velocity of motor cyclist with respect to the car is

$V_{me}=\sqrt{V_m^2+V_c^2}=\sqrt{{15}^2+{20}^2}=25m/s$

$\theta ={\tan }^{-1}\frac{V_c}{V_m}={\tan }^{-1}\frac{20}{15}={53}^o$ with x-axis.

$\therefore$ motorcyclist will more along the line MP with respect to the car as shown in the diagram below [ref. image]

The shortest distance will be the perpendicular distance from $C$ to $MP=d=50\cos {53}^o=30m$

Time taken to come closet $=$ time taken by motorcyclist to reach B

$\therefore t=\frac{MB}{|V_{mc}|}=\frac{50\sin {53}^o}{25}=2\times \frac{4}{5}=1.6second$.

$\therefore$ Option (C) is correct.