A car is moving towards south with a speed of `20 m s^(-1)`. A motorcyclist is moving towards east with a speed of `15 s^(-1)`. At a certain instant, the motorcyclist is due south of the car and is at a distance of `50 m`from the car.
The shortest distance between the motorcyclist and the car is.

Options

A. 10 m
B. 20 m
C. 30 m
D. 40 m

Answer

C

Explanation

( c) Taking `N as + y -` axis and `E as + x` Axis.
Imagine yourself as an observer sitting inside the car. You will regard the car as being at rest (at C). Relative to you, the speed of the motorcyclist is obtained by imposing the reversed velocity of the car motorcyclist as shown in (Fig.).
`v_m = 15 m s^-1, v_c = 20 m s^-1`
`v_(m C) = sqrt(15^2 + 20^2) = 25 m s^-1`
`theta = tan^-1 ((20)/(15)) = 53^@` with X - axis
The motorcyclist appears to move along the line `MP` with speed `25 m s^-1`.
The shortest distance = perpendicular distance of `MP` from
`rArr C = d rArr d = 50 cos 536@ rArr d = 30 m`
Time taken to come closet = time taken by motorcyclist to reach `B`
`t = (MB)/v_(m c) = (50 sin 53^@)/(25) rArr t = 1.6 s`.
.