Question

A car travels with a speed of \(40\ \text{km h}^{-1}\). Rain drops are falling at a constant speed vertically. The traces of the rain on the side windows of the car make an angle of \(30^\circ\) with the vertical. The magnitude of the velocity of the rain with respect to the car is

• A. \(40\sqrt3\ \text{km h}^{-1}\)
• B. \(\dfrac{40}{\sqrt3}\ \text{km h}^{-1}\)
• C. \(80\ \text{km h}^{-1}\)
• D. \(\dfrac{80}{\sqrt3}\ \text{km h}^{-1}\)

Hint:

In relative velocity problems involving rain and moving vehicles, the apparent direction of rain is determined by combining the horizontal velocity of the vehicle and vertical velocity of rain.

Answer 1

The Correct Option is C

Step 1: Understand the relative motion.
The rain is falling vertically downward with speed \(v\).
The car moves horizontally with speed
\[ 40\ \text{km h}^{-1} \]
Relative to the car, the rain appears inclined.

Step 2: Use the given angle.
The traces on the side window make an angle of \(30^\circ\) with the vertical.
Thus, for the relative velocity triangle,
\[ \tan30^\circ = \frac{\text{horizontal component}}{\text{vertical component}} \]
Horizontal component of relative velocity is the speed of the car:
\[ 40 \]
Vertical component is the speed of rain:
\[ v \]
Therefore,
\[ \tan30^\circ=\frac{40}{v} \]
\[ \frac1{\sqrt3}=\frac{40}{v} \]
\[ v=40\sqrt3 \]

Step 3: Find the relative velocity of rain with respect to the car.
Magnitude of relative velocity is
\[ V=\sqrt{(40)^2+(40\sqrt3)^2} \]
\[ =\sqrt{1600+4800} \]
\[ =\sqrt{6400} \]
\[ =80\ \text{km h}^{-1} \]

Step 4: Final conclusion.
Hence, the magnitude of velocity of rain with respect to the car is
\[ \boxed{80\ \text{km h}^{-1}} \]