Question

A man starts driving in an open gypsy at $t=0$ is rainy weather. Assume that speed of gypsy varies with time as $v=kt$, at $t=1sec$, he observes rain is falling vertically to him. At $t=2sec$, the finds rain drops hitting him at an angle of $45^\circ$ with vertical. Assuming velocity of rain to be constant, the angle with vertical at which rain is actually falling is $tan^{-1}(x)$. The value of x is:

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (B)

The value of x is 1.

Answer 2

Let the rain’s velocity be $\vec{u} = (u_x, u_y)$ (constant) and the man’s velocity be $\vec{v} = (kt,0)$.

1. At $t=1$:

   • Man’s velocity: $k$.
   • Rain appears vertical, so the relative horizontal component must vanish: $$u_x - k = 0 \implies u_x = k.$$

2. At $t=2$:

   • Man’s velocity: $2k$.
   • Relative velocity: $\vec{u} - (2k,0) = (k-2k,\, u_y) = (-k,\, u_y)$.
   • The rain appears at $45^\circ$ with the vertical. Thus: $$\tan 45^\circ = \frac{|u_x-2k|}{|u_y|} = \frac{k}{|u_y|} = 1 \implies |u_y| = k.$$
   • Since rain falls downward, $u_y = -k$.

Thus, the actual rain velocity is $(k, -k)$. The angle with the vertical $\theta$ satisfies:

$$\tan\theta = \frac{|u_x|}{|u_y|} = \frac{k}{k} = 1.$$

So, $\theta = \tan^{-1}(1)$.

The question states the angle as $\tan^{-1}(x)$, hence $x = 1$.