Question

Rain is falling vertically with a speed of \(2\,\text{ms}^{-1}\). A boy from rest starts moving with a constant acceleration of \(2\,\text{ms}^{-2}\) along a straight road holding an umbrella. For the rain to be always parallel to the axis of umbrella, the rate at which the angle of axis of umbrella with the vertical should be changing at time \(t\) is

• A. \[ \frac{1+t^2}{t^3} \]
• B. \[ \frac{1}{1+t^2} \]
• C. \[ \frac{t^2}{1+t^4} \]
• D. \[ \frac{t}{1+t^3} \]

Hint:

For rain-umbrella problems, always use relative velocity. The umbrella should be directed along the relative velocity of rain with respect to the observer: \[ \tan\theta = \frac{v_{\text{horizontal}}}{v_{\text{vertical}}}. \] Then differentiate \(\theta\) if the rate of change of angle is required.

Answer 1

The Correct Option is B

Concept: The umbrella must always be aligned with the relative velocity of rain with respect to the boy. If \(\theta\) is the angle made by the umbrella with the vertical, then \[ \tan\theta = \frac{\text{horizontal component of relative velocity}} {\text{vertical component of relative velocity}}. \]

Step 1: Find the velocity of the boy at time \(t\). The boy starts from rest with acceleration \[ a=2\,\text{ms}^{-2}. \] Hence, \[ v=at=2t. \]

Step 2: Determine the relative velocity of rain. Rain falls vertically downward with speed \[ 2\,\text{ms}^{-1}. \] Relative velocity of rain with respect to the boy has components \[ 2t \] horizontally and \[ 2 \] vertically downward. Therefore, \[ \tan\theta = \frac{2t}{2} = t. \] Thus, \[ \theta=\tan^{-1}t. \]

Step 3: Find the rate of change of \(\theta\). Differentiating, \[ \frac{d\theta}{dt} = \frac{d}{dt}\left(\tan^{-1}t\right). \] \[ \frac{d\theta}{dt} = \frac{1}{1+t^2}. \] Therefore, \[ \boxed{\frac{d\theta}{dt}=\frac{1}{1+t^2}} \] \[ \boxed{\text{Answer = (B)}} \]