Question

Rain, pouring down at an angle \( \alpha \) with the vertical, has a constant speed of 10 m/s. A woman runs against the rain with a speed of 8 m/s and sees that the rain makes an angle \( \beta \) with the vertical. The relation between \( \alpha \) and \( \beta \) is given by:

• A. \( \tan \beta = \frac{8 + 10 \sin \alpha}{10 + 8 \cos \alpha} \)
• B. \( \tan \beta = \frac{8 \cos \alpha}{10 + 8 \sin \alpha} \)
• C. \( \tan \beta = \frac{8 + 10 \cos \alpha}{10 \sin \alpha} \)
• D. \( \tan \beta = \frac{8 + 10 \sin \alpha}{10 \cos \alpha} \)

Hint:

Relative velocity is useful in solving problems involving motion of objects against each other, like this case with rain and a moving woman.

Answer 1

The Correct Option is A

By using relative velocity concepts, we can derive the relation between \( \alpha \) and \( \beta \). The resultant velocity of the rain relative to the woman will be: \[ v_{\text{rel}} = \sqrt{(v_{\text{rain}} \cos \alpha - v_{\text{woman}})^2 + (v_{\text{rain}} \sin \alpha)^2} \] The angle \( \beta \) is given by: \[ \tan \beta = \frac{v_{\text{rain}} \sin \alpha}{v_{\text{rain}} \cos \alpha - v_{\text{woman}}} \] Thus, the relation between \( \alpha \) and \( \beta \) is \( \tan \beta = \frac{8 + 10 \sin \alpha}{10 + 8 \cos \alpha} \).