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

To solve the problem of finding the relationship between the angles \( \alpha \) and \( \beta \) when rain is pouring at angle \( \alpha \) and a woman runs against it making angle \( \beta \) with the vertical, we need to consider both the horizontal and vertical components of the velocities involved.

The rain's velocity \( \mathbf{v}_r \) can be decomposed into horizontal and vertical components as follows:

• Horizontal component: \( v_{r,h} = 10 \sin \alpha \)
• Vertical component: \( v_{r,v} = 10 \cos \alpha \)

The woman's velocity \( \mathbf{v}_w = 8 \, \text{m/s} \) is purely horizontal.

From the woman's perspective, the rain appears to fall at an angle \( \beta \) with the vertical. The effective horizontal and vertical velocities as seen by the woman are:

• Effective horizontal velocity: \( v_{e,h} = 10 \sin \alpha + 8 \)
• Effective vertical velocity: \( v_{e,v} = 10 \cos \alpha \)

The angle \( \beta \) with the vertical can be determined using the tangent function, which is the ratio of the effective horizontal velocity to the effective vertical velocity:

\(\tan \beta = \frac{v_{e,h}}{v_{e,v}} = \frac{10 \sin \alpha + 8}{10 \cos \alpha}\)

Thus, the correct relationship is:

\(\tan \beta = \frac{8 + 10 \sin \alpha}{10 + 8 \cos \alpha}\)

Answer 2

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 \( \tan \beta = \frac{8 + 10 \sin \alpha}{10 + 8 \cos \alpha} \).