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

Answer 3

Take the vertical as the reference direction, since both angles are measured from it. The rain falls at 10 m/s at angle \(\alpha\) to the vertical, so its vertical and horizontal components are \(10\cos\alpha\) and \(10\sin\alpha\) respectively. The woman runs horizontally at 8 m/s against the rain's horizontal drift, so relative to her the rain's horizontal component becomes \(10\sin\alpha+8\), while its vertical component, \(10\cos\alpha\), is unaffected because her motion is purely horizontal. The apparent angle \(\beta\) with the vertical then satisfies \(\tan\beta = \dfrac{10\sin\alpha+8}{10\cos\alpha}\). Each option is checked against this relation.

1. Option A \(\left(\tan\beta=\dfrac{8+10\sin\alpha}{10+8\cos\alpha}\right)\): The denominator here adds a term in \(\cos\alpha\) scaled by the woman's speed, but her motion is purely horizontal and cannot change the rain's vertical component, so the denominator should stay as \(10\cos\alpha\) alone, not \(10+8\cos\alpha\).
2. Option B \(\left(\tan\beta=\dfrac{8\cos\alpha}{10+8\sin\alpha}\right)\): This places the woman's speed in the numerator multiplied by \(\cos\alpha\) and mixes it into the denominator with a sine term, which does not follow from adding two velocities along fixed vertical/horizontal axes.
3. Option C \(\left(\tan\beta=\dfrac{8+10\cos\alpha}{10\sin\alpha}\right)\): This swaps the roles of sine and cosine between the numerator and denominator, which would only be correct if the angles were measured from the horizontal instead of the vertical.
4. Option D \(\left(\tan\beta=\dfrac{8+10\sin\alpha}{10\cos\alpha}\right)\): This matches exactly the ratio of the combined horizontal component, \(8+10\sin\alpha\), to the unchanged vertical component, \(10\cos\alpha\).

Only option D correctly keeps the vertical component of the rain's velocity unaffected by the woman's purely horizontal motion while adding the two horizontal components together.

Hence, the correct answer is \(\tan\beta = \dfrac{8+10\sin\alpha}{10\cos\alpha}\).