Q.A girl riding a bicycle with a speed of 5 m/s towards north direction, observes rain falling vertically down. If she increases her speed to 10 m/s, rain appears to meet her at $45^\circ$ to the vertical. What is the speed of the rain? In what direction does rain fall as observed by a ground based observer? (Hint: Assume north to be $\hat{i}$ direction and vertically downward to be $-\hat{j}$. Let the rain velocity $\vec{v}_r$ be $a\hat{i} + b\hat{j}$. The velocity of rain as observed by the girl is always $\vec{v}_r - \vec{v}_{girl}$. Draw the vector diagram/s for the information given and find $a$ and $b$. You may draw all vectors in the reference frame of ground based observer.)

Concept: Relative velocity. The rain's velocity relative to the girl is $\vec{v}_{r/g} = \vec{v}_r - \vec{v}_g$.

Let $\vec{v}_r = a\hat{i} + b\hat{j}$, with north as $+\hat{i}$ and up as $+\hat{j}$. The girl's velocity is $\vec{v}_g = v_g\hat{i}$.

1. At $v_g = 5\ \text{m/s}$, rain appears vertical, so the horizontal part of $\vec{v}_{r/g}$ vanishes: $a - 5 = 0 \implies a = 5\ \text{m/s}$.
2. At $v_g = 10\ \text{m/s}$, rain appears at $45^\circ$ to the vertical. The relative velocity is $(5-10)\hat{i} + b\hat{j} = -5\hat{i} + b\hat{j}$, and at $45^\circ$ the two components are equal in magnitude: $|b| = 5$, so $b = -5\ \text{m/s}$ (rain falls downward). …

Setting up

Take north as $\hat{i}$ and vertically downward as $-\hat{j}$ (so up is $+\hat{j}$). Let the rain's velocity relative to the ground be $\vec{v}_r = a\hat{i} + b\hat{j}$. The velocity of rain as seen by the girl, moving at $\vec{v}_{\text{girl}}$, is $\vec{v}_r - \vec{v}_{\text{girl}}$.

Step 1 — First observation (girl at $5\ \text{m/s}$ north)

The rain appears to fall vertically, so the horizontal part of the relative velocity is zero:

Step 2 — Second observation (girl at $10\ \text{m/s}$ north)

Now the relative velocity is:

This makes $45^\circ$ with the vertical, so its horizontal and vertical parts must have equal magnitude:

Since the rain falls downward, $b = -5\ \text{m/s}$.

Step 3 — Speed of the rain

With $\vec{v}_r = 5\hat{i} - 5\hat{j}$:

$|\vec{v}_r| = \sqrt{5^2 + 5^2} = 5\sqrt{2} \approx 7.07\ \text{m/s}$ …

Concept: Subtract the Two Observation Equations to Eliminate the Unknown Rain Velocity First

Method: Difference-Vector Shortcut

The direct method solves for the rain's own velocity components $(a,b)$ using each observation as a separate equation. This method instead subtracts the two apparent-velocity equations from each other first -- since the (unknown, but fixed) rain velocity cancels out of the difference entirely, this immediately gives a relation between the known girl-velocity change and the known change in apparent direction, without solving for $\vec v_r$ at all until the final step.

Steps

1. Write the apparent velocity in each case as $\vec v_{r/\text{girl}} = \vec v_r - \vec v_{\text{girl}}$. Since $\vec v_r$ is the same, fixed vector in both observations, subtracting the two equations eliminates it:

This holds no matter what $\vec v_r$ actually is -- it's a pure consequence of both observations being of the same object.

2. Use the first observation to fix $\vec v_{r/\text{girl},1}$ completely. "Rain appears to fall vertically" at $5\text{ m/s}$ girl-speed means the apparent velocity is purely vertical:

3. Apply the difference-vector result from Step 1 to get the second apparent velocity directly, without any new unknowns:

4. Use the second observation's stated angle ($45^\circ$ to the vertical) to pin down $c$. At $45^\circ$, the horizontal and vertical magnitudes must be equal: $|-5| = |-c| \implies c = 5\text{ m/s}.$ …