1. Set up the rain velocity

When the car is at rest, the driver sees the rain falling vertically. Hence, in the ground frame, the rain has only a vertical velocity component.

Let the vertical speed of rain be $u$ downward.

So, rain velocity in ground frame is: $\vec v_r = -u\hat j$

2. When the car moves with speed $v$

Let the car move horizontally to the right with speed $v$.

Then the velocity of rain relative to the car is: $\vec v_{r/c} = \vec v_r - \vec v_c = -v\hat i - u\hat j$

So, relative to the driver, the rain appears to come with:

• horizontal component $v$
• vertical component $u$

The rain is seen at an angle $60^\circ$ from the horizontal.

Thus, $\tan 60^\circ = \frac{u}{v}$ $\sqrt{3} = \frac{u}{v}$ $u = \sqrt{3}v$

3. When the car speed becomes $(1+\beta)v$

Now horizontal component of rain relative to car becomes: $(1+\beta)v$

Vertical component remains $u$.

The apparent angle is now $45^\circ$ from the horizontal, so: $\tan 45^\circ = \frac{u}{(1+\beta)v}$ $1 = \frac{u}{(1+\beta)v}$ $u = (1+\beta)v$

4. Substitute $u = \sqrt{3}v$

From above, $\sqrt{3}v = (1+\beta)v$

Cancelling $v$, $1+\beta = \sqrt{3}$ $\beta = \sqrt{3} - 1$

Now, $\sqrt{3} \approx 1.732$ $\beta \approx 1.732 - 1 = 0.732$

So, $\beta \approx 0.73$

5. Match with options

The closest option is:

• B: 0.73

Hence, the correct answer is B.