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Step-by-Step Solution
Step 1: Understand the scenario
When the car is at rest, the rain appears to fall vertically downward to the driver. This tells us the rain’s velocity is vertically downward. Once the car moves forward with some speed, the driver perceives the raindrops to be coming at an angle from the horizontal due to relative motion.
Step 2: Represent the velocities
Let:
$v_r$ be the speed of the rain relative to the ground (vertically downward).
$v$ be the initial speed of the car (horizontally).
$\beta$ be the constant in the expression $(1 + \beta)\,v$ when the car’s speed is increased.
Angles ($\theta$) are measured from the horizontal as observed by the driver.
The relative velocity of rain with respect to the car is
$ \vec{V}_{\text{rain/car}} = \vec{V}_r - \vec{V}_c .$
Step 3: First observation (angle = 60°)
When the car’s speed is $v$, the rain appears to fall at 60° from the horizontal. From relative motion:
$ \tan 60^\circ = \frac{v_r}{v} = \sqrt{3}. $
Hence,
$ v_r = v \sqrt{3}. $
Step 4: Second observation (angle = 45°)
Now the speed of the car is increased to $(1 + \beta)\,v$, and the rain appears to fall at 45° from the horizontal. So:
$ \tan 45^\circ = \frac{v_r}{(1 + \beta)\,v} = 1. $
Therefore,
$ v_r = (1 + \beta)\,v. $
Step 5: Solve for β
From Step 3, we have $v_r = v\sqrt{3}$. Equating this to the expression from Step 4 gives:
$ v \sqrt{3} = (1 + \beta)\,v. $
Dividing both sides by $v$:
$ \sqrt{3} = 1 + \beta. $
Hence,
$ \beta = \sqrt{3} - 1 \approx 1.732 - 1 = 0.732. $
Which is close to 0.73.
Step 6: Final Answer
Therefore, the value of $ \beta $ is approximately 0.73.
