When a car is at rest, its driver sees rain drops falling on it vertically. When driving the car wit

Q: Why do we subtract the car's velocity?

Because we are finding the relative velocity of the rain *with respect to* the car. We must bring the observer (the car) to rest by applying an equal and opposite velocity vector to both the car and the rain.

Q: The question says the angle is from the horizontal. Why did we use the angle with the vertical?

You can use either! If the angle with the horizontal is $60^\circ$, the angle with the vertical is $90^\circ - 60^\circ = 30^\circ$. Using the vertical angle often makes the tangent ratio directly equal to $v_{car} / v_{rain}$, which is slightly more intuitive to visualize.

Q: Does the true velocity of the rain change when the car speeds up?

No, the true velocity of the rain $\vec{v}_r$ relative to the ground remains perfectly constant. Only the *apparent* velocity $\vec{v}_{r/m}$ changes because the observer's frame of reference is moving faster.

Q: What if the rain was already falling at an angle due to wind?

Then the initial $\vec{v}_r$ vector would not be perfectly vertical. You would have to resolve it into horizontal and vertical components before applying the relative velocity formula and drawing your vector triangles.

Q: Why did the angle decrease from $60^\circ$ to $45^\circ$ when the car sped up?

As the car moves faster, the horizontal component of the relative velocity increases, making the rain appear to come more horizontally. A smaller angle with the horizontal means it's hitting the windshield more directly!

Question

When a car is at rest, its driver sees rain drops falling on it vertically. When driving the car with speed $v$, he sees that rain drops are coming at an angle $60^\circ$ from the horizontal. On further increasing the speed of the car to $(1+ \beta)v$, this angle changes to $45^\circ$. The value of $\beta$ is close to

Options:

(A) 0.50

(B) 0.41

(C) 0.37

(D) 0.73

Sigma Percentile · Accepted Answer

### The Illusion of Falling Rain

Imagine you are sitting in a stationary car. The rain drops are falling perfectly vertically downwards. At this moment, your velocity is zero, and the rain's true velocity is $\vec{v}_r$.

Now, as soon as you start driving, the rain feels like it's hitting your windshield at an angle. This is the classic concept of relative velocity. To find the velocity of rain with respect to the man (you), we subtract the man's velocity from the rain's velocity:

$$\vec{v}_{r/m} = \vec{v}_r - \vec{v}_m$$

### Constructing the Vector Triangle

Let's look at the first scenario. You are driving with a speed $v$. The question states that the rain appears to come at an angle of $60^\circ$ from the horizontal. This means the angle it makes with the vertical will be $90^\circ - 60^\circ = 30^\circ$.

If we draw a vector diagram, we form a right-angled triangle where the opposite side is the magnitude of your velocity $v$, and the adjacent side is the true vertical speed of the rain $v_r$. Using basic trigonometry:

$$	an 30^\circ = \frac{v}{v_r}$$

Since $	an 30^\circ = \frac{1}{\sqrt{3}}$, we can easily solve for the true speed of the rain:

$$v_r = \sqrt{3}v$$

### Shifting Gears: The Second Scenario

Now you press the accelerator. Your new speed is $(1+\beta)v$. Because you are moving faster, the rain now appears to hit at a shallower angle of $45^\circ$ from the horizontal. Consequently, the angle from the vertical is also $45^\circ$.

Let's update our vector triangle. In this new triangle, we apply the tangent function again:

$$	an 45^\circ = \frac{(1+\beta)v}{v_r}$$

We know that $	an 45^\circ = 1$. So, from this equation, we find another expression for the true speed of the rain:

$$v_r = (1+\beta)v$$

### The Final Equivalence

We now have two distinct equations for $v_r$. Since the actual speed of the rain falling from the sky hasn't changed, we can safely equate both expressions:

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

The velocity $v$ cancels out beautifully on both sides, leaving us with a simple algebraic equation for $\beta$:

$$1+\beta = \sqrt{3}$$

$$\beta = \sqrt{3} - 1$$

Knowing that the square root of 3 is approximately $1.732$, we get:

**$\beta \approx 1.732 - 1 = 0.732$**

Rounding to two decimal places, we arrive at our final answer of **$0.73$**. See how elegantly vector triangles simplify relative motion problems!

Select Answer:

Visualized Solution

$Initial State$

$Relative Velocity Concept$

$Case 1: Speed v$

$Calculating v_{r}$

$Case 2: Speed (1 + β) v$

$Calculating v_{r} again$

$Equating and Solving$

$Final Answer$

The Sigma Insight: Relative Velocity

The Illusion of Falling Rain

Constructing the Vector Triangle

Shifting Gears

The Final Equivalence

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial State

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Relative Velocity Concept

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case 1: Speed v

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating vr​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case 2: Speed (1+β)v

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating vr​ again

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating and Solving

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer