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A car is moving on horizontal ground wit...

A car is moving on horizontal ground with a velocity vm `s^(-1)` with respect to earth. The driver of the car observes the rain to be falling vertically downwards with velocity 2vm `s^(-1)` . What is the velocity of the rain with respect to the driver ?

A

3v

B

`vsqrt(3)`

C

`vsqrt(5)`

D

none of these

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we need to analyze the situation involving the car, the rain, and their respective velocities. ### Step 1: Understand the given information - The car is moving horizontally with a velocity \( v_m \) (let's denote it as \( v \)). - The rain is observed to be falling vertically downwards with a velocity of \( 2v_m \) (denote it as \( 2v \)) from the driver's perspective. ### Step 2: Establish the reference frames - The velocity of the car with respect to the ground is \( \vec{V}_{cg} = v \hat{i} \) (where \( \hat{i} \) is the unit vector in the horizontal direction). - The velocity of the rain with respect to the driver is \( \vec{V}_{rd} = 0 \hat{i} - 2v \hat{j} \) (where \( \hat{j} \) is the unit vector in the vertical direction). ### Step 3: Use the relative velocity concept The velocity of the rain with respect to the ground can be expressed using the relative velocity formula: \[ \vec{V}_{rg} = \vec{V}_{rd} + \vec{V}_{cg} \] Where: - \( \vec{V}_{rg} \) is the velocity of the rain with respect to the ground. - \( \vec{V}_{rd} \) is the velocity of the rain with respect to the driver. - \( \vec{V}_{cg} \) is the velocity of the car with respect to the ground. ### Step 4: Substitute the known values Substituting the values we have: \[ \vec{V}_{rg} = (0 \hat{i} - 2v \hat{j}) + (v \hat{i} + 0 \hat{j}) \] This simplifies to: \[ \vec{V}_{rg} = v \hat{i} - 2v \hat{j} \] ### Step 5: Calculate the magnitude of the rain's velocity with respect to the ground To find the magnitude of the velocity of the rain with respect to the ground, we calculate: \[ |\vec{V}_{rg}| = \sqrt{(v)^2 + (-2v)^2} \] \[ |\vec{V}_{rg}| = \sqrt{v^2 + 4v^2} = \sqrt{5v^2} = v\sqrt{5} \] ### Step 6: Conclusion Thus, the velocity of the rain with respect to the driver is \( v\sqrt{5} \) m/s.

To solve the problem step by step, we need to analyze the situation involving the car, the rain, and their respective velocities. ### Step 1: Understand the given information - The car is moving horizontally with a velocity \( v_m \) (let's denote it as \( v \)). - The rain is observed to be falling vertically downwards with a velocity of \( 2v_m \) (denote it as \( 2v \)) from the driver's perspective. ### Step 2: Establish the reference frames - The velocity of the car with respect to the ground is \( \vec{V}_{cg} = v \hat{i} \) (where \( \hat{i} \) is the unit vector in the horizontal direction). ...
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