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Rain is falling vertically downwards with a speed of \(4 \mathrm{kmh}^{-1}\). A girl moves on a straight road with a velocity of \(3 \mathrm{kmh}^{-1}\). The apparent velocity of rain with respect to the girl is a. \(3 \mathrm{kmh}^{-1}\) b. \(4 \mathrm{kmh}^{-1}\) c. \(5 \mathrm{kmh}^{-1}\) d. \(7 \mathrm{kmh}^{-1}\)

Short Answer

Expert verified
The apparent velocity of the rain with respect to the girl is \(5 \mathrm{kmh}^{-1}\).

Step by step solution

01

Understanding the Given Problems

We are given two velocities: the speed of the rain falling vertically, which is \(4 \mathrm{kmh}^{-1}\), and the velocity of the girl moving on the road, which is \(3 \mathrm{kmh}^{-1}\). We are tasked with finding the apparent velocity of the rain as perceived by the girl.
02

Determine Relative Velocity Components

The rain is moving vertically downward with a speed of \(4 \mathrm{kmh}^{-1}\). This is along the vertical (y) axis. The girl is moving horizontally with a speed of \(3 \mathrm{kmh}^{-1}\), which is along the horizontal (x) axis.
03

Calculate Apparent Velocity Magnitude

To determine the apparent velocity (combined effect of both movements), apply the Pythagorean theorem. The velocity's magnitude is given by:\[ v_r = \sqrt{v_{rx}^2 + v_{ry}^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \mathrm{kmh}^{-1} \]
04

Conclusion on Apparent Velocity

The apparent velocity of the rain with respect to the girl is \(5 \mathrm{kmh}^{-1}\). Therefore, option c is the correct answer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Apparent Velocity
Apparent velocity is the perceived speed of an object as observed from another vantage point that is itself in motion. Imagine you are inside a car, watching raindrops cascading down your side window. How fast do those raindrops appear to be moving relative to your motion? This is a fundamental example of apparent velocity.

In the exercise, the girl perceives the rain's speed relating to her own movement on the road. This perceived motion combines the girl's horizontal velocity with the rain's vertical velocity. To find out how the girl sees the rain's motion, we calculate the apparent velocity by considering these two velocities together.
Pythagorean Theorem
The Pythagorean theorem is a mathematical tool used to relate dimensions in geometry. It lets us find the length of the hypotenuse of a right triangle once we know the lengths of the other two sides. This is essential for solving problems like the aforementioned exercise involving two perpendicular velocity components.

As per the theorem, for any right triangle with legs of lengths "a" and "b" and hypotenuse of length "c", you can calculate the relationship using:
  • \( c^2 = a^2 + b^2 \)
This formula is utilized directly in our inquiry of apparent velocity. The girl's movement forms one leg, and the rain's vertical downpour forms the other, creating a right triangle. The apparent velocity, which is the hypotenuse, measures the combined effect of these independent motions.
Vector Components
Understanding vector components helps us separate and analyze complex motions in physics. Vectors allow us to resolve movements into perpendicular components that can be considered individually. In our example, we have two motions: the rain falling straight down and the girl moving along a straight path.

These movements can be observed as vector components:
  • Rain's vertical velocity: downward direction with magnitude \(4 \mathrm{kmh}^{-1}\).
  • Girl's horizontal velocity: left or right with magnitude \(3 \mathrm{kmh}^{-1}\).
By breaking down the overall velocity into these separate parts, we can employ mathematical tools like the Pythagorean theorem to solve for combined effects, like apparent velocity. Understanding vector components is key to calculating how readers see such two-dimensional motions seamlessly combined.

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Most popular questions from this chapter

Ship \(\mathrm{A}\) is travelling with a velocity of \(5 \mathrm{kmh}^{-1}\) due east. The second ship is heading \(30^{\circ}\) east of north. What should be the speed of second ship if it is to remain always due north with respect to the first ship? a. \(10 \mathrm{kmh}^{-1}\) b. \(9 \mathrm{kmh}^{-1}\) c. \(8 \mathrm{kmh}^{-1}\) d. \(7 \mathrm{kmh}^{-1}\)

A car is going in south with a speed of \(5 \mathrm{~m} / \mathrm{s}\). To a man sitting in car a bus appears to move towards west with a speed of \(2 \sqrt{6} \mathrm{~m} / \mathrm{s}\). What is the actual speed of the bus? a. \(4 \mathrm{~ms}^{-1}\) b. \(3 \mathrm{~ms}^{-1}\)c. \(7 \mathrm{~ms}^{-1}\) d. None of these

A man can swim in still water with a speed of \(2 \mathrm{~m} / \mathrm{s}\). If he wants to cross a river of water current speed \(\sqrt{3} \mathrm{~m} / \mathrm{s}\) along the shortest possible path, then in which direction should he swim? a. At an angle \(120^{\circ}\) to the water current. b. At an angle \(150^{\circ}\) to the water current. c. At an angle \(90^{\circ}\) to the water current. d. None of these

The speed of a projectile at its maximum height is \(\sqrt{3} / 2\) times its initial speed. If the range of the projectile is \(P\) times the maximum height attained by it, \(P\) is equal to a. \(4 / 3\) b. \(2 \sqrt{3}\) c. \(4 \sqrt{3}\) d. \(3 / 4\)

At a height \(0.4 \mathrm{~m}\) from the ground, the velocity of a projectile in vector form is \(\vec{v}=(6 \hat{i}+2 \hat{j}) \mathrm{m} / \mathrm{s}\). The angle of projection is a. \(45^{\circ}\) b. \(60^{\circ}\) c. \(30^{\circ}\) d. \(\tan ^{-1}(3 / 4)\)

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