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Chapter 5: Problem 20

Rain is falling vertically downwards with a speed of \(4 \mathrm{~km} \mathrm{~h}^{-1} .\) A girl moves on a straight road with a velocity of \(3 \mathrm{~km} \mathrm{~h}^{-1} .\) The apparent velocity of rain with respect to the girl is (1) \(3 \mathrm{~km} \mathrm{~h}^{-1}\) (2) \(4 \mathrm{kmh}^{-1}\) (3) \(5 \mathrm{~km} \mathrm{~h}^{-1}\) (4) \(7 \mathrm{kmh}^{-1}\)

Short Answer

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

Step by step solution

01

Identify Given Information

The rain is falling vertically downwards at a speed of \(4 \text{ km h}^{-1}\) and the girl is moving on a straight road at a velocity of \(3 \text{ km h}^{-1}\).
02

Understand Relative Velocity

Apparent velocity is the velocity of one object as observed from another. Here, we want to find the apparent velocity of the rain with respect to the girl.
03

Apply Relative Velocity Formula

Since the girl is moving horizontally and the rain is falling vertically, use Pythagorean theorem to find the apparent velocity: \[ v_{ ext{apparent}} = \sqrt{v_{ ext{rain}}^2 + v_{ ext{girl}}^2} = \sqrt{4^2 + 3^2} \]
04

Calculate the Speed

Substitute the given values into the formula:\[ v_{ ext{apparent}} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ km h}^{-1} \]
05

Select the Correct Option

Based on the calculation, the apparent velocity of the rain with respect to the girl is \(5 \text{ km h}^{-1}\), which corresponds to option (3).

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Key Concepts

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

Understanding Apparent Velocity
Apparent velocity refers to how fast one object seems to be moving from the perspective of another object.
This concept is crucial when considering the motion of objects from different reference frames.
In the given exercise, we're observing the rain from the perspective of the moving girl. While the rain falls vertically downward, the girl's motion adds a horizontal perspective to the observation. Imagine watching a movie - the trees out a car window seem to zip by quickly, even if they're not actually moving.
That's similar to apparent velocity, as it depends on your point of view. To find the apparent velocity, we need to understand both how the rain falls and how the girl moves.
  • Velocity of rain: vertical direction at 4 km/h
  • Velocity of girl: horizontal direction at 3 km/h
This situation creates a combined movement, where both vertical and horizontal motions are considered.
The Role of the Pythagorean Theorem
The Pythagorean theorem is a mathematical relationship between the sides of a right triangle.
It's extremely useful in problems involving perpendicular motions, like ours. Here, the girl's movement (horizontal) and the rain's fall (vertical) form the two perpendicular sides of a right triangle.
The apparent velocity of the rain, which combines these motions, corresponds to the hypotenuse. According to the Pythagorean theorem:
\[ c^2 = a^2 + b^2 \]
Where:
  • \( c \) is the hypotenuse (apparent velocity).
  • \( a \) is the downward velocity of the rain (4 km/h).
  • \( b \) is the horizontal velocity of the girl (3 km/h).
This theorem allows us to calculate the apparent velocity as the straight-line distance resulting from these two separate motions.
Step-by-Step Velocity Calculation
Now, let's perform the actual calculation. Using the Pythagorean theorem, we'll find the hypotenuse to get the apparent velocity. Plug the velocities into the formula:
\[ v_{\text{apparent}} = \sqrt{v_{\text{rain}}^2 + v_{\text{girl}}^2} = \sqrt{4^2 + 3^2} \]
This gives us:
  • \( v_{\text{apparent}} = \sqrt{16 + 9} \)
Continuing the calculation,
  • \( v_{\text{apparent}} = \sqrt{25} = 5 \text{ km h}^{-1} \)
Thus, the apparent velocity of the rain with respect to the girl is 5 km/h.
This information helps us select the correct answer to the original problem, which is option (3).

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Most popular questions from this chapter

Two particles are projected simultaneously from the same point. with the same speed, in the same vertical plane, and at different angles with the horizontal in a uniform gravitational field acting vertically downwards. A frame of reference is fixed to one particle. The position vector of the other particle, as observed from this frame, is \(\vec{r}\). Which of the following statements is correct? (1) \(\bar{r}\) is a constant vector. (2) \(\vec{r}\) changes in magnitude as well as direction with time. (3) The magnitude of \(\bar{r}\) increases linearly with time; its direction does not change. (4) The direction of \(\vec{r}\) changes with time; its magnitude may or may not change, depending on the angles of projection.

A particle moves along positive branch of the curre \(y=\frac{x}{2}\), where \(x=\frac{t^{3}}{3}, x\) and \(y\) are measured in meters an \(t\) in seconds, then (1) The velocity of particle at \(t=1 \mathrm{~s}\) is \(\hat{i}+\frac{1}{2} \hat{j}\). (2) The velocity of particle at \(t=1 \mathrm{~s}\) is \(\frac{1}{2} \hat{i}+\hat{j}\). (3) The acceleration of particle at \(t=2 \mathrm{~s}\) is \(2 \hat{i}+\hat{j}\). (4) The acceleration of particle at \(t=2 \mathrm{~s}\) is \(\hat{i}+2 \hat{j}\).

A helicopter is flying at \(200 \mathrm{~m}\) and flying at \(25 \mathrm{~m} \mathrm{~s}^{-1}\) at an angle \(37^{\circ}\) above the horizontal when a package is dropped from it. The distance of the point from point \(O\) where the package lands is (1) \(80 \mathrm{~m}\) (2) \(100 \mathrm{~m}\) (3) \(200 \mathrm{~m}\) (4) \(160 \mathrm{~m}\)

A river is flowing towards with a velocity of \(5 \mathrm{~m} \mathrm{~s}^{-1}\). The boat velocity is \(10 \mathrm{~m} \mathrm{~s}^{-1}\). The boat crosses the river by shortest path. Hence, (1) The direction of boat's velocity is \(30^{\circ}\) west of north. (2) The direction of boat's velocity is north-west. (3) Resultant velocity is \(5 \sqrt{3} \mathrm{~m} \mathrm{~s}^{-1}\). (4) Resultant velocity of boat is \(5 \sqrt{2} \mathrm{~m} \mathrm{~s}^{-1}\).

An aeroplane is flying vertically upwards. When it is at a height of \(1000 \mathrm{~m}\) above the ground and moving at a speed of \(367 \mathrm{~m} / \mathrm{s}\), a shot is fired at it with a speed of \(567 \mathrm{~m} \mathrm{~s}^{-1}\) from a point directly below it. What should be the acceleration of aeroplane so that it may escape from being hit? (1) \(>5 \mathrm{~ms}^{-2}\) (2) \(>10 \mathrm{~ms}^{-2}\) (3) \(<10 \mathrm{~ms}^{-2}\) (4) Not possible
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