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Chapter 5: Problem 23

Rain, driven by the wind, falls on a railway compartment with a velocity of \(20 \mathrm{~m} / \mathrm{s}\), at an angle of \(30^{\circ}\) to the vertical. The train moves, along the direction of wind flow, at a speed of \(108 \mathrm{~km} / \mathrm{hr}\). Determine the apparent velocity of rain for a person sitting in the train. a. \(20 \sqrt{7} \mathrm{~m} / \mathrm{s}\) b. \(10 \sqrt{7} \mathrm{~m} / \mathrm{s}\) c. \(15 \sqrt{7} \mathrm{~m} / \mathrm{s}\) d. \(10 \sqrt{7} \mathrm{~km} / \mathrm{h}\)

Short Answer

Expert verified
The apparent velocity of rain is \( 10\sqrt{7} \mathrm{~m/s} \), matching option b.

Step by step solution

01

Convert train speed to m/s

Begin by converting the train's speed from kilometers per hour to meters per second. The train moves at 108 km/hr. Use the conversion factor where 1 km/hr is equivalent to \( \frac{1}{3.6} \) m/s. Thus, the train's speed in m/s is \( 108 \times \frac{1}{3.6} = 30 \mathrm{~m/s} \).
02

Resolve rain velocity into components

The velocity of rain is given as 20 m/s at an angle of 30 degrees to the vertical. Break this into its horizontal and vertical components:For the vertical component: \( v_{r,v} = 20 \cos(30^{\circ}) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \mathrm{~m/s} \).For the horizontal component: \( v_{r,h} = 20 \sin(30^{\circ}) = 20 \times \frac{1}{2} = 10 \mathrm{~m/s} \).
03

Calculate apparent velocity components

The apparent velocity of rain for the observer in the train is affected by the train's speed. The horizontal velocity of the rain relative to the train is the difference between the horizontal component of rain's velocity and the speed of the train:\( v_{a,h} = v_{r,h} - v_{train} = 10 - 30 = -20 \mathrm{~m/s} \).The vertical component remains unchanged as the train moves only horizontally, so:\( v_{a,v} = v_{r,v} = 10\sqrt{3} \mathrm{~m/s} \).
04

Calculate magnitude of apparent velocity

Find the magnitude of the apparent velocity vector by using the Pythagorean theorem on the components:\[ v_a = \sqrt{(v_{a,h})^2 + (v_{a,v})^2} = \sqrt{(-20)^2 + (10\sqrt{3})^2} \]\[ v_a = \sqrt{400 + 300} = \sqrt{700} = 10\sqrt{7} \mathrm{~m/s} \]
05

Choose the correct answer

Based on our calculation, the apparent velocity of rain for a person sitting in the train is \( 10\sqrt{7} \mathrm{~m/s} \). The correct answer is option b.

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Key Concepts

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

Rain movement
When rain falls, it's often influenced by additional factors like wind. In this exercise, the rain itself has a velocity of 20 meters per second at a 30-degree angle relative to the vertical.
This means that the rain doesn't fall straight down but is moved by the wind to some extent. To better understand this, think about how on a windy day, raindrops don't hit you vertically but rather diagonally when walking outside. The path the rain follows becomes a combination of straight-down motion, due to gravity, and sideways motion, caused by the wind.
Apparent velocity
Apparent velocity is the speed of an object as observed from a particular frame of reference. In this case, the person on the train sees the rain differently than someone standing still.
The train is moving forward at 30 meters per second, so it's almost like the backdrop against which the rain's motion is perceived has shifted.

  • The rain's horizontal component is affected by the speed of the train, as the observer (person in the train) moves along with it.
  • The train moving at high-speed changes the horizontal velocity of the rain based on the observer's motion.
Thus, apparent velocity shows how speed can look different depending on how both the observer and the object are moving.
Vector resolution
Vector resolution is breaking down a vector into its constituent components, usually horizontal and vertical.
This is an essential tool for understanding how different forces acting at angles can be simplified into straightforward, right-angle vectors.
  • The rain's velocity was separated into vertical and horizontal parts because this allows for easier comparison with the train's horizontal motion.
  • The horizontal component of the rain is affected directly by the train's movement, while the vertical component stays unchanged since the train only moves horizontally.
Using vector resolution, we simplify complex motions into simpler, calculable components, helping clarify which parts interact with each other.
Pythagorean theorem
To find the overall apparent velocity of the rain, we apply the Pythagorean theorem, which relates the lengths of the sides of a right triangle.
In our scenario, each velocity component (horizontal and vertical) can be thought of as a side of a right triangle.

The formula is:\[ v_a = \sqrt{(v_{a,h})^2 + (v_{a,v})^2} \]
  • Here, \( v_{a,h} \) is the apparent horizontal velocity, and \( v_{a,v} \) is the vertical velocity of the rain as observed from inside the train.
  • Solving the equation helps determine the true magnitude of the rain's apparent velocity as experienced from the moving train.
The use of the Pythagorean theorem allows us to calculate how the combination of two perpendicular movements results in a diagonal motion, which in this case is the combined speed of the rain as seen from within the train.

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Most popular questions from this chapter

A projectile is thrown at an angle of \(40^{\circ}\) with the horizontal and its range is \(R_{1}\). Another projectile is thrown at an angle \(40^{\circ}\) with the vertical and its range is \(R_{2}\). What is the relation between \(R_{1}\) and \(R_{2}\) ? a. \(R_{1}=R_{2}\) b. \(R_{1}=2 R_{2}\) c. \(2 R_{1}=R_{2}\) d. \(R R_{1}=4 R_{2} / 5\)

At what angle with the horizontal should a ball be thrown so that the range \(R\) is related to the time of flight as \(R=5 T^{2} .\) (Take \(g=10 \mathrm{~ms}^{-2}\) ) a. \(30^{\circ}\) b. \(45^{\circ}\) c. \(60^{\circ}\) d. \(90^{\circ}\)

i. If air resistance is not considered in a projectile motion, the horizontal motion takes place with a. constant velocity b. constant acceleration c. constant retardation d. variable velocityii. When a projectile is fired at an angle \(\theta\) ith horizontal with velocity \(u\), then its vertical component a. remains the sameb. goes on increasing with the height c. first decreases and then increases with the height d. first increases then decreases with the height iii. Range of a projectile is \(R\), when the angle of projection is \(30^{\circ}\). Then, the value of the other angle of projection for the same range is a. \(45^{\circ}\) b. \(60^{\circ}\) c. \(50^{\circ}\) d. \(40^{\circ}\) iv. A ball is thrown upwards and it returns to ground describing a parabolic path. Which of the following quantitics remains constant throughout the motion? a. kinetic energy of the ball b. speed of the ball c. horizontal component of velocity d. vertical component of velocity

What is the angular velocity in \(\mathrm{rad} / \mathrm{s}\) of a fly wheel making 300 r.p.m? a. \(600 \pi\) b. \(20 \pi\) c. \(10 \pi\) d. 30

A projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of \(3 \mathrm{~ms}^{-2}\) for \(0.5 \mathrm{~min}\). If the maximum height reached by it is \(80 \mathrm{~m}\), then the angle of projection is \(\left(g=10 \mathrm{~ms}^{-2}\right)\) a. \(\tan ^{-1} 3\) b. \(\tan ^{-1}(3 / 2)\) c. \(\tan ^{-1}(4 / 9)\) d. \(\sin ^{-1}(4 / 9)\)
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