/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 227 A passenger in an automobile obs... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 227

A passenger in an automobile observes that raindrops make an angle of \(30^{\circ}\) with the horizontal as the auto travels forward with a speed of \(60 \mathrm{~km} / \mathrm{h}\). Compute the terminal (constant) velocity \(\mathbf{v}_{r}\) of the rain if it is assumed to fall vertically.

Short Answer

Expert verified
The terminal velocity of the rain is \(9.64 \mathrm{~m/s}\)

Step by step solution

01

Understanding Relative Motion

Recognize that the observed angle of the rain by the passenger is due to the forward motion of the car. Since the passenger in the car is in motion, it seems like the rain is falling at an angle. If the car was stationary, the rain would look like it is falling vertically.
02

Setting Up the Geometric Relationship

The angle \(30^{\circ}\), the car's speed, and the rain's terminal velocity form a right triangle. The horizontal component is the car's speed, and the vertical component is the rain's velocity. The relationship can be derived from the tangent of the angle, which is the opposite side (rain's velocity) divided by the adjacent side (car's speed). Therefore, \(\tan(30^{\circ}) = \frac{{v_r}}{{v_c}}\), where \(v_r\) is rain's velocity and \(v_c\) is the car's speed.
03

Solve for the Rain's Velocity

Rearrange the equation from Step 2 to solve for \(v_r\): \(v_r = v_c \times \tan(30^{\circ})\). Then, plug in the given values and solve: \(v_r = 60 \mathrm{~km/h} \times \tan(30^{\circ})\). For convenience, convert the car's speed from kilometers per hour to meters per second: \(v_c = 60 \mathrm{~km/h} \times \frac{1000 \mathrm{~m}}{1 \mathrm{~km}} \times \frac{1 \mathrm{~h}}{3600 \mathrm{~s}} = 16.7 \mathrm{~m/s}\). Hence, \(v_r = 16.7 \mathrm{~m/s} \times \tan(30^{\circ})\). Since \(\tan(30^{\circ}) = 0.577\), therefore, \(v_r = 9.64 \mathrm{~m/s}\). This is the terminal velocity of the rain.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Terminal Velocity
Terminal velocity is a fascinating concept often encountered in the study of physics, specifically when dealing with falling objects. When an object falls through a fluid, like air, it accelerates until the net force acting on it is zero. This means the force due to gravity is balanced by the drag force of the air.
  • At this point, the object moves at a constant speed, known as terminal velocity, because the forces are balanced.
  • In our exercise, the rain hitting the moving car is already at terminal velocity if it is assumed to fall vertically with a constant speed.
Understanding this concept helps explain why a passenger in the car sees the rain at an angle and why the rain doesn鈥檛 speed up infinitely but instead settles into this steady state called terminal velocity.
Trigonometry in Physics
Trigonometry is vital in physics for solving problems related to angles and lengths. It is especially useful when analyzing problems of relative motion, like in our exercise where the rain's angle is observed from a moving car.
  • This problem forms a right triangle, with the rain's vertical velocity and the car's horizontal speed creating the triangle's sides.
  • The tangent function helps us relate these sides because \( an(30^\circ)\) gives the ratio of the rain's vertical speed to the car's speed.
By understanding trigonometry, we can calculate the unknown component of the rain's velocity that isn't immediately apparent just by observation. This makes trigonometry an invaluable tool in predicting how objects move with respect to each other.
Rainfall Dynamics
Rainfall dynamics explores how rain appears when observed from different frames of reference. From a stationary point, rain seems to fall straight down. However, when observed from a moving vehicle, rain may appear to fall at an angle because of relative motion.
  • The way we perceive rain is an excellent practical example of relative velocity, a fundamental concept in dynamics.
  • In our exercise, the passenger's perspective introduces a unique viewpoint on the rain's movement, making it an interesting application of relative motion.
The understanding of how rain appears in various scenarios aids in practical applications, such as designing windshield wipers and improving rain gear, making everyday life more convenient.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If the position of a particle is defined as \(s=\) \(\left(5 t-3 t^{2}\right) \mathrm{m},\) where \(t\) is in seconds, construct the \(s-t, v-t,\) and \(a-t\) graphs for \(0 \leq t \leq 2.5 \mathrm{~s}\).

At the instant shown, the watersprinkler is rotating with an angular speed \(\dot{\theta}=2 \mathrm{rad} / \mathrm{s}\) and an angular acceleration \(\ddot{\theta}=3 \mathrm{rad} / \mathrm{s}^{2}\). If the nozzle lies in the vertical plane and water is flowing through it at a constant rate of \(3 \mathrm{~m} / \mathrm{s}\), determine the magnitudes of the velocity and acceleration of a water particle as it exits the open end, \(r=0.2 \mathrm{~m}\).

Small packages traveling on the conveyor belt fall off into a l-m-long loading car. If the conveyor is running at a constant speed of \(v_{C}=2 \mathrm{~m} / \mathrm{s},\) determine the smallest and largest distance \(R\) at which the end \(A\) of the car may be placed from the conveyor so that the packages enter the car.

A rocket is fired from rest at \(x=0\) and travels along a parabolic trajectory described by \(y^{2}=\left[120\left(10^{3}\right) x\right] \mathrm{m}\). If the \(x\) component of acceleration is \(a_{x}=\left(\frac{1}{4} t^{2}\right) \mathrm{m} / \mathrm{s}^{2}\), where \(t\) is in seconds, determine the magnitude of the rocket's velocity and acceleration when \(t=10 \mathrm{~s}\).

A man walks at \(5 \mathrm{~km} / \mathrm{h}\) in the direction of a \(20 \mathrm{~km} / \mathrm{h}\) wind. If raindrops fall vertically at \(7 \mathrm{~km} / \mathrm{h}\) in still air, determine direction in which the drops appear to fall with respect to the man.
See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Medical Physics

Read Explanation

Space Physics

Read Explanation

Electrostatics

Read Explanation

Atoms and Radioactivity

Read Explanation

Geometrical and Physical Optics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.