/*! 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 77 Snow is falling vertically at a ... [FREE SOLUTION] | 91Ó°ÊÓ

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Snow is falling vertically at a constant speed of \(8.0 \mathrm{~m} / \mathrm{s}\). At what angle from the vertical do the snowflakes appear to be falling as viewed by the driver of a car traveling on a straight, level road with a speed of \(50 \mathrm{~km} / \mathrm{h} ?\)

Short Answer

Expert verified
The angle is approximately \(60.07^\circ\) from the vertical.

Step by step solution

01

Convert Car's Speed to Meters per Second

The car's speed is given as \(50 \text{ km/h}\). To convert this to meters per second, use the conversion factor \(1 \text{ km/h} = \frac{1}{3.6} \text{ m/s}\). Thus, \(50 \times \frac{1}{3.6} = 13.89 \text{ m/s}\).
02

Find the Relative Velocity Components

The vertical speed of the snow is \(8.0 \text{ m/s}\) and the horizontal speed of the car is \(13.89 \text{ m/s}\). The snowflakes' relative speed in the horizontal direction from the car's perspective is \(13.89 \text{ m/s}\), and in the vertical direction, it remains \(8.0 \text{ m/s}\).
03

Use Trigonometry to Find the Angle

The angle \(\theta\) from the vertical at which the snow appears to fall can be found using the tangent function: \(\tan \theta = \frac{\text{horizontal speed of car}}{\text{vertical speed of snow}} = \frac{13.89}{8.0}\).
04

Calculate the Angle

Solve for \(\theta\) using the inverse tangent function: \(\theta = \tan^{-1}\left(\frac{13.89}{8.0}\right)\). Calculating this gives \(\theta \approx 60.07^\circ\).

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Key Concepts

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

Relative Velocity
Imagine yourself moving in a car while observing the world around you. You might notice that static objects seem to "move" backward at the speed you're traveling, creating an illusion of motion. This phenomenon connects to the idea of relative velocity.
The term "relative velocity" describes how the velocity of one object appears from the perspective of another moving object. In our exercise, the snowflakes falling vertically appear to have a different motion when observed from the moving car.
Here's how it works: the car is moving horizontally at a speed of 13.89 m/s, while the snow is falling vertically at 8.0 m/s. The apparent movement of snowflakes combines these two motions. Consequently, this combination results in a diagonal path relative to the car's driver rather than the vertical descent perceived by a stationary observer.
Trigonometry
Trigonometry is a branch of mathematics that explores the relationships in triangles, making it highly useful in physics problems involving angles. In our scenario, we leverage it to determine how the path of snowflakes changes with motion.
When we deal with vectors, particularly in two dimensions like our snow and car scenario, trigonometry becomes especially helpful. We can break down complex motion into simpler components. For the car and snowflakes, we observe two perpendicular components: horizontal and vertical. Then, we use these to form a right triangle.
The horizontal side of the triangle represents the car's speed, and the vertical side signifies the falling snow's speed. Trigonometry helps us understand how these components are related through functions like tangent, sine, and cosine, serving as tools to calculate angles and other key parameters.
Angle Calculation
Calculating angles using trigonometry involves knowing a bit about functions like sine, cosine, or tangent. In this exercise, to find out how the snowflakes appear to an observer in the moving car, we rely on the tangent function.
Here's a quick breakdown: The tangent of an angle in a right triangle is the ratio of the side opposite the angle to the adjacent side. For our exercise, the tangent function lets us relate the horizontal movement (car's speed) to the vertical movement (snow's speed).
By using the formula \[\tan \theta = \frac{\text{horizontal speed of car}}{\text{vertical speed of snow}}\]we find the angle \(\theta\). Employing an inverse tangent function gives us the precise angle, here calculated as approximately 60.07 degrees. This angle represents how much the path of the snowflakes deviates from the vertical line, as seen from the moving car.

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Most popular questions from this chapter

A certain airplane has a speed of \(290.0 \mathrm{~km} / \mathrm{h}\) and is diving at an angle of \(\theta=30.0^{\circ}\) below the horizontal when the pilot releases a radar decoy (Fig. \(4-33\) ) . The horizontal distance between the release point and the point where the decoy strikes the ground is \(d=700 \mathrm{~m} .\) (a) How long is the decoy in the air? (b) How high was the release point?

An airport terminal has a moving sidewalk to speed passengers through a long corridor. Larry does not use the moving sidewalk; he takes \(150 \mathrm{~s}\) to walk through the corridor. Curly, who simply stands on the moving sidewalk, covers the same distance in 70 s. Moe boards the sidewalk and walks along it. How long does Moe take to move through the corridor? Assume that Larry and Moe walk at the same speed.

A particle \(P\) travels with constant speed on a circle of radius \(r=3.00 \mathrm{~m}\) (Fig. \(4-56\) ) and completes one revolution in \(20.0 \mathrm{~s}\). The particle passes through \(O\) at time \(t=0 .\) State the following vectors in magnitudeangle notation (angle relative to the positive direction of \(x\) ). With respect to \(O,\) find the particle's position vector at the times \(t\) of (a) \(5.00 \mathrm{~s}\), (b) \(7.50 \mathrm{~s},\) and (c) \(10.0 \mathrm{~s}\). (d) For the \(5.00 \mathrm{~s}\) interval from the end of the fifth second to the end of the tenth second, find the particle's displacement. For that interval, find (e) its average velocity and its velocity at the (f) beginning and (g) end. Next, find the acceleration at the (h) beginning and (i) end of that interval.

For women's volleyball the top of the net is \(2.24 \mathrm{~m}\) above the floor and the court measures \(9.0 \mathrm{~m}\) by \(9.0 \mathrm{~m}\) on each side of the net. Using a jump serve, a player strikes the ball at a point that is \(3.0 \mathrm{~m}\) above the floor and a horizontal distance of \(8.0 \mathrm{~m}\) from the net. If the initial velocity of the ball is horizontal, (a) what minimum magnitude must it have if the ball is to clear the net and (b) what maximum magnitude can it have if the ball is to strike the floor inside the back line on the other side of the net?

A rifle that shoots bullets at \(460 \mathrm{~m} / \mathrm{s}\) is to be aimed at a target \(45.7 \mathrm{~m}\) away. If the center of the target is level with the rifle, how high above the target must the rifle barrel be pointed so that the bullet hits dead center?

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