/*! 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 5 A train at a constant \(60.0 \ma... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A train at a constant \(60.0 \mathrm{~km} / \mathrm{h}\) moves east for \(40.0 \mathrm{~min}\), then in a direction \(50.0^{\circ}\) east of due north for \(20.0 \mathrm{~min}\), and then west for \(50.0 \mathrm{~min}\). What are the (a) magnitude and (b) angle of its average velocity during this trip?

Short Answer

Expert verified
Magnitude: 8.5 km/h; Angle: 79.5° from east towards north.

Step by step solution

01

Convert Time to Hours

To perform calculations, we need time in hours. Convert the time from minutes to hours.- Moving east: \(40.0 \text{ minutes} = \frac{40.0}{60} \text{ hours} = 0.667 \text{ hours}\)- Moving northeast: \(20.0 \text{ minutes} = \frac{20.0}{60} \text{ hours} = 0.333 \text{ hours}\)- Moving west: \(50.0 \text{ minutes} = \frac{50.0}{60} \text{ hours} = 0.833 \text{ hours}\)
02

Calculate Displacement Vectors

Find the displacement for each segment:1. Eastward: \[ d_1 = 60.0 \text{ km/h} \times 0.667 \text{ h} = 40.0 \text{ km} \] Displacement vector: \((40.0 \hat{i}, 0 \hat{j}) \text{ km}\)2. Northeast: \[ d_2 = 60.0 \text{ km/h} \times 0.333 \text{ h} = 20.0 \text{ km} \] Decompose into x and y components using the angle: \[ x_2 = 20.0 \cos(50.0^{\circ}) \text{ km} y_2 = 20.0 \sin(50.0^{\circ}) \text{ km} \] Displacement vector: \((x_2 \hat{i}, y_2 \hat{j}) \text{ km}\)3. Westward: \[ d_3 = 60.0 \text{ km/h} \times 0.833 \text{ h} = 50.0 \text{ km} \] Displacement vector: \((-50.0 \hat{i}, 0 \hat{j}) \text{ km}\)
03

Combine Displacement Vectors

Add components of all displacement vectors:- x-component: \[x_{\text{total}} = 40.0 + x_2 - 50.0\]- y-component: \[y_{\text{total}} = y_2\]Calculate each component:- \(x_2 = 20.0 \cos(50.0^{\circ}) \approx 12.86\) km- \(y_2 = 20.0 \sin(50.0^{\circ}) \approx 15.32\) kmCombine:- \(x_{\text{total}} = 40.0 + 12.86 - 50.0 = 2.86\) km- \(y_{\text{total}} = 15.32\) km
04

Calculate Total Displacement and Average Velocity

Compute the magnitude of total displacement:\[d_{\text{total}} = \sqrt{(x_{\text{total}})^2 + (y_{\text{total}})^2}\]\[d_{\text{total}} = \sqrt{2.86^2 + 15.32^2} \approx 15.58 \text{ km}\]Total time is \(0.667 + 0.333 + 0.833 = 1.833\) hours.Average velocity magnitude:\[v_{\text{avg}} = \frac{d_{\text{total}}}{\text{Total time}} = \frac{15.58}{1.833} \approx 8.5 \text{ km/h}\]
05

Calculate Average Velocity Angle

The angle \(\theta\) of average velocity can be determined using:\[\theta = \tan^{-1}\left( \frac{y_{\text{total}}}{x_{\text{total}}} \right)\]\[\theta = \tan^{-1}\left( \frac{15.32}{2.86} \right) \approx 79.5^{\circ}\]The angle is measured from the east direction, moving towards north.

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 91Ó°ÊÓ!

Key Concepts

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

vector addition
In physics, vector addition plays a crucial role when dealing with quantities that have both magnitude and direction. In our exercise, vector addition helps us determine the total displacement by combining several segments of motion. Each segment is represented as a vector, characterized by its components.

For vector addition, you need to sum up the individual components of each vector. In this example:
  • The eastward motion contributes a vector of \((40.0 \hat{i}, 0 \hat{j})\) km.
  • The northeast motion needs decomposition into x and y components. These are calculated using trigonometry as \((x_2 \hat{i}, y_2 \hat{j})\) km.
  • The westward motion involves only the x-axis, leading to a vector of \((-50.0 \hat{i}, 0 \hat{j})\) km.
By strategically adding these components, you compute the resultant vector that represents the train's total path. This final vector provides the data needed for further calculations, such as the magnitude and direction of the average velocity.
displacement vectors
Displacement vectors give a clear picture of a journey's net effect from start to finish. In contrast to distance, which only accounts for how much ground is covered, displacement focuses on how far and in what direction the object ends up.

To determine displacement, you start by breaking down each motion segment into vectors that show direction and length. In this exercise:
  • The displacement eastward was straight along the x-axis, yielding a simple vector: \((40.0 \hat{i}, 0 \hat{j})\) km.
  • The second segment had a north-eastern trajectory. This required calculating the x and y components using angles, resulting in \((x_2, y_2)\).
  • Finally, the westward motion is shown as a vector \((-50.0 \hat{i}, 0 \hat{j})\) km.
Displacement vectors are then combined, giving a thorough overview of the complete journey. Understanding this allows you to solve for the total distance covered in a particular direction, facilitating the calculation of average velocity.
angle calculation
Calculating angles in physics, especially involving displacement vectors, is critical for understanding direction. With vectors having both magnitude and direction, knowing the angle helps in defining vector orientation accurately.

To find the angle of a resultant vector like average velocity, use the trigonometric function tangent. It correlates the opposite and adjacent sides of a right triangle:
  • Given this trip scenario, the cumulative vectors create a right triangle.
  • The x and y components of the final vector are known: \(x_{\text{total}} = 2.86\) km and \(y_{\text{total}} = 15.32\) km.
  • Utilize the formula: \[\theta = \tan^{-1}\left( \frac{y_{\text{total}}}{x_{\text{total}}} \right)\] to calculate the angle.
From this, you discover the angle related to the direction, which in this case was approximately \(79.5^{\circ}\), indicating northeast orientation from the east direction. Mastery of these calculations helps unravel complex problems involving vector orientations.
trigonometry in physics
Trigonometry is indispensable in physics for analyzing and solving problems involving angles and dimensions. In this exercise, trigonometry simplifies how vectors are decomposed and analyzed.

Trigonometric functions such as sine and cosine are key to breaking down a vector into its horizontal and vertical projections. When a vector points in a non-standard direction, such as northeast, trigonometry helps find components along the cartesian coordinates.
  • To find x-component: \[x_2 = 20.0 \cos(50.0^{\circ})\]
  • To find y-component: \[y_2 = 20.0 \sin(50.0^{\circ})\]
Physics often requires transforming a vector into convenient x and y parts—trigonometry is what makes this feasible. Understanding these principles bolsters your ability to tackle physical problems where direction impacts the solution profoundly.

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

A cameraman on a pickup truck is traveling westward at \(20 \mathrm{~km} / \mathrm{h}\) while he records a cheetah that is moving westward \(30 \mathrm{~km} / \mathrm{h}\) faster than the truck. Suddenly, the cheetah stops, turns, and then runs at \(45 \mathrm{~km} / \mathrm{h}\) eastward, as measured by a suddenly nervous crew member who stands alongside the cheetah's path. The change in the animal's velocity takes \(2.0 \mathrm{~s}\). What are the (a) magnitude and (b) direction of the animal's acceleration according to the cameraman and the (c) magnitude and (d) direction according to the nervous crew member?

The position vector \(\vec{r}\) of a particle moving in the \(x y\) plane is \(\vec{r}=2 \hat{i}+2 \sin [(\pi / 4 \mathrm{rad} / \mathrm{s}) t] \hat{\mathrm{j}}, \quad\) with \(\vec{r}\) in meters and \(t\) in seconds. (a) Calculate the \(x\) and \(y\) components of the particle's position at \(t=0,1.0,2.0,3.0\), and \(4.0 \mathrm{~s}\) and sketch the particle's path in the \(x y\) plane for the interval \(0 \leq t \leq\) \(4.0 \mathrm{~s}\). (b) Calculate the components of the particle's velocity at \(t=1.0,2.0\), and \(3.0 \mathrm{~s}\). Show that the velocity is tangent to the path of the particle and in the direction the particle is moving at each time by drawing the velocity vectors on the plot of the particle's path in part (a). (c) Calculate the components of the particle's acceleration at \(t=1.0,2.0\), and \(3.0 \mathrm{~s}\).

A moderate wind accelerates a pebble over a horizontal \(x y\) plane with a constant acceleration \(\vec{a}=\left(5.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(7.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). At time \(t=0\), the velocity is \((4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathbf{i}}\). What are the (a) magnitude and (b) angle of its velocity when it has been displaced by \(12.0 \mathrm{~m}\) parallel to the \(x\) axis?

You are to throw a ball with a speed of \(12.0 \mathrm{~m} / \mathrm{s}\) at a target that is height \(h=5.00 \mathrm{~m}\) above the level at which you release the ball (Fig. 4-58). You want the ball's velocity to be horizontal at the instant it reaches the target. (a) At what angle \(\theta\) above the horizontal must you throw the ball? (b) What is the horizontal distance from the release point to the target? (c) What is the speed of the ball just as it reaches the target?

A helicopter is flying in a straight line over a level field at a constant speed of \(6.20 \mathrm{~m} / \mathrm{s}\) and at a constant altitude of \(9.50 \mathrm{~m}\). A package is ejected horizontally from the helicopter with an initial velocity of \(12.0 \mathrm{~m} / \mathrm{s}\) relative to the helicopter and in a direction opposite the helicopter's motion. (a) Find the initial speed of the package relative to the ground. (b) What is the horizontal distance between the helicopter and the package at the instant the package strikes the ground? (c) What angle does the velocity vector of the package make with the ground at the instant before impact, as seen from the ground?

See all solutions

Recommended explanations on Physics Textbooks

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.