/*! 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 12 A truck is traveling due north a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 12

A truck is traveling due north at \(40 \mathrm{~km} / \mathrm{hr}\) approaching a crossroad. On a perpendicular road a police car is traveling west toward the intersection at \(35 \mathrm{~km} / \mathrm{hr}\). Both vehicles will reach the crossroad in exactly one hour. Find the vector currently representing the displacement of the truck with respect to the police car. displacement \(\vec{d}=\)_________

Short Answer

Expert verified
The displacement vector of the truck with respect to the police car is: \(\vec{d} = 40 \, km \, \hat{i} + 35 \, km \, \hat{j}\)

Step by step solution

01

Find the distance each vehicle will travel

In this problem, we know the velocity of both vehicles and the time it takes to reach the intersection. We can use the formula: distance = velocity * time to find the total distance traveled by each vehicle in one hour. For the truck: Distance = Velocity * Time Distance = 40 km/hr * 1 hr = 40 km (North) For the police car: Distance = Velocity * Time Distance = 35 km/hr * 1 hr = 35 km (West)
02

Write the vectors for each vehicle

Now that we have the distance both vehicles will travel within one hour, we can write their displacement vectors. Truck vector: \(\vec{A} = 40 \, km \, \hat{i}\) (since the truck is traveling north) Police car vector: \(\vec{B} = -35 \, km \, \hat{j}\) (since the police car is traveling west)
03

Calculate the displacement vector

To find the displacement vector of the truck with respect to the police car, we can subtract the police car vector from the truck vector. \(\vec{d} = \vec{A} - \vec{B}\) \(\vec{d} = (40 \, km \, \hat{i} + 0 \, km \, \hat{j}) - (0 \, km \, \hat{i} - 35 \, km \, \hat{j})\) \(\vec{d} = 40 \, km \, \hat{i} + 35 \, km \, \hat{j}\) The displacement vector of the truck with respect to the police car is: \(\vec{d} = 40 \, km \, \hat{i} + 35 \, km \, \hat{j}\)

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.

Multivariable Calculus
Multivariable calculus is an extension of single-variable calculus where functions of more than one variable are involved. When dealing with vector displacement in the context of multivariable calculus, we explore the displacement in a space defined by two or more dimensions, typically on an 'xy' plane for two-dimensional space or in 'xyz' space for three-dimensional analyses.

For example, in the given exercise, the movement of the truck and police car can be illustrated on a two-dimensional plane with positions and displacements represented by vectors. These vectors account for movement in multiple directions鈥攏orth and west鈥攚hich correspond to the 'i' and 'j' unit vectors, respectively, in a Cartesian coordinate system. Students can visualize problems like these by plotting vectors on a graph and applying vector operations to determine the resulting displacement.
Vector Addition and Subtraction
Understanding vector addition and subtraction is crucial for solving physics and engineering problems where forces, velocities, and displacements are often represented as vectors. Vectors are mathematical objects with both magnitude and direction. They can be added or subtracted by aligning them head-to-tail and drawing a resulting vector.

In the exercise, the displacement of the truck with respect to the police car is found by subtracting the displacement vector of the police car from that of the truck. This is a straightforward application where we simply take the corresponding components in each direction鈥攏orth and west鈥攁nd perform the subtraction algebraically. The exercise demonstrates how the resulting vector gives us the relative position from one point to another, essentially showing how far and in what direction one must travel to get from the police car's position to the truck's location.
Relative Motion in Physics
The concept of relative motion refers to the motion of an object as observed from a particular frame of reference. In physics, it's often important to compare the motion of two or more objects with respect to each other to understand their interactions and trajectories.

For the given problem, we are interested in the truck's displacement relative to the police car. By calculating the vector displacement from the police car to the truck, we are effectively switching to a frame of reference where the police car is at rest, and the truck is moving. This shifts the problem from absolute motion (how each vehicle is moving in relation to the crossroads) to relative motion (how the truck's position is changing in relation to the police car). It's a foundational concept for understanding how different observers may perceive motion differently, depending on their own state of motion.

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

Consider the line which passes through the point \(\mathrm{P}(3,-5,-1),\) and which is parallel to the line \(x=1+6 t, y=2+2 t, z=3+6 t\) Find the point of intersection of this new line with each of the coordinate planes: xy-plane:_________,_________,_________ xz-plane:_________,_________,_________ yz-plane:_________,_________,_________

Find a parametrization of the circle of radius 6 in the xy-plane, centered at the origin, oriented clockwise. The point (6,0) should correspond to \(t=0 .\) Use \(t\) as the parameter for all of your answers. \(x(t)=\)_________ \(y(t)=\)_________

This exercise explores key relationships between a pair of lines. Consider the following two lines: one with parametric equations \(x(s)=4-2 s\), \(y(s)=-2+s, z(s)=1+3 s,\) and the other being the line through (-4,2,17) in the direction \(\mathbf{v}=\langle-2,1,5\rangle\) a. Find a direction vector for the first line, which is given in parametric form. b. Find parametric equations for the second line, written in terms of the parameter \(t\) c. Show that the two lines intersect at a single point by finding the values of \(s\) and \(t\) that result in the same point. Then find the point of intersection. d. Find the acute angle formed where the two lines intersect, noting that this angle will be given by the acute angle between their respective direction vectors. e. Find an equation for the plane that contains both of the lines described in this problem.

Resolve the following vectors into components: (a) The vector \(\vec{v}\) in 2 -space of length 5 pointing up at an angle of \(\pi / 4\) measured from the positive \(x\) -axis. \vec{v}=________\(\vec{i}+\)________\(\vec{j}\) (b) The vector \(\vec{w}\) in 3 -space of length 3 lying in the \(x z\) -plane pointing upward at an angle of \(2 \pi / 3\) measured from the positive \(x\) -axis. \vec{v}=________\(\vec{i}+\)________\(\vec{j}\)+________\(\vec{k}\)

A force (like gravity) has both a magnitude and a direction. If two forces \(\mathbf{u}\) and \(\mathbf{v}\) are applied to an object at the same point, the resultant force on the object is the vector sum of the two forces. When a force is applied by a rope or a cable, we call that force tension. Vectors can be used to determine tension. As an example, suppose a painting weighing 50 pounds is to be hung from wires attached to the frame as illustrated in Figure \(9.2 .10 .\) We need to know how much tension will be on the wires to know what kind of wire to use to hang the picture. Assume the wires are attached to the frame at point \(O\). Let \(\mathbf{u}\) be the vector emanating from point \(O\) to the left and \(\mathbf{v}\) the vector emanating from point \(O\) to the right. Assume \(\mathbf{u}\) makes a \(60^{\circ}\) angle with the horizontal at point \(O\) and \(\mathbf{v}\) makes a \(45^{\circ}\) angle with the horizontal at point \(O\). Our goal is to determine the vectors \(\mathbf{u}\) and \(\mathbf{v}\) in order to calculate their magnitudes. a. Treat point \(O\) as the origin. Use trigonometry to find the components \(u_{1}\) and \(u_{2}\) so that \(\mathbf{u}=u_{1} \mathbf{i}+u_{2} \mathbf{j} .\) Since we don't know the magnitude of \(\mathbf{u},\) your components will be in terms of \(|\mathbf{u}|\) and the cosine and sine of some angle. Then find the components \(v_{1}\) and \(v_{2}\) so that \(\mathbf{v}=v_{1} \mathbf{i}+v_{2} \mathbf{j}\). Again, your components will be in terms of \(|\mathbf{v}|\) and the cosine and sine of some angle. b. The total force holding the picture up is given by \(\mathbf{u}+\mathbf{v}\). The force acting to pull the picture down is given by the weight of the picture. Find the force vector \(\mathbf{w}\) acting to pull the picture down. c. The picture will hang in equilibrium when the force acting to hold it up is equal in magnitude and opposite in direction to the force acting to pull it down. Equate these forces to find the components of the vectors \(\mathbf{u}\) and \(\mathbf{v}\).
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Applied Mathematics

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.