/*! 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 85 You are kidnapped by political-s... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

You are kidnapped by political-science majors (who are upset because you told them political science is not a real science). Although blindfolded, you can tell the speed of their car (by the whine of the engine), the time of travel (by mentally counting off seconds), and the direction of travel (by turns along the rectangular street system). From these clues, you know that you are taken along the following course: \(50 \mathrm{~km} / \mathrm{h}\) for \(2.0 \mathrm{~min}\), turn \(90^{\circ}\) to the right, \(20 \mathrm{~km} / \mathrm{h}\) for \(4.0 \mathrm{~min}\), turn \(90^{\circ}\) to the right, \(20 \mathrm{~km} / \mathrm{h}\) for \(60 \mathrm{~s}\), turn \(90^{\circ}\) to the left, \(50 \mathrm{~km} / \mathrm{h}\) for \(60 \mathrm{~s}\), turn \(90^{\circ}\) to the right, \(20 \mathrm{~km} / \mathrm{h}\) for \(2.0 \mathrm{~min}\), turn \(90^{\circ}\) to the left, \(50 \mathrm{~km} / \mathrm{h}\) for \(30 \mathrm{~s}\). At that point, (a) how far are you from your starting point, and (b) in what direction relative to your initial direction of travel are you?

Short Answer

Expert verified
a) 0.67 km from start; b) ~6.8° left of initial direction.

Step by step solution

01

Understanding the problem

To determine your final position and direction, you need to follow the path based on the given speed, time, and turns. This involves calculating distances, considering turns and calculating the resulting displacement from the starting point.
02

Convert time to hours

For each segment of travel, convert the given time from minutes or seconds to hours by dividing minutes by 60 and seconds by 3600. This is necessary because speed is given in kilometers per hour.
03

Calculate distances traveled

Multiply the speed by the time (in hours) for each segment to find the distance traveled. For instance, for the first segment: \[ \text{Distance} = 50 \text{ km/h} \times \frac{2}{60} \text{ h} = 1.67 \text{ km} \]
04

Track position changes with directional turns

Start from the origin (0,0). For each segment, adjust your position on a coordinate grid relative to direction changes and the calculated distances. Assume starting direction along the positive x-axis.
05

Plot each segment and update position

1. First segment: Go 1.67 km along +x. 2. Right turn: Go 1.33 km along -y. 3. Right turn: Go 0.33 km along -x. 4. Left turn: Go 0.83 km along +y. 5. Right turn: Go 0.67 km along -x. 6. Left turn: Go 0.42 km along +y.
06

Use Pythagorean theorem to find distance from the starting point

Calculate the net displacement in x and y direction from the starting point and use these to find the distance:\[\text{Net } x = (1.67 - 0.33 - 0.67) \text{ km} \approx 0.67 \text{ km} \\text{Net } y = (-1.33 + 0.83 + 0.42) \text{ km} \approx -0.08 \text{ km} \\text{Distance} = \sqrt{(0.67)^2 + (-0.08)^2} \approx 0.67 \text{ km}\]
07

Determine final direction

Calculate the angle using the arc tangent of the ratio of y-displacement to x-displacement to find the direction relative to the original direction:\[\theta = \tan^{-1} \left(\frac{-0.08}{0.67}\right) \approx -6.8^\circ\]This indicates the direction is approximately 6.8 degrees to the left of the x-axis.

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.

Coordinate System
A coordinate system acts as a reference framework that helps us describe the precise location of a point in a given space. In this exercise, when tracking movement along the streets, we create a coordinate plane where:
  • The starting point is typically designated as the origin ( (0,0) ) .
  • The positive x-axis can be thought of as your initial direction or path.
  • Turns which are 90 degrees to the right or left shift your path to align with the y-axis either in positive or negative direction.
Whenever there is motion, updating positions within this grid becomes crucial for calculating displacement effectively. In practical terms, the coordinate system assists in keeping track of how each movement and turn translates into coordinate changes, allowing you to figure out where you end up relative to where you began. Thus, it's a central concept for understanding spatial relationships in vector problems like this one.
Including such a logical structure is critical since it provides clarity in visualizing shifts in direction and distance, making problem-solving more straightforward.
Vector Mathematics
Vector mathematics is essential for calculating physical quantities along defined directions. In this displacement problem, each segment of movement forms a vector with two main characteristics: magnitude and direction:
  • Magnitude: This is the length of the vector, which represents the distance the car travels in each segment. You compute it by multiplying the speed by the time (both converted to consistent units such as kilometers per hour and hours).
  • Direction: Vectors have directional properties, which change with each 90-degree turn. These direction changes can be graphically envisioned as rotating the vector 90 degrees around the coordinate plane.
Tracking the journey along different vectors and updating each on the coordinate grid through their horizontal and vertical components is what gives the resultant path. Adding or subtracting these vectors (based on their respective directions) computes the overall journey's net vector.
This makes vector mathematics extremely powerful for solving multidimensional problems where direction changes occur.
Pythagorean Theorem
The Pythagorean theorem is employed to compute the straight-line distance from the starting point to the final location, even when the path taken is convoluted. Once the net changes in the x and y directions are known, apply the theorem as follows:
  • Consider the net x-distance and y-distance as the two shorter sides (legs) of a right triangle.
  • The hypotenuse of this triangle represents the direct line distance (or displacement) from the start to the finish.
  • Calculate it using the formula: \[\text{Distance} = \sqrt{(\text{Net } x)^2 + (\text{Net } y)^2}\]
For example, in the exercise, using the calculated values of approximately 0.67 km for the net x and -0.08 km for the net y, the computed displacement or the straight-line path distance turned out to be approximately 0.67 km. This method efficiently provides a real understanding of how far you've moved regardless of the route's complexity.

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 baseball leaves a pitcher's hand horizontally at a speed of \(161 \mathrm{~km} / \mathrm{h}\). The distance to the batter is \(18.3 \mathrm{~m}\). (a) How long does the ball take to travel the first half of that distance? (b) The second half? (c) How far does the ball fall freely during the first half? (d) During the second half? (e) Why aren't the quantities in (c) and (d) equal?

An Earth satellite moves in a circular orbit \(640 \mathrm{~km}\) (uniform circular motion) above Earth's surface with a period of \(98.0 \mathrm{~min}\). What are (a) the speed and (b) the magnitude of the centripetal acceleration of the satellite?

A woman can row a boat at \(6.40 \mathrm{~km} / \mathrm{h}\) in still water. (a) If she is crossing a river where the current is \(3.20 \mathrm{~km} / \mathrm{h}\), in what direction must her boat be headed if she wants to reach a point directly opposite her starting point? (b) If the river is \(6.40 \mathrm{~km}\) wide, how long will she take to cross the river? (c) Suppose that instead of crossing the river she rows \(3.20 \mathrm{~km}\) down the river and then back to her starting point. How long will she take? (d) How long will she take to row \(3.20 \mathrm{~km} u p\) the river and then back to her starting point? (e) In what direction should she head the boat if she wants to cross in the shortest possible time, and what is that time?

Suppose that a shot putter can put a shot at the worldclass speed \(v_{0}=15.00 \mathrm{~m} / \mathrm{s}\) and at a height of \(2.160 \mathrm{~m}\). What horizontal distance would the shot travel if the launch angle \(\theta_{0}\) is (a) \(45.00^{\circ}\) and (b) \(42.00^{\circ}\) ? The answers indicate that the angle of \(45^{\circ}\), which maximizes the range of projectile motion, does not maximize the horizontal distance when the launch and landing are at different heights.

A rifle is aimed horizontally at a target \(30 \mathrm{~m}\) away. The bullet hits the target \(1.9 \mathrm{~cm}\) below the aiming point. What are (a) the bullet's time of flight and (b) its speed as it emerges from the rifle?

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.