/*! 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 7 A motorist drives north for \(35... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 7

A motorist drives north for \(35.0\) minutes at \(85.0 \mathrm{~km} / \mathrm{h}\) and then stops for \(15.0\) minutes. He then continues north, traveling \(130 \mathrm{~km}\) in \(2.00 \mathrm{~h}\). (a) What is his total displacement? (b) What is his average velocity?

Short Answer

Expert verified
The total displacement is 179.6 km and the average velocity is 63.5 km/h.

Step by step solution

01

Calculate the distance traveled in the first half of the journey

In the first section of the journey, the motorist travels for \(35.0\) minutes at \(85.0 \mathrm{~km} / \mathrm{h}\). We convert the time to hours (since the velocity is in km/h) by dividing \(35.0\) minutes by \(60.0\). The distance is obtained by multiplying the speed by the time, so the distance traveled in the first half of the journey, \( d_1 \), is \(85.0 \mathrm{~km} / \mathrm{h} \times \frac{35.0}{60.0} \mathrm{~h} = 49.6 \mathrm{~km}\).
02

Determine the total displacement

The total displacement is the sum of the distances covered in each section of the journey, which is, \( d_1 \) in the first half and \( 130 \mathrm{~km} \) in the second half. So, this gives a total displacement \(d_{\text{total}} = 49.6 \mathrm{~km} + 130.0 \mathrm{~km} = 179.6 \mathrm{~km}\).
03

Calculate the total time of the journey

The total time of the journey is given by the sum of the driving time and the resting time. In this case, the motorist drives for \(35.0\) minutes, rests for \(15.0\) minutes and then drives for \(2.00\) hours. Converting all times to hours gives a total time of \(t_{\text{total}} = \frac{35.0}{60} + \frac{15.0}{60} + 2.00 = 2.83 \mathrm{~h}\).
04

Determine the average velocity

The average velocity is given by the total distance traveled divided by the total time taken. Thus, the average velocity \(v_{\text{avg}} = \frac{d_{\text{total}}}{t_{\text{total}}} = \frac{179.6 \mathrm{~km}}{2.83 \mathrm{~h}} = 63.5 \mathrm{~km/h}\).

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!

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 jet plane lands with a speed of \(100 \mathrm{~m} / \mathrm{s}\) and can accelerate at a maximum rate of \(-5.00 \mathrm{~m} / \mathrm{s}^{2}\) as it comes to rest. (a) From the instant the plane touches the runway, what is the minimum time needed before it can come to rest? (b) Can this plane land on a small tropi\(\mathrm{cal}\) island airport where the runway is \(0.800 \mathrm{~km}\) long?

A mountain climber stands at the top of a \(50.0-\mathrm{m}\) cliff that overhangs a calm pool of water. She throws two stones vertically downward \(1.00 \mathrm{~s}\) apart and observes that they cause a single splash. The first stone had an initial velocity of \(-2.00 \mathrm{~m} / \mathrm{s}\). (a) How long after release of the first stone did the two stones hit the water? (b) What initial velocity must the second stone have had, given that they hit the water simultaneously? (c) What was the velocity of each stone at the instant it hit the water?

A \(50.0-\mathrm{g}\) Super Ball travel- ing at \(25.0 \mathrm{~m} / \mathrm{s}\) bounces off a brick wall and rebounds at \(22.0 \mathrm{~m} / \mathrm{s} .\) A high-speed camera records this event. If the ball is in contact with the wall for \(3.50 \mathrm{~ms}\), what is the magnitude of the aver- age acceleration of the ball during this time interval?

An object moves with constant acceleration \(4.00 \mathrm{~m} / \mathrm{s}^{2}\) and over a time interval reaches a final velocity of \(12.0 \mathrm{~m} / \mathrm{s}\). (a) If its original velocity is \(6.00 \mathrm{~m} / \mathrm{s}\), what is its displacement during the time interval? (b) What is the distance it travels during this interval? (c) If its original velocity is \(-6.00 \mathrm{~m} / \mathrm{s}\), what is its displacement during this interval? (d) What is the total distance it travels during the interval in part (c)?

A model rocket is launched straight upward with an initial speed of \(50.0 \mathrm{~m} / \mathrm{s}\). It accelerates with a constant upward acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\) until its engines stop at an altitude of \(150 \mathrm{~m}\). (a) What can you say about the motion of the rocket after its engines stop? (b) What is the maximum height reached by the rocket? (c) How long after liftoff does the rocket reach its maximum height? (d) How long is the rocket in the air?
See all solutions

Recommended explanations on Physics Textbooks

Dynamics

Read Explanation

Particle Model of Matter

Read Explanation

Physics of Motion

Read Explanation

Medical Physics

Read Explanation

Engineering Physics

Read Explanation

Solid State Physics

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.