Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 29

A runner times his speed around a circular track with a circumference of \(0.478 \mathrm{mi} .\) At the start he is running toward the east and the track starts bending toward the north. If he goes halfway around, he will be running toward the west. He finds that he has run a distance of \(0.750 \mathrm{mi}\) in 4.00 min. What is his (a) average speed and (b) average velocity in \(\mathrm{m} / \mathrm{s} ?\)

Short Answer

Expert verified
Question: A runner takes 4.00 minutes to run around half of a circular track with a circumference of 0.478 miles. Given that he runs a total distance of 0.750 miles, calculate (a) his average speed in meters per second (m/s), and (b) his average velocity in m/s. Answer: (a) The runner's average speed is \(5.03\,\mathrm{m/s}\). (b) The runner's average velocity is \(1.02\,\mathrm{m/s}\).

Step by step solution

01

Find the Conversion Factor (from miles to meters)

In order to convert the distance and circumference from miles into meters, find out how many meters are in 1 mile. 1 mile = 1609.34 meters.
02

Convert Distance and Circumference to Meters

Now, use the conversion factor to convert the given distance traveled, 0.750 miles, and track circumference, 0.478 miles, into meters. Distance in meters: \(0.750 \mathrm{mi} * 1609.34 \mathrm{m/mi} = 1207 \mathrm{m}\) (rounded to nearest meter) Circumference in meters: \(0.478 \mathrm{mi} * 1609.34 \mathrm{m/mi} = 769 \mathrm{m}\) (rounded to nearest meter)
03

Calculate Displacement

Because the runner goes halfway around the circular track, the displacement is a straight line from start to end (known as the diameter of the circle). To find the displacement, use the circumference as follows: Diameter (displacement) = Circumference / π Displacement = \(769\textrm{m} / \pi \approx 245\textrm{m}\) (rounded to nearest meter)
04

Convert Time to Seconds

The given time is 4.00 minutes. To convert this into seconds, multiply the time by 60: Time in seconds: \(4.00 \textrm{min} * 60 \textrm{s/min} = 240\textrm{s}\)
05

Calculate Average Speed

Now that we have the distance in meters and time in seconds, we can find the average speed using the formula: average speed = distance traveled/time taken. Average speed = \(\frac{1207\mathrm{m}}{240\mathrm{s}} = 5.03\,\mathrm{m/s}\)
06

Calculate Average Velocity

To find the average velocity, use the formula: average velocity = displacement/time taken Average velocity = \(\frac{245\mathrm{m}}{240\mathrm{s}} = 1.02\,\mathrm{m/s}\) So, the runner's (a) average speed is \(5.03\,\mathrm{m/s}\) and (b) average velocity is \(1.02\,\mathrm{m/s}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 car travels east at \(96 \mathrm{km} / \mathrm{h}\) for \(1.0 \mathrm{h}\). It then travels \(30.0^{\circ}\) east of north at \(128 \mathrm{km} / \mathrm{h}\) for \(1.0 \mathrm{h} .\) (a) What is the average speed for the trip? (b) What is the average velocity for the trip?
Vector \(\overrightarrow{\mathbf{B}}\) has magnitude 7.1 and direction \(14^{\circ}\) below the \(+x\) -axis. Vector \(\quad \overrightarrow{\mathbf{C}}\) has \(x\) -component \(C_{x}=-1.8\) and \(y\) -component \(C_{y}=-6.7 .\) Compute \((a)\) the \(x-\) and \(y\) -components of \(\overrightarrow{\mathbf{B}} ;\) (b) the magnitude and direction of \(\overrightarrow{\mathbf{C}} ;\) (c) the magnitude and direction of \(\overrightarrow{\mathbf{C}}+\overrightarrow{\mathbf{B}} ;\) (d) the magnitude and direction of \(\overrightarrow{\mathbf{C}}-\overrightarrow{\mathbf{B}} ;\) (e) the \(x\) - and \(y\) -components of \(\overrightarrow{\mathbf{C}}-\overrightarrow{\mathbf{B}}\)
To get to a concert in time, a harpsichordist has to drive \(122 \mathrm{mi}\) in \(2.00 \mathrm{h} .\) (a) If he drove at an average speed of \(55.0 \mathrm{mi} / \mathrm{h}\) in a due west direction for the first $1.20 \mathrm{h}\( what must be his average speed if he is heading \)30.0^{\circ}$ south of west for the remaining 48.0 min? (b) What is his average velocity for the entire trip?
In each of these, the \(x\) - and \(y\) -components of a vector are given. Find the magnitude and direction of the vector. (a) $A_{x}=-5.0 \mathrm{m} / \mathrm{s}, A_{y}=+8.0 \mathrm{m} / \mathrm{s} .\( (b) \)B_{x}=+120 \mathrm{m}$ $B_{y}=-60.0 \mathrm{m} .(\mathrm{c}) C_{x}=-13.7 \mathrm{m} / \mathrm{s}, C_{y}=-8.8 \mathrm{m} / \mathrm{s} .(\mathrm{d}) D_{x}=$ \(2.3 \mathrm{m} / \mathrm{s}^{2}, D_{y}=6.5 \mathrm{cm} / \mathrm{s}^{2}\)
A runner times his speed around a track with a circumference of $0.50 \mathrm{mi} .\( He finds that he has run a distance of \)1.00 \mathrm{mi}$ in 4.0 min. What is his (a) average speed and (b) average velocity magnitude in \(\mathrm{m} / \mathrm{s}\) ?
See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Turning Points in Physics

Read Explanation

Space Physics

Read Explanation

Kinematics Physics

Read Explanation

Physics of Motion

Read Explanation

Waves 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.

Sign-up for free