/*! 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 64 In the problems, please assume t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 64

In the problems, please assume the free-fall acceleration $g=9.80 \mathrm{m} / \mathrm{s}^{2}$ unless a more precise value is given in the problem statement. Ignore air resistance. To pass a physical fitness test, Marcella must run \(1000 \mathrm{m}\) at an average speed of \(4.00 \mathrm{m} / \mathrm{s} .\) She runs the first $500 \mathrm{m}\( at an average of \)4.20 \mathrm{m} / \mathrm{s} .$ (a) How much time does she have to run the last \(500 \mathrm{m} ?\) (b) What should be her average speed over the last \(500 \mathrm{m}\) in order to finish with an overall average speed of \(4.00 \mathrm{m} / \mathrm{s} ?\)

Short Answer

Expert verified
Answer: (a) Marcella has 130.95 seconds to run the last 500 meters, and (b) she should have an average speed of approximately 3.82 m/s for the last 500 meters to achieve the desired overall average speed.

Step by step solution

01

Calculate the time taken for the first 500m

We are given that Marcella runs the first 500m at an average speed of 4.20 m/s. We will use the formula for average speed to find the time taken for the first 500m. Average speed = Distance / Time 4.20 m/s = 500m / Time Time = 500m / 4.20 m/s = 119.05 s
02

Calculate the total time for 1000m

Since Marcella needs to run the full 1000m at an average speed of 4.00 m/s, we can find the total time for the run using the same formula: 4.00 m/s = 1000m /Total Time Total Time = 1000m / 4.00 m/s = 250.00 s
03

Calculate the time allowed for the last 500m

Now that we know the total time and the time taken for the first 500m, we can find the time allowed for the last 500m: Time allowed for the last 500m = Total Time - Time taken for the first 500m Time allowed for the last 500m = 250.00 s - 119.05 s = 130.95 s
04

Calculate the required average speed for the last 500m

To find the required average speed for the last 500m, we will use the formula for average speed: Required average speed = Distance / Time Required average speed = 500m / 130.95 s 鈮� 3.82 m/s So, the answers are: (a) Marcella has 130.95 seconds to run the last 500 meters, and (b) She should have an average speed of approximately 3.82 m/s during the last 500 meters to finish the race with the desired overall average speed of 4.00 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.

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

Two cars, a Toyota Yaris and a Jeep, are traveling in the same direction, although the Yaris is \(186 \mathrm{m}\) behind the Jeep. The speed of the Yaris is \(24.4 \mathrm{m} / \mathrm{s}\) and the speed of the Jeep is $18.6 \mathrm{m} / \mathrm{s}$. How much time does it take for the Yaris to catch the Jeep? [Hint: What must be true about the displacement of the two cars when they meet?]
You are driving your car along a country road at a speed of $27.0 \mathrm{m} / \mathrm{s} .$ As you come over the crest of a hill, you notice a farm tractor \(25.0 \mathrm{m}\) ahead of you on the road, moving in the same direction as you at a speed of \(10.0 \mathrm{m} / \mathrm{s} .\) You immediately slam on your brakes and slow down with a constant acceleration of magnitude $7.00 \mathrm{m} / \mathrm{s}^{2} .$ Will you hit the tractor before you stop? How far will you travel before you stop or collide with the tractor? If you stop, how far is the tractor in front of you when you finally stop?
In the problems, please assume the free-fall acceleration $g=9.80 \mathrm{m} / \mathrm{s}^{2}$ unless a more precise value is given in the problem statement. Ignore air resistance. A cheetah can accelerate from rest to \(24 \mathrm{m} / \mathrm{s}\) in $2.0 \mathrm{s}$ Assuming the acceleration is constant over the time interval, (a) what is the magnitude of the acceleration of the cheetah? (b) What is the distance traveled by the cheetah in these \(2.0 \mathrm{s} ?\) (c) A runner can accelerate from rest to \(6.0 \mathrm{m} / \mathrm{s}\) in the same time, $2.0 \mathrm{s} .$ What is the magnitude of the acceleration of the runner? By what factor is the cheetah's average acceleration magnitude greater than that of the runner?
A car is speeding up and has an instantaneous velocity of $1.0 \mathrm{m} / \mathrm{s}\( in the \)+x\( -direction when a stopwatch reads \)10.0 \mathrm{s} .$ It has a constant acceleration of \(2.0 \mathrm{m} / \mathrm{s}^{2}\) in the \(+x\) -direction. (a) What change in speed occurs between \(t=10.0 \mathrm{s}\) and \(t=12.0 \mathrm{s} ?(\mathrm{b})\) What is the speed when the stopwatch reads \(12.0 \mathrm{s} ?\)
In the problems, please assume the free-fall acceleration $g=9.80 \mathrm{m} / \mathrm{s}^{2}$ unless a more precise value is given in the problem statement. Ignore air resistance. A streetcar named Desire travels between two stations \(0.60 \mathrm{km}\) apart. Leaving the first station, it accelerates for \(10.0 \mathrm{s}\) at $1.0 \mathrm{m} / \mathrm{s}^{2}$ and then travels at a constant speed until it is near the second station, when it brakes at \(2.0 \mathrm{m} / \mathrm{s}^{2}\) in order to stop at the station. How long did this trip take? [Hint: What's the average velocity?]
See all solutions

Recommended explanations on Physics Textbooks

Oscillations

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Geometrical and Physical Optics

Read Explanation

Energy Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Dynamics

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.