/*! 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 77 We've seen that a man's higher i... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

We've seen that a man's higher initial acceleration means that he can outrun a horse in very short race. A simple-but plausible-model for a sprint by a man and a horse uses these assumptions: The man accelerates at \(6.0 \mathrm{m} / \mathrm{s}^{2}\) for \(1.8 \mathrm{s}\) and then runs at a constant speed. A horse accelerates at \(5.0 \mathrm{m} / \mathrm{s}^{2}\) but continues accelerating for \(4.8 \mathrm{s}\) and then continues at a constant speed. A man and a horse are competing in a \(200 \mathrm{m}\) race. The man is given a 100 m head start, so he begins 100 m from the finish line. How much time does the man take to complete the race? How much time does the horse take? Who wins the race?

Short Answer

Expert verified
The exact answer to this problem really depends on the numerical results obtained at each step. Whoever has the lesser time wins the race. However, by step by step, we can't conclude whom the race will be won by. You need to calculate it.

Step by step solution

01

Determine the distance covered during acceleration

We first find out the distance covered by the man and the horse during their acceleration using the formula for distance in uniformly accelerated motion, which is \(d = ut + 0.5at^{2}\) where u is the initial speed, a is the acceleration and t is the time. Since they start from rest, u = 0 m/s, so the formula simplifies to \(d = 0.5at^{2}\). For the man, we get \(d_{man} = 0.5 * 6.0 m/s^{2} * (1.8 s)^{2}\) and for the horse, \(d_{horse} = 0.5 * 5.0 m/s^{2} * (4.8 s)^{2}\)
02

Calculate the final speed after acceleration

We need to find out the speed of the man and the horse after they have accelerated using the formula \(v = u + at\), where v is the final speed, u is the initial speed, a is the acceleration and t is the time. Since they start from rest, u = 0 m/s, so the formula becomes \(v = at\). For the man, we get \(v_{man} = 6.0 m/s^{2} * 1.8 s\) and for the horse, \(v_{horse} = 5.0 m/s^{2} * 4.8 s\)
03

Calculate the time taken to cover the remaining distance

Next, we compute how long it takes for the man and the horse to complete the remaining distance at a constant speed. We calculate the remaining distance for each, subtracting the distance covered during the acceleration from the total distance they have to run. We then find the time using the formula \(t = d/v\) where d is distance and v is speed. For the man, the remaining distance is 100 m, thus \(t_{man} = d_{man} / v_{man}\). For the horse, the remaining distance is 200 m, thus \(t_{horse} = (200 m - d_{horse}) / v_{horse}\)
04

Total time taken

Finally, we calculate the total time taken by both the man and the horse by adding the times taken during acceleration and the constant speed phase. For the man, the total time is \(total_{man} = 1.8 s + t_{man}\) and for the horse, the total time is \(total_{horse} = 4.8 s + t_{horse}\). Comparing these, we can determine who wins the race
05

Comparing the results

Compare the total time taken by the horse and man. If the man's time is less than horse's time, then the man wins. If the horse's time is less, then the horse wins.

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.

Uniformly Accelerated Motion
In physics, the concept of uniformly accelerated motion describes an object that is speeding up or slowing down at a steady rate. When the acceleration is constant, the motion of the object can be analyzed using a set of equations known as the kinematic equations.

One of these is the equation for the displacement during a period of constant acceleration, which is given by \(d = ut + \frac{1}{2}at^2\), where \(d\) is the displacement, \(u\) is the initial velocity, \(t\) is the time, and \(a\) is the acceleration. Since both the man and the horse in the given problem are starting from rest (u = 0 m/s), the formula simplifies to \(d = \frac{1}{2}at^2\). This equation allows us to calculate the distance each competitor covers during their acceleration phase before reaching a constant speed.

The uniform acceleration gives us a predictable pattern of motion, making it easier to calculate various aspects of motion, which is valuable for solving kinematics problems.
Final Speed Calculation
Calculating the final speed of an object after a period of uniform acceleration is crucial for understanding the dynamics of its motion. The formula to find the final velocity \(v\) after acceleration is \(v = u + at\), where \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time during which the object accelerated.

When the motion starts from rest, as with the man and the horse in our race scenario, the initial velocity \(u\) is zero and the equation simplifies to \(v = at\). By inputting the respective acceleration and time values for the man and the horse into this equation, we determine their final speeds after the period of acceleration. This is the speed they will maintain after they stop accelerating and continue with a constant velocity towards the finish line.
Time-Distance Relationship
The relationship between time and distance is a fundamental aspect of kinematics. Once we know the speed at which an object is moving (assuming it's constant), we can calculate the time it takes to cover a certain distance with the simple equation \(t = \frac{d}{v}\), where \(t\) is the time, \(d\) is the distance, and \(v\) is the velocity. When either the distance or the time is unknown, rearranging this equation can solve for the missing variable.

Looking at our race, after the man and the horse have each reached their constant speeds, we can apply this equation to find out how long it takes them to complete the remaining distance to the finish line. In practice, we use the speeds calculated after acceleration and the remaining distances, which we obtained by subtracting the distances they covered while accelerating from the total of 200 meters for the horse and 100 meters for the man, allowing us to calculate the time each will take to reach the finish, and thus, determine who wins the race.

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. What constant acceleration, in SI units, must a car have to go from zero to 60 mph in 10 s? b. What fraction of \(g\) is this? c. How far has the car traveled when it reaches \(60 \mathrm{mph} ?\) Give your answer both in \(\mathrm{SI}\) units and in feet.

Heather and Jerry are standing on a bridge \(50 \mathrm{m}\) above a river. Heather throws a rock straight down with a speed of 20 \(\mathrm{m} / \mathrm{s} .\) Jerry, at exactly the same instant of time, throws a rock straight up with the same speed. Ignore air resistance. a. How much time elapses between the first splash and the second splash? b. Which rock has the faster speed as it hits the water?

A driver has a reaction time of \(0.50 \mathrm{s}\), and the maximum deceleration of her car is \(6.0 \mathrm{m} / \mathrm{s}^{2}\). She is driving at \(20 \mathrm{m} / \mathrm{s}\) when suddenly she sees an obstacle in the road \(50 \mathrm{m}\) in front of her. Can she stop the car in time to avoid a collision?

Jenny and Alyssa are members of the cross-country team. On a training run, Jenny starts off and runs at a constant \(3.8 \mathrm{m} / \mathrm{s}\). Alyssa starts \(15 \mathrm{s}\) later and runs at a constant \(4.0 \mathrm{m} / \mathrm{s} .\) At what time after Jenny's start does Alyssa catch up with Jenny?

A small propeller airplane can comfortably achieve a high enough speed to take off on a runway that is \(1 / 4\) mile long. A large, fully loaded passenger jet has about the same acceleration from rest, but it needs to achieve twice the speed to take off. What is the minimum runway length that will serve? Hint: You can solve this problem using ratios without having any additional information.

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.