/*! 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 15 To qualify for the finals in a r... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 15

To qualify for the finals in a racing event, a race car must achieve an average speed of \(250 \mathrm{~km} / \mathrm{h}\) on a track with a total length of \(1600 \mathrm{~m}\). If a particular car covers the first half of the track at an average speed of \(230 \mathrm{~km} / \mathrm{h}\), what minimum average speed must it have in the second half of the event in order to qualify?

Short Answer

Expert verified
The car must maintain a minimum average speed of \(280.5 \mathrm{~km/h}\) for the second half of the race to qualify.

Step by step solution

01

Calculate time taken for the first half of the race

Average speed = distance/time. So, we can calculate the time taken for the first half of the race: \(230 \mathrm{~km/h} = 1150 \mathrm{~m} / t_1\). Solving this equation gives us \(t_1= 18 \mathrm{~s}\).
02

Calculate total time for the race

To maintain an average speed of \(250 \mathrm{~km/h}\) for the whole race, we calculate the total time required: \(250 \mathrm{~km/h}\) = \(1600 \mathrm{~m} / t_{total}\). Solving this equation gives us \(t_{total}= 23.04 \mathrm{~s}\).
03

Calculate time for the second half of the race

Now, to find the time taken for the second half of the race, subtract the time taken for the first half from the total time: \(t_{total} - t_1\). This gives us \(23.04 \mathrm{~s} - 18 \mathrm{~s}\), which equals to \(5.04 \mathrm{~s}\)
04

Calculate minimum average speed for the second half

Now, using this time for the second half, we can calculate the minimum average speed: average speed = distance/time. So, \(speed_{min}\) = \(1150 \mathrm{~m}\) / \(5.04 \mathrm{~s}\), which gives us \(minimum speed = 280.5 \mathrm{~km/h}\).

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 certain freely falling object, released from rest, requires \(1.50 \mathrm{~s}\) to travel the last \(30.0 \mathrm{~m}\) before it hits the ground. (a) Find the velocity of the object when it is \(30.0 \mathrm{~m}\) above the ground. (b) Find the total distance the object travels during the fall.

In Bosnia, the ultimate test of a young man's courage used to be to jump off a 400 -year-old bridge (destroyed in 1993; rebuilt in 2004) into the River Neretva, \(23 \mathrm{~m}\) below the bridge. (a) How long did the jump last? (b) How fast was the jumper traveling upon impact with the river? (c) If the speed of sound in air is \(340 \mathrm{~m} / \mathrm{s}\), how long after the jumper took off did a spectator on the bridge hear the splash?

Two boats start together and race across a \(60-\mathrm{km}\)-wide lake and back. Boat A goes across at \(60 \mathrm{~km} / \mathrm{h}\) and returns at \(60 \mathrm{~km} / \mathrm{h}\). Boat B goes across at \(30 \mathrm{~km} / \mathrm{h}\), and its crew, realizing how far behind it is getting, returns at \(90 \mathrm{~km} / \mathrm{h}\). Turnaround times are negligible, and the boat that completes the round trip first wins. (a) Which boat wins and by how much? (Or is it a tie?) (b) What is the average velocity of the winning boat?

Two students are on a balcony a distance \(h\) above the street. One student throws a ball vertically downward at a speed \(v_{0}\); at the same time, the other student throws a ball vertically upward at the same speed. Answer the following symbolically in terms of \(v_{0}, g\), \(h\), and \(t\). (a) Write the kinematic equation for the \(y\)-coordinate of each ball. (b) Set the equations found in part (a) equal to height 0 and solve each for \(t\) symbolically using the quadratic formula. What is the difference in the two balls' time in the air? (c) Use the time-independent kinematics equation to find the velocity of each ball as it strikes the ground. (d) How far apart are the balls at a time \(t\) after they are released and before they strike the ground?

A certain cable car in San Francisco can stop in \(10 \mathrm{~s}\) when traveling at maximum speed. On one occasion, the driver sees a dog a distance \(d\) in front of the car and slams on the brakes instantly. The car reaches the dog \(8.0 \mathrm{~s}\) later, and the dog jumps off the track just in time. If the car travels \(4.0 \mathrm{~m}\) beyond the position of the dog before coming to a stop, how far was the car from the dog? (Hint: You will need three equations.)
See all solutions

Recommended explanations on Physics Textbooks

Nuclear Physics

Read Explanation

Solid State Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Mechanics and Materials

Read Explanation

Electrostatics

Read Explanation

Rotational 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.