/*! 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 16 When you normally drive the free... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 16

When you normally drive the freeway between Sacramento and San Francisco at an average speed of \(105 \mathrm{~km} / \mathrm{h}\) (65 \(\mathrm{mi} / \mathrm{h}\) ), the trip takes \(1.0 \mathrm{~h}\) and \(20 \mathrm{~min}\). On a Friday afternoon, however, heavy traffic slows you down to an average of \(70 \mathrm{~km} / \mathrm{h}\) (43 \(\mathrm{mi} / \mathrm{h}\) ) for the same distance. How much longer does the trip take on Friday than on the other days?

Short Answer

Expert verified
The trip takes 40 minutes longer on Friday.

Step by step solution

01

Convert Time to Hours

First, convert the trip time from hours and minutes to only hours. The trip normally takes 1 hour and 20 minutes, which is equivalent to \( 1 + \frac{20}{60} = \frac{80}{60} = \frac{4}{3} \) hours or approximately 1.33 hours.
02

Calculate Normal Distance

Calculate the distance of the trip using the normal speed. The formula for distance is \( \, \text{Distance} = \text{Speed} \times \text{Time} \, \). Using the normal speed of 105 km/h and time of \( \frac{4}{3} \) hours, the distance is \( 105 \text{ km/h} \times \frac{4}{3} \text{ h} = 140 \text{ km} \).
03

Calculate Friday Travel Time

Now, calculate the travel time on Friday using the slower speed of 70 km/h. Use the distance of 140 km: \( \, \text{Time} = \frac{\text{Distance}}{\text{Speed}} \, \). The time in slower traffic is \( \frac{140 \text{ km}}{70 \text{ km/h}} = 2 \text{ hours} \).
04

Compute Time Difference

Find the difference between the slow Friday travel time and the normal travel time. Subtract the normal time from the Friday time: \( 2 \text{ hours} - \frac{4}{3} \text{ hours} = \frac{6}{3} \text{ hours} - \frac{4}{3} \text{ hours} = \frac{2}{3} \text{ hours} \), which is approximately 40 minutes.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Speed and Distance Calculations
Understanding how to calculate speed and distance is key to solving many physics problems. Essentially, speed defines how fast something moves over a period, while distance tells you how far it moves.
The formula to remember is:
  • Distance = Speed 脳 Time
In our problem, when traveling at an average speed of 105 km/h, the distance was calculated as 140 km using:
  • 140 km = 105 km/h 脳 1.33 h
This demonstrates using speed to calculate how far you have traveled during a specific time frame. Knowing either two of these three elements鈥攕peed, distance, or time鈥攁llows you to calculate the third.
Time Conversion
Time conversion is the process of expressing time in different units to make calculations easier, often needed in physics problems. In our scenario, we converted the trip time of 1 hour and 20 minutes into hours.
To convert minutes to hours, use:
  • 1 hour 20 minutes = 1 + \(\frac{20}{60}\) hours = \(\frac{4}{3}\) hours or approximately 1.33 hours
This conversion is crucial for consistency in unit calculations, as most physics equations require consistent units.
Travel Time Analysis
Analyzing travel time allows us to understand how long a journey takes under different conditions. In our problem, you calculate travel time by considering changes in speed.
On a normal day, traveling at 105 km/h, the journey takes 1.33 hours. By slowing down to 70 km/h on a busy Friday, the time of travel increases:
Time for Friday:
  • Time = \( \frac{140\text{ km}}{70\text{ km/h}} = 2\text{ hours} \)
This analysis compares travel times to highlight the effect of slower speeds on overall journey duration.
Traffic Impact on Travel
Traffic can significantly alter travel times by reducing the speed at which you travel. In this case, on a Friday afternoon, heavy traffic reduced the speed from 105 km/h to 70 km/h.
As a result, the trip took longer. Here, the Friday travel time increased to 2 hours. In comparison, the same trip without traffic would take only 1.33 hours.
Thus, the impact of traffic was an additional \(\frac{2}{3}\) hour or roughly 40 minutes, showing how crucially traffic affects travel efficiency.
Step-by-Step Physics Solutions
Learning to solve physics problems involves following a step-by-step approach for clarity and accuracy. Breaking down the problem into manageable steps ensures understanding and minimizes errors.
For this exercise:
  • First, convert all time to a consistent unit, such as hours.
  • Then, calculate distance using speed and time.
  • Next, find the new travel time using slower speeds.
  • Finally, determine the time difference caused by traffic delay.
Following each step methodically makes even complex problems easier to tackle and understand.

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 hot-air balloonist, rising vertically with a constant speed of \(5.00 \mathrm{~m} / \mathrm{s},\) releases a sandbag at the instant the balloon is \(40.0 \mathrm{~m}\) above the ground. (See Figure \(2.54 .)\) After it is released, the sandbag encounters no appreciable air drag. (a) Compute the position and velocity of the sandbag at \(0.250 \mathrm{~s}\) and \(1.00 \mathrm{~s}\) after its release. (b) How many seconds after its release will the bag strike the ground? (c) How fast is it moving as it strikes the ground? (d) What is the greatest height above the ground that the sandbag reaches? (e) Sketch graphs of this bag's acceleration, velocity, and vertical position as functions of time.

A healthy heart pumping at a rate of 72 beats per minute increases the speed of blood flow from 0 to \(425 \mathrm{~cm} / \mathrm{s}\) with each beat. Calculate the acceleration of the blood during this process.

A jetliner has a cruising air speed of \(600 \mathrm{mi} / \mathrm{h}\) relative to the air. How long does it take this plane to fly round trip from San Francisco to Chicago, an east-west flight of \(2000 \mathrm{mi}\) each way, (a) if there is no wind blowing and (b) if the wind is blowing at \(150 \mathrm{mi} / \mathrm{h}\) from the west to the east?

On March \(27,2004,\) the United States successfully tested the hypersonic X-43A scramjet, which flew at Mach 7 (seven times the speed of sound) for 11 s. (a) At this rate, how many minutes would it take such a scramjet to carry passengers the approximately \(5000 \mathrm{~km}\) from San Francisco to New York? (Use the speed of sound at \(\left.0^{\circ} \mathrm{C}, 331 \mathrm{~m} / \mathrm{s} .\right)\) (b) How many kilometers did the scramjet travel during its 11 s test?

Nerve impulses travel at different speeds, depending on the type of fiber through which they move. The impulses for touch travel at \(76.2 \mathrm{~m} / \mathrm{s},\) while those registering pain move at \(0.610 \mathrm{~m} / \mathrm{s} .\) If a person stubs his toe, find (a) the time for each type of impulse to reach his brain, and (b) the time delay between the pain and touch impulses. Assume that his brain is \(1.85 \mathrm{~m}\) from his toe and that the impulses travel directly from toe to brain.
See all solutions

Recommended explanations on Physics Textbooks

Famous Physicists

Read Explanation

Medical Physics

Read Explanation

Quantum Physics

Read Explanation

Nuclear Physics

Read Explanation

Turning Points in Physics

Read Explanation

Astrophysics

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.