/*! 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 3 Trip Home. You normally drive on... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Trip Home. You normally drive on the freeway between San Diego and Los Angeles at an average speed of 105 \(\mathrm{km} / \mathrm{h}\) (65 \(\mathrm{mi} / \mathrm{h} ),\) and the trip takes 2 \(\mathrm{h}\) and 20 min. On a Friday afternoon, however, heavy traffic slows you down and you drive the same distance at an average speed of only 70 \(\mathrm{km} / \mathrm{h}(43 \mathrm{mi} / \mathrm{h})\) How much longer does the trip take?

Short Answer

Expert verified
The trip takes 1 hour and 10 minutes longer.

Step by step solution

01

Understand the Problem

We need to find how much longer the trip takes at a slower speed compared to the normal trip time.
02

Convert Trip Time to Hours

The normal trip time is given as 2 hours and 20 minutes. Convert 20 minutes to hours: \( 20 \text{ minutes} = \frac{20}{60} = \frac{1}{3} \text{ hours} \). Thus, the total time is \( 2 + \frac{1}{3} = \frac{7}{3} \text{ hours} \).
03

Calculate Distance

Use the formula for distance \( \text{Distance} = \text{Speed} \times \text{Time} \) with normal conditions: Distance \( = 105 \frac{\text{km}}{\text{h}} \times \frac{7}{3} \text{ h} = 245 \text{ km} \).
04

Calculate Time at Slower Speed

Use the distance and the new speed to find the time: \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{245 \text{ km}}{70 \frac{\text{km}}{\text{h}}} = 3.5 \text{ hours} \).
05

Find the Additional Time

Subtract the normal trip time from the slower trip time to find the additional time taken: \( 3.5 \text{ hours} - \frac{7}{3} \text{ hours} = 3.5 \text{ hours} - 2.333 \text{ hours} = 1.167 \text{ hours} \) or 1 hour and 10 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 91Ó°ÊÓ!

Key Concepts

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

Average Speed Calculation
Understanding how to calculate average speed is crucial in solving this physics problem. Average speed is defined as the total distance traveled divided by the total time taken. It can be expressed using the formula: \[ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} \] In the problem, you're given two scenarios with different speeds for the same distance. Initially, the average speed is 105 km/h for the journey, allowing the calculation of total distance when multiplied by the total travel time of 2 hours and 20 minutes, converted into hours. On the second occasion, the average speed is reduced to 70 km/h due to heavy traffic.
  • Average speed helps understand how consistent velocity is throughout the trip.
  • It's useful for predicting travel times under different conditions.
Recognizing the role of average speed is essential to find out how slower speeds impact travel time.
Time Conversion
In physics problems like this, time conversion is a key skill. We often need to convert time given in minutes to hours for consistent use in formulas. Here's a step-by-step explanation of how we achieve this:Given the original time of 2 hours and 20 minutes, first convert the 20 minutes into a fractional hour. Since 1 hour equals 60 minutes, you convert minutes to hours by dividing the number of minutes by 60. Thus,\[ 20 \text{ minutes} = \frac{20}{60} = \frac{1}{3} \text{ hours} \]Adding this to the 2 hours results in:\[ 2 + \frac{1}{3} = \frac{7}{3} \text{ hours} \]Moving all time measures to the same unit simplifies calculations.
  • This method is especially useful when calculating with speeds given in terms of km/h or mi/h.
  • Ensures consistency and accuracy in further travel time or distance computation calculations.
Proper time conversion prevents errors and facilitates seamless calculations.
Distance Calculation
Distance calculation is an essential part of solving this problem. We use the formula: \[ \text{Distance} = \text{Speed} \times \text{Time} \] In our scenario, the distance is calculated during the first journey using the known average speed of 105 km/h and the total converted time of \(\frac{7}{3}\) hours:\[ \text{Distance} = 105 \frac{\text{km}}{\text{h}} \times \frac{7}{3} \text{h} = 245 \text{ km} \]Then, when the speed is reduced to 70 km/h due to traffic, the same distance of 245 km is used in our formula to determine the new travel time:\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{245 \text{ km}}{70 \frac{\text{km}}{\text{h}}} = 3.5 \text{ hours} \]
  • Ensures accurate travel predictions by relating speed and time.
  • Facilitates comparison between different travel speeds for the same route.
Accurately calculating distance is crucial for resolving time-related inquiries and understanding journey dynamics.

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 bird is flying due east. Its distance from a tall building is given by \(x(t)=28.0 \mathrm{m}+(12.4 \mathrm{m} / \mathrm{s}) t-\left(0.0450 \mathrm{m} / \mathrm{s}^{3}\right) t^{3}\) What is the instantaneous velocity of the bird when \(t=8.00 \mathrm{s}?\)

Collision. The engineer of a passenger train traveling at 25.0 \(\mathrm{m} / \mathrm{s}\) sights a freight train whose caboose is 200 \(\mathrm{m}\) ahead onthe same track (Fig. P2.70). The freight train is traveling at 15.0 \(\mathrm{m} / \mathrm{s}\) in the same direction as the passenger train. The engineer of the passenger train immediately applies the brakes, causing a constant acceleration of 0.100 \(\mathrm{m} / \mathrm{s}^{2}\) in a direction opposite to the train's velocity, while the freight train continues with constant speed. Take \(x=0\) at the location of the front of the passenger train when the engineer applies the brakes. (a) Will the cows nearby witness a collision? (b) If so, where will it take place? (c) On a single graph, sketch the positions of the front of the passenger train and the back of the freight train.

A subway train starts from rest at a station and accelerates at a rate of 1.60 \(\mathrm{m} / \mathrm{s}^{2}\) for 14.0 \(\mathrm{s}\) . It runs at constant speed for 70.0 \(\mathrm{s}\) and slows down at a rate of 3.50 \(\mathrm{m} / \mathrm{s}^{2}\) until it stops at the next station. Find the total distance covered.

A jet fighter pilot wishes to accelerate from rest at a constant acceleration of 5\(g\) to reach Mach 3 (three times the speed of sound) as quickly as possible. Experimental tests reveal that he will black out if this acceleration lasts for more than 5.0 s. Use 331 \(\mathrm{m} / \mathrm{s}\) for the speed of sound. (a) Will the period of acceleration last long enough to cause him to black out? (b) What is the greatest speed he can reach with an acceleration of 5\(g\) before blacking out?

During launches, rockets often discard unneeded parts. A certain rocket starts from rest on the launch pad and accelerates upward at a steady 3.30 \(\mathrm{m} / \mathrm{s}^{2}\) . When it is 235 \(\mathrm{m}\) above the launch pad, it discards a used fuel canister by simply disconnecting it. Once it is disconnected, the only force acting on the canister is gravity (air resistance can be ignored). (a) How high is the rocket when the canister hits the launch pad, assuming that the rocket does not change its acceleration? (b) What total distance did the canister travel between its release and its crash onto the launch pad?

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.