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Chapter 2: Problem 3

Alan leaves Los Angeles at 8: 00 a.m. to drive to San Francisco, 400 mi away. He travels at a steady 50 mph. Beth leaves Los Angeles at 9: 00 a.m. and drives a steady 60 mph. a. Who gets to San Francisco first? b. How long does the first to arrive have to wait for the second?

Short Answer

Expert verified
Beth arrives first in San Francisco. She has to wait 20 minutes for Alan.

Step by step solution

01

Calculate Alan's Travel Time

Alan covers 400 miles at a steady speed of 50 miles per hour. To calculate his time of travel, the total distance is divided by the speed. So, Alan's travel time is \( \frac{400}{50} = 8 \) hours
02

Factor in Alan's Departure Time

Alan departed at 8:00 a.m., traveling for 8 hours. This means he'll reach San Francisco at 4:00 p.m.
03

Calculate Beth's Travel Time

Beth covers 400 miles at a speed of 60 mph. To work out her time of travel, the total distance is divided by the speed. So, Beth's travel time is \( \frac{400}{60} = 6.67 \) hours, roughly 6 hours and 40 minutes.
04

Factor in Beth's Departure Time

Beth departed at 9:00 a.m., traveling for approximately 6 hours and 40 minutes. This means she will reach San Francisco at approximately 3:40 p.m.
05

Determine Who Arrives First and the Wait Time

Since Beth arrives at approximately 3:40 p.m. and Alan arrives at 4:00 p.m., Beth gets to San Francisco first. Alan arrives about 20 minutes after Beth, so she has to wait 20 minutes for him.

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Key Concepts

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

Travel time calculation
Calculating travel time is essential to determine how long it takes to get from one place to another. Understanding this concept involves simple arithmetic using the formula:
  • Travel time = Distance 梅 Speed.
In the example, Alan and Beth are traveling the same distance of 400 miles. To find out how long their respective journeys will take, we divide the distance by their speeds.
Alan, traveling at 50 mph, will have a travel time of \( \frac{400}{50} = 8 \) hours. Beth, traveling faster at 60 mph, calculates her travel time as \( \frac{400}{60} \approx 6.67 \) hours, or roughly 6 hours and 40 minutes.
It's important to accurately calculate travel times to plan arrival times effectively.
Speed and distance relation
Speed and distance are closely related and influence how quickly one can reach a destination. The formula connecting speed, distance, and time is a fundamental aspect of motion:
  • Distance = Speed 脳 Time.
In scenarios like this exercise, understanding this relationship helps us make predictions and solve problems involving travel.
For both Alan and Beth, the total distance to be covered is set at 400 miles. The differences in their journey times arise from the difference in speed: Alan travels at 50 mph and Beth at 60 mph.
This variation affects travel time, and hence, who reaches the destination first. Faster speeds reduce the travel time needed to cover the same distance, exemplified by Beth arriving earlier than Alan despite departing later.
Problem-solving strategy
To tackle problems involving relative motion, it's helpful to use a systematic approach. Here's a strategy:
  • Identify what you're solving for: In this exercise, it's the arrival times and wait time.
  • Use known formulas: Apply the distance formula to calculate travel time.
  • Consider all factors: Take into account departure times, speeds, and distance.
  • Compare results: Determine who arrives first by aligning travel times with departure times.
In the given problem, first calculate travel times for both Alan and Beth using the speed and distance formula. Then, factor in their respective start times.
Beth leaves after Alan but drives faster, so her earlier arrival is confirmed by simple subtraction of the travel hours from her starting time. Finally, compare their arrival times to find the duration Beth waits for Alan in San Francisco, which rounds out the solution.

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Most popular questions from this chapter

An object starts from rest at \(x=0\) m at time \(t=0\) s. Five seconds later, at \(t=5.0 \mathrm{s},\) the object is observed to be at \(x=40.0 \mathrm{m}\) and to have velocity \(v_{x}=11 \mathrm{m} / \mathrm{s}\) a. Was the object's acceleration uniform or nonuniform? Explain your reasoning. b. Sketch the velocity-versus-time graph implied by these data. Is the graph a straight line or curved? If curved, is it concave upward or downward?

A toy train is pushed forward and released at \(x_{0}=2.0 \mathrm{m}\) with a speed of \(2.0 \mathrm{m} / \mathrm{s} .\) It rolls at a steady speed for \(2.0 \mathrm{s}\), then one wheel begins to stick. The train comes to a stop \(6.0 \mathrm{m}\) from the point at which it was released. What is the magnitude of the train's acceleration after its wheel begins to stick?

Ann and Carol are driving their cars along the same straight road. Carol is located at \(x=2.4 \mathrm{mi}\) at \(t=0\) hours and drives at a steady 36 mph. Ann, who is traveling in the same direction, is located at \(x=0.0 \mathrm{mi}\) at \(t=0.50\) hours and drives at a steady \(50 \mathrm{mph}\) a. At what time does Ann overtake Carol? b. What is their position at this instant? c. Draw a position-versus-time graph showing the motion of both Ann and Carol.

Nicole throws a ball straight up. Chad watches the ball from a window \(5.0 \mathrm{m}\) above the point where Nicole released it. The ball passes Chad on the way up, and it has a speed of \(10 \mathrm{m} / \mathrm{s}\) as it passes him on the way back down. How fast did Nicole throw the ball?

A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at \(30 \mathrm{m} / \mathrm{s}^{2}\) for \(30 \mathrm{s}\), then runs out of fuel. Ignore any air resistance effects. a. What is the rocket's maximum altitude? b. How long is the rocket in the air before hitting the ground? c. Draw a velocity-versus-time graph for the rocket from liftoff until it hits the ground.
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