In 2005, Hurricane Rita hit several states in the southern United States. In the panic to escape her wrath, thousands of people tried to flee Houston, Texas by car. One car full of college students traveling to Tyler, Texas, 199 miles north of Houston, moved at an average speed of \(3.0 \mathrm{~m} / \mathrm{s}\) for one-fourth of the time, then at \(4.5 \mathrm{~m} / \mathrm{s}\) for another one-fourth of the time, and at \(6.0 \mathrm{~m} / \mathrm{s}\) for the remainder of the trip. a) How long did it take the students to reach their destination? b) Sketch a graph of position versus time for the trip.

Short Answer

Expert verified
Answer: It took the students 49,268.72 seconds to reach Tyler, Texas. The position vs. time graph is a continuous line with three distinct segments representing the different speeds (3.0 m/s, 4.5 m/s, and 6.0 m/s) at different time intervals. The slope of each line segment corresponds to the speed at that particular segment of the trip. The final point on the graph is at the total time of 49,268.72 seconds and the final position of 320,241.66 meters.

Step by step solution

01

Determine the time taken for each speed segment

The students traveled at 3.0 m/s for one-fourth of the time, at 4.5 m/s for another one-fourth of the time, and at 6.0 m/s for the remaining half of the time. Let t be the total time. Then, we have: - Time taken for the first segment: \(\frac{1}{4}t\). - Time taken for the second segment: \(\frac{1}{4}t\). - Time taken for the third segment: \(\frac{1}{2}t\).
02

Calculate the distance traveled in each segment

Now, we can calculate the distance traveled in each segment using the formula: distance = speed × time. - Distance covered in the first segment: \(3.0\,\text{m/s} \times \frac{1}{4}t = \frac{3}{4}t\) m. - Distance covered in the second segment: \(4.5\,\text{m/s} \times \frac{1}{4}t = \frac{9}{8}t\) m. - Distance covered in the third segment: \(6.0\,\text{m/s} \times \frac{1}{2}t = 3t\) m.
03

Add the distances to find the total distance

Add all three distances to find the total distance traveled: \(\frac{3}{4}t + \frac{9}{8}t + 3t = 199\) miles We must convert 199 miles into meters to solve for t. There are 1609.34 meters in a mile, so 199 miles is equal to \(199 \times 1609.34\,\text{m} = 320241.66\,\text{m}\). Therefore, \(\frac{3}{4}t + \frac{9}{8}t + 3t = 320241.66\,\text{m}\).
04

Solve for t

Now, we can solve for t: \(t(\frac{3}{4} + \frac{9}{8} + 3) = 320241.66\,\text{m}\) \(t(6.5) = 320241.66\,\text{m}\) Divide both sides by 6.5: \(t=\frac{320241.66\,\text{m}}{6.5}=49268.72\,\text{s}\) So, it took the students 49,268.72 seconds to reach their destination.
05

a) Answer

The students took 49,268.72 seconds to reach Tyler, Texas.
06

b) Sketch a graph of position vs. time

To sketch the position vs. time graph, we need to plot the position (distance) at each interval of time. - At \(t = 0\,\text{s}\), the position is \(0\,\text{m}\). - At \(t = \frac{1}{4}t\), the position is \(\frac{3}{4}t\) or (since \(t = 49268.72\,\text{s}\)), \(12317.18\,\text{m}\). - At \(t = \frac{1}{2}t\), the position is \(\frac{3}{4}t + \frac{9}{8}t\) or \(24634.36\,\text{m}\). - At \(t = 49268.72\,\text{s}\), the position is \(320241.66\,\text{m}\), which is the final position. Now, connect these points on the graph with straight lines to represent the distance traveled at each speed. The graph should be a continuous line, indicating that the position is changing throughout the time. The slope of each line segment represents the speed at each segment of the trip: 3.0 m/s, 4.5 m/s, and 6.0 m/s. The final point on the graph will be at the total time of 49,268.72 seconds and the final position of 320,241.66 meters.

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 Vaia!

Key Concepts

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

Average Speed
Understanding the concept of average speed is crucial for interpreting motion in physics. Average speed is defined as the total distance traveled divided by the total time taken to cover that distance. In mathematical terms, the average speed, V_avg, can be represented by the equation:
\[\begin{equation} V_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} \end{equation}\]
Consider our example with the students fleeing Hurricane Rita. Here, we want to find out how long it took to travel 199 miles. The average speed isn't constant, but changes in a piecewise manner as they traveled at different speeds during different portions of their journey. To calculate the total time of travel, one approach is to determine the distance covered in each segment at the given speeds and then sum these distances before converting to total time using the average speed equation above. This is an essential skill in kinematics, as it allows one to analyze motion over varying speed segments and still determine overall time for travel.
Distance-Time Graph
A distance-time graph is an invaluable tool in kinematics, enabling us to visualize an object's motion over time. Each point on the graph represents the position of an object at a specific time. The steeper the graph's slope, the faster the object is moving. Conversely, a flat line indicates the object is stationary.
To construct a distance-time graph for our hurricane evacuation scenario, you would plot time on the horizontal axis and distance on the vertical axis. For each speed segment, you would calculate the distance using the average speed and the time taken. After plotting these calculated distances against the corresponding times, straight lines would connect these points, thereby creating a piecewise linear graph. The slope of each segment corresponds to the speed at that particular part of the trip, depicting the changes in speed visually as the students traveled to their destination.
Units Conversion
In physics problems, especially those involving kinematics, it's imperative to check that all measurements are in compatible units before beginning any calculations. Converting units is a common and necessary step for working through many physics exercises.
For instance, in the problem concerning the students escaping Hurricane Rita, distances are initially given in miles, while speeds are in meters per second. To proceed with the calculations, we must convert miles to meters, since meters per second is the standard unit for speed in the International System of Units (SI). One mile is equivalent to 1609.34 meters. By multiplying the total distance in miles by this conversion factor, we obtain the distance in meters, allowing for coherent and accurate calculations. Always remember, correctly converting units can make the difference between a right and wrong answer in physics problems.

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

You drive with a constant speed of \(13.5 \mathrm{~m} / \mathrm{s}\) for 30.0 s. You then accelerate for 10.0 s to a speed of \(22.0 \mathrm{~m} / \mathrm{s}\). You then slow to a stop in \(10.0 \mathrm{~s}\). How far have you traveled?

The velocity as a function of time for a car on an amusement park ride is given as \(v=A t^{2}+B t\) with constants \(A=2.0 \mathrm{~m} / \mathrm{s}^{3}\) and \(B=1.0 \mathrm{~m} / \mathrm{s}^{2} .\) If the car starts at the origin, what is its position at \(t=3.0\) s?

A car starts from rest and accelerates at \(10.0 \mathrm{~m} / \mathrm{s}^{2}\) How far does it travel in 2.00 s?

A car travels \(22.0 \mathrm{~m} / \mathrm{s}\) north for \(30.0 \mathrm{~min}\) and then reverses direction and travels \(28.0 \mathrm{~m} / \mathrm{s}\) for \(15.0 \mathrm{~min}\). What is the car's total displacement? a) \(1.44 \cdot 10^{4} \mathrm{~m}\) b) \(6.48 \cdot 10^{4} \mathrm{~m}\) c) \(3.96 \cdot 10^{4} \mathrm{~m}\) d) \(9.98 \cdot 10^{4} \mathrm{~m}\)

You drop a water balloon straight down from your dormitory window \(80.0 \mathrm{~m}\) above your friend's head. At \(2.00 \mathrm{~s}\) after you drop the balloon, not realizing it has water in it your friend fires a dart from a gun, which is at the same height as his head, directly upward toward the balloon with an initial velocity of \(20.0 \mathrm{~m} / \mathrm{s}\). a) How long after you drop the balloon will the dart burst the balloon? b) How long after the dart hits the balloon will your friend have to move out of the way of the falling water? Assume the balloon breaks instantaneously at the touch of the dart.

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.

Sign-up for free