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Chapter 2: Problem 13

You drive on Interstate 10 from San Antonio to Houston, half the time at \(55 \mathrm{~km} / \mathrm{h}\) and the other half at \(90 \mathrm{~km} / \mathrm{h}\). On the way back you travel half the distance at \(55 \mathrm{~km} / \mathrm{h}\) and the other half at \(90 \mathrm{~km} / \mathrm{h}\). What is your average speed (a) from San Antonio to Houston, (b) from Houston back to San Antonio, and (c) for the entire trip? (d) What is your average velocity for the entire trip? (e) Sketch \(x\) versus \(t\) for (a), assuming the motion is all in the positive \(x\) direction. Indicate how the average velocity can be found on the sketch.

Short Answer

Expert verified
(a) 68.28 km/h, (b) 72.5 km/h, (c) 68.28 km/h, (d) 0 km/h.

Step by step solution

01

Understanding the Problem

To solve this, we first need to recognize that for part (a), we are given speeds for equal time segments, while for part (b), we have speeds for equal distance segments. This will affect how we calculate the average speed in each scenario.
02

Calculate Average Speed from San Antonio to Houston (Part a)

For equal time segments, the average speed can be calculated using the harmonic mean of the speeds. The formula for average speed (\(v_{avg}\)) when traveling half the time at a speed of \(v_1\) and the other half at \(v_2\) is:\[v_{avg} = \frac{2v_1v_2}{v_1 + v_2}\]Plugging in \(v_1 = 55\, \mathrm{km/h}\) and \(v_2 = 90\, \mathrm{km/h}\), we get:\[v_{avg} = \frac{2 \times 55 \times 90}{55 + 90} = \frac{9900}{145} \approx 68.28\, \mathrm{km/h}\]
03

Calculate Average Speed from Houston to San Antonio (Part b)

For equal distance segments, the average speed can be calculated using the arithmetic mean of the speeds. The average speed is simply:\[v_{avg} = \frac{v_1 + v_2}{2}\]Substituting \(v_1 = 55\, \mathrm{km/h}\) and \(v_2 = 90\, \mathrm{km/h}\), we find:\[v_{avg} = \frac{55 + 90}{2} = 72.5\, \mathrm{km/h}\]
04

Calculate Average Speed for the Entire Trip (Part c)

The average speed for the entire trip can be calculated using the total distance divided by the total time. Assuming the distance from San Antonio to Houston is \(x\), the entire trip distance is \(2x\). The time traveling from San Antonio to Houston is:\[t_1 = \frac{1}{2}(\frac{x}{55} + \frac{x}{90})\]From Houston to San Antonio it is:\[t_2 = 2(\frac{x/2}{55}) + 2(\frac{x/2}{90}) = \frac{x}{55} + \frac{x}{90}\]Total time \(T = t_1 + t_2\) simplifies:\[T = \frac{x}{55} + \frac{x}{90}\]Total average speed \(v_{avg}\):\[v_{avg} = \frac{2x}{T} = \frac{2x}{\frac{x}{55} + \frac{x}{90}} = 68.28\, \mathrm{km/h}\]
05

Calculate Average Velocity for Entire Trip (Part d)

The average velocity for the entire trip is zero because the displacement over the entire trip is zero (you start and end at the same place). Therefore, \(v_{avg} = 0\).
06

Sketch x versus t for the Motion to Houston (Part e)

To sketch an \(x\) versus \(t\) graph, plot distance (\(x\)) on the vertical axis and time (\(t\)) on the horizontal. For the trip from San Antonio to Houston, use two line segments with different slopes representing different speeds. The slope of the first segment follows \(55\, \mathrm{km/h}\) and the second \(90\, \mathrm{km/h}\). The average velocity can be found by drawing a single line from the origin to the ending point of the motion, this line鈥檚 slope represents the average speed calculated earlier.

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Key Concepts

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

Kinematics
Kinematics involves the study of motion without considering the forces that cause it. One of the fundamental aspects is understanding how objects move in terms of displacement, velocity, and acceleration. The exercise presented involves a car trip that requires computing average speeds under different conditions.
First, we differentiate between the average speed and average velocity. Average speed is a scalar quantity, reflecting the total distance traveled divided by the total time of travel. It does not consider the direction of travel. Meanwhile, average velocity is a vector that considers the change in position (displacement) over time.
In our exercise, the calculations for average speed from San Antonio to Houston and vice-versa involve constant speeds over different variables (time and distance). It's crucial to note that while average speed gives magnitude, average velocity for a roundtrip turns out to be zero since initial and final positions are the same, canceling each other out in terms of displacement.
Harmonic Mean
The harmonic mean is particularly useful when speeds are averaged over equal time segments. This concept becomes important in scenarios like part (a) of the original exercise, where travel times at different speeds need to be averaged.
The harmonic mean formula for two speeds, say and ,1/2 is given by:\[v_{avg} = \frac{2v_1v_2}{v_1 + v_2}\]
The harmonic mean is beneficial because it appropriately weights each speed according to its associated duration, hence accurately representing the average rate of travel over a period. In our solution, using speeds of 55 km/h and 90 km/h over equal time segments results in an average speed of approximately 68.28 km/h.
This concept underscores the importance of accounting for time when computing average speeds in kinematic problems, providing a balanced and representative measure of travel rate.
Distance and Displacement
These terms often cause confusion but are distinct concepts in physics. Distance is the total path covered and is a scalar quantity, meaning it has magnitude but no direction. Displacement, on the other hand, is a vector quantity that considers only the change in position from the start to the end of a trip.
In the exercise, the total distance of the round trip from San Antonio to Houston and back is twice the distance between the two cities. In contrast, the displacement for the entire trip is zero because you start and end at the same location.
It's important to remember:
  • Distance is not dependent on direction.
  • Displacement includes directional information.
For example, traveling 200 km east and then 200 km west results in a distance of 400 km, but a displacement of 0 km. Understanding the difference is crucial in solving kinematic problems correctly.
Velocity vs. Time Graph
Understanding a velocity vs. time graph is crucial for visualizing motion. The graph typically presents velocity on the vertical axis and time on the horizontal axis. For a trip with constant speeds, as seen in our exercise, the graph would have horizontal line segments representing each constant velocity phase.
In the case of traveling from San Antonio to Houston, the graph would include two segments: - A line at 55 km/h for the first time segment - A line at 90 km/h for the second time segment
The change in slope between segments indicates a change in speed. The area under the curve of this graph represents the displacement. For roundtrips, a complete return to the starting point will mean the net displacement shown by the graph is zero, evident by the equal areas under positive and negative portions of the curve.
By analyzing these graphs, we gather insights about motion patterns, helping confirm calculations like average speeds and understanding overall travel dynamics.

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Most popular questions from this chapter

If a particle's position is given by \(x=4-12 t+3 t^{2}\) (where \(t\) is in seconds and \(x\) is in meters), what is its velocity at \(t=1 \mathrm{~s} ?\) (b) Is it moving in the positive or negative direction of \(x\) just then? (c) What is its speed just then? (d) Is the speed increasing or decreasing just then? (Try answering the next two questions without further calculation.) (e) Is there ever an instant when the velocity is zero? If so, give the time \(t\); if not, answer no. (f) Is there a time after \(t=3 \mathrm{~s}\) when the particle is moving in the negative direction of \(x ?\) If so, give the time \(t ;\) if not, answer no.

To set a speed record in a measured (straight-line) distance \(d\), a race car must be driven first in one direction (in time \(t_{1}\) ) and then in the opposite direction (in time \(t_{2}\) ). (a) To eliminate the effects of the wind and obtain the car's speed \(v_{c}\) in a windless situation, should we find the average of \(d / t_{1}\) and \(d / t_{2}\) (method 1 ) or should we divide \(d\) by the average of \(t_{1}\) and \(t_{2} ?(\mathrm{~b})\) What is the fractional difference in the two methods when a steady wind blows along the car's route and the ratio of the wind speed \(v_{w}\) to the car's speed \(v_{c}\) is \(0.0240 ?\)

A rock is shot vertically upward from the edge of the top of a tall building. The rock reaches its maximum height above the top of the building \(1.60 \mathrm{~s}\) after being shot. Then, after barely missing the edge of the building as it falls downward, the rock strikes the ground \(6.00 \mathrm{~s}\) after it is launched. In SI units: (a) with what upward velocity is the rock shot, (b) what maximum height above the top of the building is reached by the rock, and (c) how tall is the building?

Shows part of a street where traffic flow is to be controlled to allow a platoon of cars to move smoothly along the street. Suppose that the platoon leaders have just reached intersection 2, where the green appeared when they were distance \(d\) from the intersection. They continue to travel at a certain speed \(v_{p}\) (the speed limit) to reach intersection 3, where the green appears when they are distance \(d\) from it. The intersections are separated by distances \(D_{23}\) and \(D_{12}\). (a) What should be the time delay of the onset of green at intersection 3 relative to that at intersection 2 to keep the platoon moving smoothly? Suppose, instead, that the platoon had been stopped by a red light at intersection \(1 .\) When the green comes on there, the leaders require a certain time \(t_{r}\) to respond to the change and an additional time to accelerate at some rate \(a\) to the cruising speed \(v_{p} .\) (b) If the green at intersection 2 is to appear when the leaders are distance \(d\) from that intersection, how long after the light at intersection 1 turns green should the light at intersection 2 turn green?

A ball of moist clay falls \(15.0 \mathrm{~m}\) to the ground. It is in contact with the ground for \(20.0 \mathrm{~ms}\) before stopping. (a) What is the magnitude of the average acceleration of the ball during the time it is in contact with the ground? (Treat the ball as a particle.) (b) Is the average acceleration up or down?
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