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Chapter 2: Problem 22

You drive in a straight line at \(20.0 \mathrm{m} / \mathrm{s}\) for \(10.0 \mathrm{miles}\), then at \(30.0 \mathrm{m} / \mathrm{s}\) for another \(10.0 \mathrm{miles}\). (a) Is your average speed \(25.0 \mathrm{m} / \mathrm{s},\) more than \(25.0 \mathrm{m} / \mathrm{s},\) or less than \(25.0 \mathrm{m} / \mathrm{s}\) ? Explain. (b) Verify your answer to part (a) by calculating the average speed.

Short Answer

Expert verified
The average speed is less than 25 m/s, calculated as approximately 24.0 m/s.

Step by step solution

01

Understand the Problem

We need to find the average speed of a car traveling two different distances at different speeds. We are given the speeds and distances, and we need to find if the average speed is equal, more, or less than 25.0 m/s.
02

Convert Distance from Miles to Meters

To perform consistent calculations, convert miles to meters. Knowing that 1 mile = 1609.34 meters, the distance for each segment in meters is:\[10.0 \text{ miles} \times 1609.34 \text{ meters/mile} = 16093.4 \text{ meters}.\]
03

Calculate Time for Each Segment

Use the formula for time, \( t = \frac{d}{v} \), where \( d \) is the distance and \( v \) is the velocity. Calculate the time for each segment:**First Segment:**\[t_1 = \frac{16093.4 \text{ meters}}{20.0 \text{ m/s}} = 804.67 \text{ seconds}\]**Second Segment:**\[t_2 = \frac{16093.4 \text{ meters}}{30.0 \text{ m/s}} = 536.45 \text{ seconds}\]
04

Calculate Total Time and Total Distance

Add the times and distances of each segment:**Total Time:**\[t_{ ext{total}} = t_1 + t_2 = 804.67 + 536.45 = 1341.12 \text{ seconds}\]**Total Distance:**\[d_{ ext{total}} = 16093.4 + 16093.4 = 32186.8 \text{ meters}\]
05

Calculate the Average Speed

Average speed is calculated using the formula:\[v_{ ext{avg}} = \frac{\text{Total Distance}}{\text{Total Time}}\]Substitute the values:\[v_{ ext{avg}} = \frac{32186.8 \text{ meters}}{1341.12 \text{ seconds}} \approx 24.0 \text{ m/s}\]
06

Interpret the Result

The calculated average speed of approximately 24.0 m/s is less than 25.0 m/s, confirming that the average speed is indeed less than 25.0 m/s.

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Key Concepts

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

Distance Conversion
To find the average speed for a journey with segments measured in different units, it鈥檚 crucial to ensure all units are consistent. In this exercise, distances were given in miles, but calculations required meters (the standard unit for such physics problems). To convert distance from miles to meters, use the conversion factor: 1 mile equals 1609.34 meters. Thus, a 10-mile journey translates to:
  • 10.0 miles 脳 1609.34 meters/mile = 16093.4 meters.
This conversion enables more straightforward calculations with other measurements, like speed measured in meters per second (m/s). Keeping units consistent across all measurements ensures that your calculations yield correct results.
Velocity Calculation
Velocity calculation involves determining the speed at which an object moves in a specific direction. In this problem, two different velocities are given. The car travels at:
  • 20.0 m/s for the first segment of 10 miles,
  • 30.0 m/s for the second segment of 10 miles.
This disparity in speed must be considered when calculating the average speed over the entire journey. A common error is assuming that average speed is simply the mean of these two velocities. However, average speed depends on both the time spent and distance covered at each speed. It's calculated with total distance over total time instead of averaging individual speeds.
Time Calculation
To calculate the average speed, we first need to compute the time for each segment of travel. The formula for time is simple: given by dividing distance (\( d \)) by velocity (\( v \)). Apply this formula to each segment:
  • First segment: \( t_1 = \frac{16093.4 \, \text{meters}}{20.0 \, \text{m/s}} = 804.67 \, \text{seconds} \).
  • Second segment: \( t_2 = \frac{16093.4 \, \text{meters}}{30.0 \, \text{m/s}} = 536.45 \, \text{seconds} \).
Adding these gives the total time for the journey, which is critical to determine the average speed accurately. Always remember, the longer distance traveled at a slower speed affects total travel time significantly.
Unit Conversion
Unit conversion is a fundamental skill in solving physics problems. In the realm of speed and distance problems, units must often be changed so they match other units used in calculations. In our example, we started with distances in miles but needed meters to match the speed units, meters per second (m/s). Why is unit consistency important? Because formulas involving distance, time, and speed (velocity) depend on having cohesive units for accurate computations. Mismatched units lead to incorrect results. Typical conversions in physics problems include:
  • Miles to meters (1 mile = 1609.34 meters)
  • Hours to seconds (1 hour = 3600 seconds)
Mastering unit conversions will make solving physics problems smoother and more reliable.

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Most popular questions from this chapter

At the 18 th green of the U.S. Open you need to make a \(20.5-f t\) putt to win the tournament. When you hit the ball, giving it an initial speed of \(1.57 \mathrm{m} / \mathrm{s},\) it stops \(6.00 \mathrm{ft}\) short of the hole. (a) Assuming the deceleration caused by the grass is constant, what should the initial speed have been to just make the putt? (b) What initial speed do you need to make the remaining \(6.00-\mathrm{ft}\) putt?

The world's highest fountain of water is located, appropriately enough, in Fountain Hills, Arizona. The fountain rises to a height of \(560 \mathrm{ft}(5\) feet higher than the Washington Monument). (a) What is the initial speed of the water? (b) How long does it take for water to reach the top of the fountain?

Two cars drive on a straight highway. At time \(t=0,\) car 1 passes mile marker 0 traveling due east with a speed of \(20.0 \mathrm{m} / \mathrm{s} .\) At the same time, car 2 is \(1.0 \mathrm{km}\) east of mile marker 0 traveling at \(30.0 \mathrm{m} / \mathrm{s}\) due west. Car 1 is speeding up with an acceleration of magnitude \(2.5 \mathrm{m} / \mathrm{s}^{2}\), and car 2 is slowing down with an acceleration of magnitude \(3.2 \mathrm{m} / \mathrm{s}^{2}\) (a) Write \(x\) -versust equations of motion for both cars, taking east as the positive direction. (b) At what time do the cars pass next to one another?

A 747 airliner reaches its takeoff speed of \(173 \mathrm{mi} / \mathrm{h}\) in \(35.2 \mathrm{s}\). What is the magnitude of its average acceleration?

An astronaut on the Moon drops a rock straight downward from a height of \(1.25 \mathrm{m}\). If the acceleration of gravity on the Moon is \(1.62 \mathrm{m} / \mathrm{s}^{2},\) what is the speed of the rock just before it lands?
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