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Chapter 2: Problem 56

On a 20 -mile bike ride, you ride the first 10 miles at an average speed of 8 \(\mathrm{mi} / \mathrm{h} .\) What must your average speed over the next 10 miles be to have your average speed for the total 20 miles be (a) 4 \(\mathrm{mi} / \mathrm{h} ?\) (b) 12 \(\mathrm{mi} / \mathrm{h} ?\) (c) Given this average speed for the first 10 miles, can you possibly attain an average speed of 16 \(\mathrm{mi} / \mathrm{h}\) for the total 20 -mile ride? Explain.

Short Answer

Expert verified
(a) 2.67 mi/h, (b) 23.81 mi/h, (c) No, 16 mi/h is unattainable.

Step by step solution

01

Understanding the Problem

We need to determine the necessary speed for the second half of a bike ride so that the overall average speed meets given criteria. We divide the problem into three parts: (a) achieving an average speed of 4 mi/h, (b) achieving 12 mi/h, and (c) determining the possibility of achieving 16 mi/h, given a constant speed initially.
02

Calculation for Average Speed of 4 mi/h

To find the desired average speed, use the formula \( \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} \).**Step A1:** Calculate the total time for 20 miles at 4 mi/h:\( \text{Total time required} = \frac{20 \text{ miles}}{4 \text{ mi/h}} = 5 \text{ hours} \).**Step A2:** Calculate time spent for the first 10 miles at 8 mi/h:\( \text{Time for 10 miles} = \frac{10 \text{ miles}}{8 \text{ mi/h}} = 1.25 \text{ hours} \).**Step A3:** Determine the time left for the remaining 10 miles:\( \text{Time left} = 5 \text{ hours} - 1.25 \text{ hours} = 3.75 \text{ hours} \).**Step A4:** Compute needed speed for the last 10 miles:\( \text{Required speed} = \frac{10 \text{ miles}}{3.75 \text{ hours}} = 2.67 \text{ mi/h} \).
03

Calculation for Average Speed of 12 mi/h

**Step B1:** Calculate the total time for 20 miles at 12 mi/h:\( \text{Total time required} = \frac{20 \text{ miles}}{12 \text{ mi/h}} \approx 1.67 \text{ hours} \).**Step B2:** Using the time for the first 10 miles previously calculated (1.25 hours), find the time left for the next 10 miles:\( \text{Time left} = 1.67 \text{ hours} - 1.25 \text{ hours} \approx 0.42 \text{ hours} \).**Step B3:** Compute the needed speed for the last 10 miles:\( \text{Required speed} = \frac{10 \text{ miles}}{0.42 \text{ hours}} \approx 23.81 \text{ mi/h} \).
04

Evaluating the Possibility of 16 mi/h Average

**Step C1:** Calculate the total time for 20 miles at 16 mi/h:\( \text{Total time required} = \frac{20 \text{ miles}}{16 \text{ mi/h}} = 1.25 \text{ hours} \).**Step C2:** Compare this with the time already spent for the first 10 miles (1.25 hours). This indicates no time is left for the second half of the journey, meaning that it is impossible to reach this speed with the given initial constraints.

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Key Concepts

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

Kinematics
Kinematics is a branch of physics that deals with the motion of objects without considering their causes. In our problem, we are interested in how different speeds over segments of a journey affect the overall average speed of a bike ride.
To understand kinematics, focus on a few core concepts:
  • Speed: How fast an object is moving, calculated as distance divided by time.
  • Average Speed: The total distance traveled divided by the total time taken for the journey.
When we talk about average speed, it's important to remember that it is concerned with the overall journey, not just individual segments. This means that even if you travel at different speeds for different parts of a trip, what ultimately matters is how far you went in total and how long it took.
This insight helps to tackle problems like figuring out what speed is needed for the second half of a bike ride to achieve a specific average for the entire trip.
Time-Distance Relationship
In physics, understanding the relationship between time and distance is crucial for solving motion problems. The formula for average speed is a great starting point for understanding this relationship:
\[\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}}\]
Let's break this down with our bike ride example. Suppose you want to achieve a specific average speed over 20 miles; the key is to manage how long you spend traveling each part of the distance.
  • For example, if it takes 1.25 hours to cover the first 10 miles at 8 mi/h, you must calculate how much time you can afford to spend on the next 10 miles and adjust your speed accordingly.
  • This calculation involves using the inverse relationship between time and speed: the faster you go, the less time it takes; conversely, the slower you go, the more time it takes.
By applying these principles, you can estimate the necessary speed adjustments required as you proceed through different sections of your journey to meet your average speed goal.
Problem-Solving in Physics
Problem-solving in physics often involves breaking a problem down into smaller, manageable parts. Carefully analyzing and following these steps can lead you to the solution more efficiently:
  • Start by understanding the problem. In our exercise, identify the goal: what average speed needs to be achieved?
  • Divide the task into parts. For the bike ride, this means considering each segment separately.
  • Apply relevant formulas. Use the average speed formula to compute required values.
For each segment, determine how the known factors (like speed and time for the first segment) influence the unknowns (the needed speed for the next segment).
Finally, evaluate the feasibility of the problem. For instance, in part c of our exercise, recognize that it's impossible to increase the average speed of 16 mi/h because the required conditions exceed limitations (i.e., zero time left for the second half).
By following this structured approach, you can tackle complex problems with greater confidence and precision.

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Most popular questions from this chapter

The acceleration of a particle is given by \(a_{x}(t)=\) \(-2.00 \mathrm{m} / \mathrm{s}^{2}+\left(3.00 \mathrm{m} / \mathrm{s}^{3}\right) t .\) (a) Find the initial velocity \(v_{0 x}\) such that the particle will have the same \(x\) -coordinate at \(t=4.00 \mathrm{s}\) as it had at \(t=0 .\) (b) What will be the velocity at \(t=4.00\) s?

A lunar lander is descending toward the moon's surface. Until the lander reaches the surface, its height above the surface of the moon is given by \(y(t)=b-c t+d t^{2},\) where \(b=800 \mathrm{m}\) is the initial height of the lander above the surface, \(c=60.0 \mathrm{m} / \mathrm{s},\) and \(d=1.05 \mathrm{m} / \mathrm{s}^{2} .\) (a) What is the initial velocity of the lander, at \(t=0 ?\) (b) What is the velocity of the lander just before it reaches the lunar surface?

Air-Bag Injuries. During an auto accident, the vehicle's air bags deploy and slow down the passengers more gently than if they had hit the windshield or steering wheel. According to safety standards, the bags produce a maximum acceleration of 60\(g\) that lasts for only 36 \(\mathrm{ms}\) (or less). How far (in meters) does a person travel in coming to a complete stop in 36 \(\mathrm{ms}\) at a constant acceleration of 60\(g?\)

An egg is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the way down and passes a point 30.0 m below its starting point 5.00 \(\mathrm{s}\) after it leaves the thrower's hand. Air resistance may be ignored. (a) What is the initial speed of the egg? (b) How high does it rise above its starting point? (c) What is the magnitude of its velocity at the highest point? (d) What are the magnitude and direction of its acceleration at the highest point? (e) Sketch \(a_{y}-t, v_{y}-t,\) and \(y-t\) graphs for the motion of the egg.

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?
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