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Chapter 2: Problem 54

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 \(\mathrm{mi} / \mathrm{h} ?(\mathrm{c})\) 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
To achieve 4 mi/h, ride next 10 miles at ~2.67 mi/h; for 12 mi/h, ~23.81 mi/h. Average 16 mi/h is impossible.

Step by step solution

01

Calculate Time for First 10 Miles

The time taken to ride the first 10 miles at an average speed of 8 mi/h is calculated using the formula \( t = \frac{d}{v} \). For the first part of the journey: \( t_1 = \frac{10}{8} = 1.25 \) hours.
02

Establish the Total Time Equation

The total time for 20 miles is the sum of the time for the first 10 miles and the time for the next 10 miles: \( T = t_1 + t_2 \), where \( t_2 \) is the time required for the next 10 miles. Our goal is to find the required average speed to meet the given conditions.
03

Calculate Time for Average Speed of 4 mi/h

To find the time when the average speed for the total journey is 4 mi/h: \( T = \frac{20}{4} = 5 \) hours. So, the time for the next 10 miles, \( t_2 = T - t_1 = 5 - 1.25 = 3.75 \) hours. The average speed required for the next 10 miles is: \( v_2 = \frac{10}{t_2} = \frac{10}{3.75} \approx 2.67 \) mi/h.
04

Calculate Time for Average Speed of 12 mi/h

When the average speed for the total ride is 12 mi/h: \( T = \frac{20}{12} \approx 1.67 \) hours. Thus, \( t_2 = T - t_1 \approx 1.67 - 1.25 = 0.42 \) hours. The required average speed over the next 10 miles is \( v_2 = \frac{10}{0.42} \approx 23.81 \) mi/h.
05

Determine Possibility of Average Speed 16 mi/h

If the average speed over 20 miles is 16 mi/h, the total time \( T = \frac{20}{16} = 1.25 \) hours. However, the time for the first 10 miles is already 1.25 hours. This means there is no remaining time for the next 10 miles, which is impossible.

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Key Concepts

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

Average Speed Calculation
Calculating the average speed involves determining the total distance traveled divided by the total time taken. It gives us an overall sense of how fast something is moving during a journey. Here鈥檚 how you calculate it:
  • First, determine the total distance covered. In our example, it鈥檚 20 miles.
  • Next, add the total time taken for the journey.
  • The average speed is then found using the formula: \( \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} \).
To illustrate, if it takes 5 hours to complete 20 miles, the average speed would be \( \frac{20}{5} = 4 \text{ mi/h} \). This value helps compare different parts of a journey or plan for specific time constraints.
Time-Distance-Velocity Relationship
The relationship between time, distance, and velocity is a fundamental concept in kinematics, essential for solving motion problems.
  • Distance: This is how far you have traveled, measured in units like miles or kilometers.
  • Velocity: It鈥檚 the speed with a direction, but for average calculations, we often focus just on speed, measured in miles per hour (mi/h) or kilometers per hour (km/h).
  • Time: The amount taken to cover a distance, measured in hours, minutes, etc.
The formula connecting these is \( t = \frac{d}{v} \). For instance, if distance \( d = 10 \) miles and velocity \( v = 8 \) mi/h, the time \( t = \frac{10}{8} = 1.25 \) hours. Understanding this relationship helps in problem-solving by rearranging the formula depending on which variable you need to find.
Velocity Problem Solving
Solving velocity problems typically involves knowing either the time, distance, or velocity, and using it to find the remaining unknown. Let鈥檚 see how this is done:
  • Use the equation \( v = \frac{d}{t} \) to solve for velocity if time and distance are known.
  • Rearrange to \( t = \frac{d}{v} \) if you need to find time but know distance and velocity.
  • If distance is unknown, rearrange to \( d = v \times t \).
Here's an example: In our exercise, knowing the average speed for a certain part of the journey (e.g., the first 10 miles at 8 mi/h) lets us solve for time. Then, using the total distance, we determine the unknown values by applying these principles to different parts of the journey.
Motion Analysis in Physics
Motion analysis is all about understanding the dynamics of objects in motion. In physics, this means applying core concepts to interpret different motion scenarios.
  • Uniform Motion: When an object moves at a constant speed, like the initial 10 miles of our journey at 8 mi/h, it simplifies analysis.
  • Non-uniform Motion: Speeds can vary, requiring more dynamic calculation. We see this in the need for different speeds to attain certain average speeds over the total distance.
  • Apply consistent logic: Use known equations and values from one part of a journey to deduce unknowns in another.
In our exercise, we evaluate total journey dynamics to understand feasible average speeds. Such analysis is crucial for real-world applications, from planning a bike trip to programming autonomous vehicles.

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Most popular questions from this chapter

Catching the Bus. A student is running at her top speed of 5.0 \(\mathrm{m} / \mathrm{s}\) to catch a bus, which is stopped at the bus stop. When the student is still 40.0 \(\mathrm{m}\) from the bus, it starts to pull away, moving with a constant acceleration of 0.170 \(\mathrm{m} / \mathrm{s}^{2}\) . (a) For how much time and what distance does the student have to run at 5.0 \(\mathrm{m} / \mathrm{s}\) before she overtakes the bus? (b) When she reaches the bus, how fast is the bus traveling? (c) Sketch an \(x-t\) graph for both the student and the bus. Take \(x=0\) at the initial position of the student. (d) The equations you used in part (a) to find the time have a second solution, corresponding to a later time for which the student and bus are again at the same place if they continue their specified motions. Explain the significance of this second solution. How fast is the bus traveling at this point? (e) If the student's top speed is 3.5 \(\mathrm{m} / \mathrm{s}\) , will she catch the bus? (t) What is the minimum speed the student must have to just catch up with the bus? For what time and what distance does she have to run in that case?

You are standing at rest at a bus stop. A bus moving at a constant speed of 5.00 m/s passes you. When the rear of the bus is 12.0 m past you, you realize that it is your bus, so you start to run toward it with a constant acceleration of 0.960 m/s. How far would you have to run before you catch up with the rear of the bus, and how fast must you be running then? Would an average college student be physically able to accomplish this?

A brick is dropped (zero initial speed) from the roof of a building. The brick strikes the ground in 2.50 s. You may ignore air resistance, so the brick is in free fall. (a) How tall, in meters, is the building? (b) What is the magnitude of the brick's velocity just before it reaches the ground? (c) Sketch \(a_{y}-t, v_{y}-t,\) and \(y-t\) graphs for the motion of the brick.

You may have noticed while driving that your car's velocity does not continue to increase, even though you keep your foot on the gas pedal. This behavior is due to air resistance and friction between the moving parts of the car. Figure 2.48 shows a qualitative \(v_{x^{-}} t\) graph for a typical car if it starts from rest at the origin and travels in a straight line (the \(x\) -axis). Sketch qualitative \(a_{x}-t\) and \(x-t\) graphs for this car.

A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by \(x(t)=b t^{2}-c t^{3},\) where \(b=240 \mathrm{m} / \mathrm{s}^{2}\) and \(c=0.120 \mathrm{m} / \mathrm{s}^{3}\) . (a) Calculate the average velocity of the car for the time interval \(t=0\) to \(t=10.0 \mathrm{s}\) . (b) Calculate the instantaneous velocity of the car at \(t=0, t=5.0 \mathrm{s},\) and \(t=10.0 \mathrm{s} .\) (c) How long after starting from rest is the car again at rest?
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