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Chapter 2: Problem 5

Starting from the front door of your ranch house, you walk 60.0 \(\mathrm{m}\) due east to your windmill, and then you turn around and slowly walk 40.0 \(\mathrm{m}\) west to a bench where you sit and watch the sunrise. It takes you 28.0 s to walk from your house to the windmill and then 36.0 s to walk from the windmill to the bench. For the entire trip from your front door to the bench, what are (a) your average velocity and (b) your average speed?

Short Answer

Expert verified
(a) 0.3125 m/s east, (b) 1.5625 m/s.

Step by step solution

01

Understand the Problem

To solve this problem, we need to find two quantities: average velocity and average speed. Velocity is a vector quantity that considers both magnitude and direction, while speed is a scalar that considers only magnitude.
02

Calculate Total Displacement

Displacement is the shortest straight-line distance from the starting point to the final position. You walk 60.0 m east to the windmill and then 40.0 m west to the bench. The total displacement is calculated as:\[ \text{Displacement} = 60.0 \, m - 40.0 \, m = 20.0 \, m \] Therefore, the total displacement is 20.0 meters east.
03

Calculate Total Distance Traveled

The total distance you walked is the sum of the distance to the windmill and to the bench, regardless of direction:\[ \text{Total Distance} = 60.0 \, m + 40.0 \, m = 100.0 \, m \] Thus, the total distance is 100.0 meters.
04

Calculate Total Time

Add the two time intervals: 28.0 seconds to the windmill and 36.0 seconds to the bench. The total time for the trip is:\[ \text{Total Time} = 28.0 \, s + 36.0 \, s = 64.0 \, s \]
05

Calculate Average Velocity

Average velocity is calculated by dividing the total displacement by the total time:\[ \text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Total Time}} = \frac{20.0 \, m}{64.0 \, s} \approx 0.3125 \, m/s \] The average velocity is approximately 0.3125 m/s in the eastward direction.
06

Calculate Average Speed

Average speed is calculated by dividing the total distance traveled by the total time:\[ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{100.0 \, m}{64.0 \, s} \approx 1.5625 \, m/s \] The average speed is approximately 1.5625 m/s.

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Key Concepts

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

Displacement
When we talk about displacement, we're focusing on the change in position from the starting point to the final destination in a straight line. This is irrespective of the path taken. Think of displacement as the direct distance "as the crow flies".
In the example exercise, you start at your house, walk 60 meters east, and then turn back 40 meters west. The displacement ends up being only the difference between the east walk and the return west walk: 60 meters - 40 meters = 20 meters east.
  • This means the shortest route from start to end is 20 meters east.
  • Displacement is a vector quantity, meaning it has both magnitude (20 meters) and direction (east).
Distance Traveled
Distance traveled is all about the entire journey, regardless of direction. It鈥檚 the overall ground covered during your trip. Think of it as adding up all the steps you've taken without worrying about where you end up.
In the exercise, you walk 60 meters to the windmill and another 40 meters back to the bench. Your total distance traveled is therefore the sum of both legs of the walk: 60 meters + 40 meters = 100 meters.
  • Distance is a scalar quantity, so it doesn鈥檛 take direction into account, only the magnitude.
  • It's the total walk your feet have done, which is why it's larger than displacement.
Velocity
Velocity gives you the rate at which you change your position, accounting for direction. Average velocity helps describe the entire motion with a single vector.To find average velocity, divide your total displacement by the total time taken. In this case, with a displacement of 20 meters east over 64 seconds, the average velocity is:\[\text{Average Velocity} = \frac{20 \text{ m}}{64 \text{ s}} \approx 0.3125 \text{ m/s east}\]
  • Velocity is a vector, so it includes direction. Here, it's 0.3125 m/s east.
  • The direction helps in understanding the kind of motion involved 鈥� whether it鈥檚 moving forward, backward, etc.
Speed
Speed measures how fast an object is moving along a path, regardless of direction. It鈥檚 the measure of distance over time.Average speed is found by dividing the total distance by the total time taken. Here, covering a distance of 100 meters in 64 seconds results in:\[\text{Average Speed} = \frac{100 \text{ m}}{64 \text{ s}} \approx 1.5625 \text{ m/s}\]
  • Speed is a scalar quantity, meaning direction isn't part of the calculation.
  • Unlike velocity, speed tells us only "how much ground" you've covered per second.

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Most popular questions from this chapter

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 1.00 s later. You may ignore air resistance. (a) If the height of the building is \(20.0 \mathrm{m},\) what must the initial speed of the first ball be if both are to hit the ground at the same time? On the same graph, sketch the position of each ball as a function of time, measured from when the first ball is thrown. Consider the same situation, but now let the initial speed \(v_{0}\) of the first ball be given and treat the height \(h\) of the building as an unknown. (b) What must the height of the building be for both balls to reach the ground at the same time (i) if \(v_{0}\) is 6.0 \(\mathrm{m} / \mathrm{s}\) and (ii) if \(v_{0}\) is 9.5 \(\mathrm{m} / \mathrm{s} ?\) (c) If \(v_{0}\) is greater than some value \(v_{\text { max }},\) a value of \(h\) does not exist that allows both balls to hit the ground at the same time. Solve for \(v_{\text { max. }}\) . The value \(v_{\text { max }}\) has a simple physical interpretation. What is it? (d) If \(v_{0}\) is less than some value \(v_{\text { min }}\) , a value of \(h\) does not exist that allows both balls to hit the ground at the same time. Solve for \(v_{\text { min. }}\) The value \(v_{\min }\) also has a simple physical interpretation. What is it?

Dan gets on Interstate Highway \(\mathrm{I}-80\) at Seward, Nebraska, and drives due west in a straight line and at an average velocity of magnitude 88 \(\mathrm{km} / \mathrm{h}\) . After traveling 76 \(\mathrm{km}\) , he reaches the Aurora exit (Fig. \(\mathrm{P} 2.63 ) .\) Realizing he has gone too far, he turns around and drives due east 34 \(\mathrm{km}\) back to the York exit at an average velocity of magnitude 72 \(\mathrm{km} / \mathrm{h}\) . For his whole trip from Seward to the York exit, what are (a) his average speed and (b) the magnitude of his average velocity?

From Pillar to Post. Starting from a pillar, you run 200 \(\mathrm{m}\) east (the \(+x\) -direction) at an average speed of \(5.0 \mathrm{m} / \mathrm{s},\) and then run 280 \(\mathrm{m}\) west at an average speed of 4.0 \(\mathrm{m} / \mathrm{s}\) to a post. Calculate (a) your average speed from pillar to post and (b) your average velocity from pillar to post.

A Simple Reaction-Time Test. A meter stick is held vertically above your hand, with the lower end between your thumb and first finger. On seeing the meter stick released, you grab it with these two fingers. You can calculate your reaction time from the distance the meter stick falls, read directly from the point where your fingers grabbed it. (a) Derive a relationship for your reaction time in terms of this measured distance, \(d\) (b) If the measured distance is \(17.6 \mathrm{cm},\) what is the reaction time?

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.
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