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

Short Answer

Expert verified
Average velocity: -0.3125 m/s; Average speed: 1.5625 m/s.

Step by step solution

01

Understand the Problem

We need to calculate both the average velocity and average speed for the trip described. Remember, average velocity is concerned with the net displacement over time, while average speed considers the total distance traveled over time.
02

Determine Total Displacement and Total Time

Calculate the overall displacement by finding the difference between the ending and starting positions. The starting position is the front door of the house; the ending position is the bench to the west of the starting point. Total displacement is \(-60 \, \text{m} + 40 \, \text{m} = -20 \, \text{m}\).Total time is the sum of the time for each leg of the journey,28.0 s + 36.0 s = 64.0 s.
03

Calculate Average Velocity

Average velocity is given by the formula: \(\text{Average velocity} = \frac{\text{Total displacement}}{\text{Total time}}\).Substitute the known values: \(\text{Average velocity} = \frac{-20 \, \text{m}}{64.0 \, \text{s}} = -0.3125 \, \text{m/s}\).
04

Calculate Total Distance Traveled

The total distance traveled is the sum of both distances: 60.0 m east to the windmill plus 40.0 m west to the bench, which equals 100.0 m.
05

Calculate Average Speed

Average speed is given by the formula:\(\text{Average speed} = \frac{\text{Total distance traveled}}{\text{Total time}}\).Substitute the known values: \(\text{Average speed} = \frac{100.0 \, \text{m}}{64.0 \, \text{s}} = 1.5625 \, \text{m/s}\).

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Key Concepts

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

Average Velocity
Average velocity is an important concept in kinematics. It tells you how fast you are moving in a specific direction over a period of time. It's all about displacement, which is the change in position from the start to the end of a journey.
To calculate average velocity, use the formula:
  • \( \text{Average velocity} = \frac{\text{Total displacement}}{\text{Total time}} \)
In this exercise, you start at a ranch house and end at a bench, resulting in a total displacement of -20 meters. Even though you walked 100 meters in total, the displacement is only 20 meters west (negative means westward here), because east and west cancel each other out.
Divide this displacement by the total time of 64 seconds, and you find an average velocity of -0.3125 meters per second. The negative value indicates the westward direction of the net movement.
Average Speed
Average speed, unlike average velocity, focuses on the total ground covered without considering direction. It provides the overall pace of your movement regardless of where you ended up.Calculate average speed by applying:
  • \( \text{Average speed} = \frac{\text{Total distance traveled}}{\text{Total time}} \)
Here, you walked 60 meters east and then 40 meters back west. So, the total distance is a straightforward sum of these two paths, equaling 100 meters.
Dividing 100 meters by the total time of 64 seconds gives you an average speed of 1.5625 meters per second. Unlike average velocity, average speed is always a positive number because it does not account for direction.
Displacement
Displacement is a vector quantity, which means it has both magnitude and direction. It tells you how far out of place an object is; it's the object's overall change in position.In our exercise, displacement is the net result of moving 60 meters east and then 40 meters west. Instead of worrying about the pathway taken, you only consider where you started and where you ended up.
  • Calculation: \(-60 \, \text{m} + 40 \, \text{m} = -20 \, \text{m}\)
This means you ended up 20 meters west of your starting point. The negative sign indicates that the ending point is west of the starting point.
Total Distance Traveled
Total distance traveled is a scalar quantity, focusing simply on how much ground was covered during the entire journey. It's about the actual path taken without worrying about direction. To find this, you count every meter walked during the trip:
  • You walked 60 meters east to reach the windmill.
  • Then, you walked another 40 meters west to get back to the bench.
Add these two distances together to get a total of 100 meters traveled. This total respects every step you took, embodying the complete journey, not just where you ended up relative to the start.

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Most popular questions from this chapter

A typical male sprinter can maintain his maximum acceleration for 2.0 s, and his maximum speed is 10 m/s. After he reaches this maximum speed, his acceleration becomes zero, and then he runs at constant speed. Assume that his acceleration is constant during the first 2.0 s of the race, that he starts from rest, and that he runs in a straight line. (a) How far has the sprinter run when he reaches his maximum speed? (b) What is the magnitude of his average velocity for a race of these lengths: (i) 50.0 m; (ii) 100.0 m; (iii) 200.0 m?

The driver of a car wishes to pass a truck that is traveling at a constant speed of 20.0 m/s (about 45 mi/h). Initially, the car is also traveling at 20.0 m/s, and its front bumper is 24.0 m behind the truck's rear bumper. The car accelerates at a constant 0.600 m/s\(^2\), then pulls back into the truck's lane when the rear of the car is 26.0 m ahead of the front of the truck. The car is 4.5 m long, and the truck is 21.0 m long. (a) How much time is required for the car to pass the truck? (b) What distance does the car travel during this time? (c) What is the final speed of the car?

You are standing at rest at a bus stop. A bus moving at a constant speed of \(5.00 \mathrm{~m} / \mathrm{~s}\) passes you. When the rear of the bus is \(12.0 \mathrm{~m}\) past you, you realize that it is your bus, so you start to run toward it with a constant acceleration of \(0.960 \mathrm{~m} / \mathrm{~s}^{2}\). 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?

An object's velocity is measured to be \(v_x(t) =\alpha - \beta{t}^2\), where \(\alpha\) = 4.00 m/s and \(\beta\) = 2.00 m/s\(^3\). At \(t =\) 0 the object is at \(x =\) 0. (a) Calculate the object's position and acceleration as functions of time. (b) What is the object's maximum \(positive\) displacement from the origin?

At launch a rocket ship weighs 4.5 million pounds. When it is launched from rest, it takes 8.00 s to reach 161 km/h; at the end of the first 1.00 min, its speed is 1610 km/h. (a) What is the average acceleration (in m/s\(^2\)) of the rocket (i) during the first 8.00 s and (ii) between 8.00 s and the end of the first 1.00 min? (b) Assuming the acceleration is constant during each time interval (but not necessarily the same in both intervals), what distance does the rocket travel (i) during the first 8.00 s and (ii) during the interval from 8.00 s to 1.00 min?

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