Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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: \(\text{Average speed} = \frac{100.0 \, \text{m}}{64.0 \, \text{s}} = 1.5625 \, \text{m/s}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 20 m/s (45 mi/h) when it reaches the end of the 120-m-long ramp. (a) What is the acceleration of the car? (b) How much time does it take the car to travel the length of the ramp? (c) The traffic on the freeway is moving at a constant speed of 20 m/s. What distance does the traffic travel while the car is moving the length of the ramp?

A ball is thrown straight up from the ground with speed \(v_0\). At the same instant, a second ball is dropped from rest from a height \(H\), directly above the point where the first ball was thrown upward. There is no air resistance. (a) Find the time at which the two balls collide. (b) Find the value of \(H\) in terms of \(v_0\) and \(g\) such that at the instant when the balls collide, the first ball is at the highest point of its motion.

A Honda Civic travels in a straight line along a road. The car's distance \(x\) from a stop sign is given as a function of time \(t\) by the equation \(x(t) = \alpha{t^2} - \beta{t^3}\), where \(\alpha =\) 1.50 m/s\(^2\) and \(\beta =\) 0.0500 m/s\(^3\). Calculate the average velocity of the car for each time interval: (a) \(t =\) 0 to \(t =\) 2.00 s; (b) \(t =\) 0 to \(t =\) 4.00 s; (c) \(t =\) 2.00 s to \(t =\) 4.00 s.

The human body can survive an acceleration trauma incident (sudden stop) if the magnitude of the acceleration is less than 250 m/s\(^{2}\). If you are in an automobile accident with an initial speed of 105 km/h (65 mi/h) and are stopped by an airbag that inflates from the dashboard, over what distance must the airbag stop you for you to survive the crash?

A 7500-kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.25 m/s\(^2\) and feels no appreciable air resistance. When it has reached a height of 525 m, its engines suddenly fail; the only force acting on it is now gravity. (a) What is the maximum height this rocket will reach above the launch pad? (b) How much time will elapse after engine failure before the rocket comes crashing down to the launch pad, and how fast will it be moving just before it crashes? (c) Sketch \(a_y-t, v_y-t\), and \(y-t\) graphs of the rocket's motion from the instant of blast-off to the instant just before it strikes the launch pad.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free