/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 From Pillar to Post. Starting fr... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

Short Answer

Expert verified
Average speed: 4.36 m/s, Average velocity: -0.73 m/s.

Step by step solution

01

Understand the Concepts

Average speed is calculated as the total distance traveled divided by the total time taken. Average velocity, on the other hand, is the total displacement from the starting point to the final point divided by the total time taken.
02

Calculate Total Distance and Displacement

The total distance is the sum of all distances regardless of direction: 200 m east + 280 m west = 480 m. Displacement is the net change in position: 200 m east - 280 m west = -80 m (since west is in the opposite direction to east).
03

Calculate Total Time

First, calculate the time taken for each segment of the journey. For the 200 m east at 5.0 m/s, time = distance/speed = 200 m / 5.0 m/s = 40 s. For the 280 m west at 4.0 m/s, time = distance/speed = 280 m / 4.0 m/s = 70 s. Total time = 40 s + 70 s = 110 s.
04

Calculate Average Speed

Average speed is the total distance divided by the total time. Therefore, average speed = 480 m / 110 s ≈ 4.36 m/s.
05

Calculate Average Velocity

Average velocity is the total displacement divided by the total time. Therefore, average velocity = -80 m / 110 s ≈ -0.73 m/s.

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Key Concepts

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

Displacement
Displacement is a fundamental concept in kinematics that refers to the change in position of an object. Contrary to distance, displacement takes into account the direction of movement. It is calculated as the difference between the final and the initial positions of an object. In physics problems, displacement can be positive, negative, or zero depending on the direction of movement.

For example, during a journey from a pillar to a post, if you initially travel 200 meters east and then 280 meters west, your displacement is the final distance from the starting point considering direction: 200 meters (east) minus 280 meters (west), which results in a displacement of -80 meters. This reflects that the endpoint is 80 meters west of the starting point.

Understanding displacement is essential when solving physics problems involving vectors, as it provides insight into the net change in position.
Distance
Distance measures the total path traveled by an object without considering the direction of movement. It accumulates all the lengths of the paths covered during the motion.

In the example given, the runner first covers 200 meters towards the east and then 280 meters towards the west. The total distance is the sum of both segments: 200 meters + 280 meters = 480 meters. Distance is always positive and does not convey any information about the direction of travel.

Knowing the total distance is useful in calculating average speed and understanding the extent of travel, but it doesn't provide directional context like displacement does.
Time Calculation
Calculating time is a crucial part of solving kinematic problems. It involves determining how long an object takes to travel a certain distance given its speed.

For instance, if you know you're running 200 meters at a speed of 5.0 meters per second, you can calculate the time it takes by rearranging the formula for speed: time = distance / speed. So, for the 200 meters east, time is 40 seconds, and for the 280 meters west at 4.0 meters per second, time is 70 seconds.
  • First part: 200 m / 5.0 m/s = 40 s
  • Second part: 280 m / 4.0 m/s = 70 s
The total journey time is the sum of individual times: 40 seconds + 70 seconds = 110 seconds.

Mastering time calculations is important for analyzing motion and predicting future positions.
Physics Problem-Solving
Physics problem-solving involves applying the correct principles and formulas to understand real-world phenomena. When faced with a kinematic problem, it is essential to:

  • Identify all given variables and the unknowns you need to find.
  • Apply relevant formulas, such as those for speed, velocity, and displacement.
  • Evaluate the signs and directions of vectors, especially when dealing with displacement and velocity.
In our exercise, correctly determining average speed involved calculating total distance divided by total time, while average velocity required considering displacement. Paying attention to details like direction (east vs. west) ensures accurate solutions.

Consistent practice with these types of problems strengthens your ability to approach more complex physics scenarios with confidence.
Kinematics
Kinematics is a branch of physics that studies motion without considering the forces that cause it. It focuses on different aspects of motion such as displacement, velocity, and acceleration.

In kinematics, understanding the difference between scalar quantities (like distance and speed) that do not consider direction, and vector quantities (like displacement and velocity) that do, is crucial.
  • Scalar: Only magnitude, no direction (e.g., distance).
  • Vector: Magnitude and direction (e.g., displacement, velocity).
The exercise of running from a pillar to a post illustrates kinematic concepts by involving calculations of distance, speed, displacement, and velocity. Recognizing how these concepts interrelate helps in grasping more comprehensive physics principles, such as dynamics.

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Most popular questions from this chapter

(a) If a flea can jump straight up to a height of 0.440 \(\mathrm{m}\) , what is its initial speed as it leaves the ground? (b) How long is it in the air?

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.

The catapult of the aireraft carrier USS Abraham Lincoln accelerates an FIA-18 Homet jet fighter from rest to a takeoff speed of 173 \(\mathrm{mi} / \mathrm{h}\) in a distance of 307 \(\mathrm{ft}\) . Assume constant acceleration. (a) Calculate the acceleration of the fighter in \(\mathrm{m} / \mathrm{s}^{2} .(\mathrm{b})\) Calculate the time required for the fighter to accelerate to takeoff speed.

A \(15-k g\) rock is dropped from rest on the carth and reaches the ground in 1.75 s. When it is dropped from the same height on Saturn's satellite Enceladus, it reaches the ground in 18.6 \(\mathrm{s}\) . What is the acceleration due to gravity on Enceladus?

The position of a particle between \(t=0\) and \(t=2.00 \mathrm{s}\) is given by \(x(t)=\left(3.00 \mathrm{m} / \mathrm{s}^{3}\right) t^{3}-\left(10.0 \mathrm{m} / \mathrm{s}^{2}\right) t^{2}+(9.00 \mathrm{m} / \mathrm{s}) t\) (a) Draw the \(x-t, v_{x}-t,\) and \(a_{x}-t\) graphs of this particle. (b) At what time(s) between \(t=0\) and \(t=2.00 \mathrm{s}\) is the particle instantaneously at rest? Does your numerical result agree with the \(v_{x}-t\) graph in part \((a) ?\) At each time calculated in part \((b)\) is the acceleration of the particle positive or negative? Show that in each case the same answer is deduced from \(a_{x}(t)\) and from the \(v_{x}-t\) graph. (d) At what time(s) between \(t=0\) and \(t=2.00 \mathrm{s}\) is the velocity of the particle instantancously not changing? Locate this point on the \(v_{x} t\) and \(a_{x}-t\) graphs of part (a). (e) What is the particle's greatest distance from the origin \((x=0)\) between \(t=0\) and \(t=2.00 \mathrm{s} ?(\mathrm{f})\) At what time(s) between \(t=0\) and \(t=2.00 \mathrm{s}\) is the particle speeding \(u p\) at the greatest rate? At what time(s) between \(t=0\) and \(t=2.00 \mathrm{s}\) is the particle slowing down at the greatest rate? Locate these points on the \(v_{x}-t\) and \(a_{x}-t\) graphs of part (a).

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