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Chapter 2: Problem 4

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

Calculate Total Distance

First, calculate the total distance traveled. You ran 200 meters east and then 280 meters west. Therefore, the total distance traveled is:\[\text{Total distance} = 200 \text{ m} + 280 \text{ m} = 480 \text{ m}\]
02

Calculate Total Time

Next, calculate the total time for the entire journey. Use the formula \( \text{time} = \frac{\text{distance}}{\text{speed}} \). For the first part of your run, time is:\[\text{Time}_{\text{east}} = \frac{200 \text{ m}}{5.0 \text{ m/s}} = 40 \text{ s}\]For the second part of your run, time is:\[\text{Time}_{\text{west}} = \frac{280 \text{ m}}{4.0 \text{ m/s}} = 70 \text{ s}\]Thus, the total time taken is:\[\text{Total time} = 40 \text{ s} + 70 \text{ s} = 110 \text{ s}\]
03

Calculate Average Speed

The average speed is calculated by dividing the total distance by the total time. Thus:\[\text{Average speed} = \frac{480 \text{ m}}{110 \text{ s}} \approx 4.36 \text{ m/s}\]
04

Determine Displacement

Displacement is defined as the change in position. You ran 200 meters east and then 280 meters west, so the total displacement is:\[\text{Displacement} = 200 \text{ m} - 280 \text{ m} = -80 \text{ m}\]The negative sign indicates a final position west of the starting point.
05

Calculate Average Velocity

Average velocity is the displacement divided by the total time. Using the previously calculated displacement and total time:\[\text{Average velocity} = \frac{-80 \text{ m}}{110 \text{ s}} \approx -0.73 \text{ m/s}\]The negative sign signifies that the direction is towards the west.

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Key Concepts

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

Average Velocity
In the world of motion, average velocity serves as a vital concept. It measures how quickly an object changes its position and in what direction. Unlike average speed, which considers only the total distance covered, average velocity accounts for the direction as well.
Consider the exercise where you run from a pillar to a post. You start by running 200 meters east, then turn back and head 280 meters west. This entire journey gives you a displacement of -80 meters, since you end up 80 meters west of your starting point.
To find the average velocity, use the formula:
  • Average velocity = Displacement / Total time
  • Here, Displacement = -80 meters and Total time = 110 seconds
So, your average velocity becomes \[-0.73 \, \text{m/s}\].
The negative sign is crucial as it indicates the direction, which is towards the west in this case. Average velocity not only tells you about speed but also incorporates the direction of your journey, offering a comprehensive picture of your motion from start to finish.
Displacement
Displacement might sound like a complex term, but it's quite straightforward. It's all about the change in position from the start point to the end point. Think of displacement as the shortest path between two points, complete with a direction.
In the scenario of running from the pillar to the post, you ran 200 meters to the east and then 280 meters back to the west. The displacement isn't the total distance you travel, instead, it's the net change in position.
Your initial move 200 meters east gets offset when you turn back and head 280 meters west. This leaves you with a displacement:
  • Displacement = 200 m (east) - 280 m (west)
  • Displacement = -80 m
The negative sign signifies that you end up west of your starting point. Displacement considers only the direct distance between start and end points, making it a vector quantity due to its directional component.
Distance-Time Relationship
Understanding the relationship between distance and time is essential as it lays the foundation for concepts like speed and velocity. When you travel, distance is the total path you cover, regardless of direction.
In the pillar-to-post exercise, you cover 480 meters total (200 m east + 280 m west). Time plays a crucial role here, as each segment of the journey involves calculating how long it takes to cover the distance at given speeds.
To find the time, use the formula:
  • Time = Distance / Speed
If you ran 200 meters at 5.0 m/s, it takes you 40 seconds. For the 280 meters at 4.0 m/s, it takes 70 seconds, totaling up to 110 seconds.
The average speed, then, uses this total distance over total time concept:
  • Average Speed = Total Distance / Total Time
  • = 480 m / 110 s 鈮� 4.36 m/s
The distance-time relationship not only influences speed but serves as a basis for further analysis of motion, helping to estimate and optimize travel times in practical scenarios.

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Most popular questions from this chapter

Two cars, \(A\) and \(B,\) travel in a straight line. The dis tance of \(A\) from the starting point is given as a function of time by \(x_{A}(t)=\alpha t+\beta t^{2},\) with \(\alpha=2.60 \mathrm{m} / \mathrm{s}\) and \(\beta=1.20 \mathrm{m} / \mathrm{s}^{2} .\) The distance of \(B\) from the starting point is \(x_{B}(t)=\gamma t^{2}-\delta t^{3},\) with \(\gamma=2.80 \mathrm{m} / \mathrm{s}^{2}\) and \(\delta=0.20 \mathrm{m} / \mathrm{s}^{3} .\) (a) Which car is ahead just after they leave the starting point? (b) At what time(s) are the cars at the same point? (c) At what time(s) is the distance from \(A\) to \(B\) neither increasing nor decreasing? (d) At what time(s) do \(A\) and \(B\) have the same acceleration?

A jet fighter pilot wishes to accelerate from rest at a constant acceleration of 5\(g\) to reach Mach 3 (three times the speed of sound) as quickly as possible. Experimental tests reveal that he will black out if this acceleration lasts for more than 5.0 s. Use 331 \(\mathrm{m} / \mathrm{s}\) for the speed of sound. (a) Will the period of acceleration last long enough to cause him to black out? (b) What is the greatest speed he can reach with an acceleration of 5\(g\) before blacking out?

Large cockroaches can run as fast as 1.50 \(\mathrm{m} / \mathrm{s}\) in short bursts. Suppose you turn on the light in a cheap motel and see one scurrying directly away from you at a constant 1.50 \(\mathrm{m} / \mathrm{s} .\) If you start 0.90 m behind the cockroach with an initial speed of 0.80 \(\mathrm{m} / \mathrm{s}\) toward it, what minimum constant acceleration would you need to catch up with it when it has traveled \(1.20 \mathrm{m},\) just short of safety under a counter?

\(\mathrm{A}\) certain volcano on earth can eject rocks vertically to a maximum height \(H .\) (a) How high (in terms of \(H\) ) would these rocks go if a volcano on Mars ejected them with the same initial velocity? The acceleration due to gravity on Mars is \(3.71 \mathrm{m} / \mathrm{s}^{2},\) and you can neglect air resistance on both planets. (b) If the rocks are in the air for a time \(T\) on earth, for how long (in terms of \(T\)) will they be in the air on Mars?

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?
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