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Chapter 1: Problem 1

Consider a car whose position, \(s\), is given by the table $$\begin{array}{|l|l|l|l|l|l|l|} \hline t(\mathrm{~s}) & 0 & 0.2 & 0.4 & 0.6 & 0.8 & 1 \\ \hline s(\mathrm{ft}) & 0 & 0.45 & 1.7 & 3.8 & 6.5 & 9.6 \\ \hline \end{array}$$ Find the average velocity over the interval \(0 \leq t \leq 0.2\). average velocity \(=\square\) help (units) Estimate the velocity at \(t=0.2\) velocity \(=\square\) help (units)

Short Answer

Expert verified
Average velocity over \(0 \leq t \leq 0.2\) is 2.25 ft/s. Estimated velocity at \(t=0.2\) is 4.25 ft/s.

Step by step solution

01

Define Average Velocity

Average velocity is the total displacement divided by the total time taken. It can be calculated using the formula: \text{Average Velocity} = \frac{\Delta s}{\Delta t}, where \(\Delta s\) is the change in position and \(\Delta t\) is the change in time.
02

Calculate the Change in Position (\(\Delta s\))

From the table, at \(t=0\) seconds, the position \(s\) is 0 feet, and at \(t=0.2\) seconds, the position \(s\) is 0.45 feet. So, the change in position, \(\Delta s\), is: \( \Delta s = s_{0.2} - s_0 = 0.45 - 0 = 0.45 \) feet.
03

Calculate the Change in Time (\(\Delta t\))

From the table, the time interval is from \(t=0\) to \(t=0.2\) seconds. So, \(\Delta t\) is: \( \Delta t = t_{0.2} - t_0 = 0.2 - 0 = 0.2 \) seconds.
04

Compute the Average Velocity

Using the formula from Step 1, substitute the values obtained in Steps 2 and 3 to get the average velocity: \( \text{Average Velocity} = \frac{0.45 \text{ feet}}{0.2 \text{ seconds}} = 2.25 \text{ ft/s} \).
05

Estimate the Velocity at \(t=0.2\)

To estimate the instantaneous velocity at \(t=0.2\) seconds, use the positions at times just before and just after \(t=0.2\). From the table, at \(t=0\) seconds, \(s=0\) feet, and at \(t=0.4\) seconds, \(s=1.7\) feet. Approximate the velocity as: Velocity at \(t=0.2\) = \frac{s_{0.4} - s_{0.0}}{ t_{0.4} - t_{0.0}} = \frac{1.7 - 0}{ 0.4 - 0} = 4.25 \text{ ft/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 key concept in understanding motion. It's the overall change in position of an object. Unlike distance, which considers the total path traveled, displacement only looks at the starting point and the ending point, connecting them in a straight line. For instance, if a car moves from 0 feet to 0.45 feet in 0.2 seconds, its displacement is just 0.45 feet.
Displacement is a vector quantity, meaning it has both magnitude and direction. In our given problem, the magnitude of displacement is 0.45 feet, and the direction is implied along a path or one-dimensional line. This simplification helps us in calculating other related terms like velocity.
Instantaneous Velocity
Instantaneous velocity is the velocity of an object at a specific point in time. It tells you how fast something is moving right at that instant. Think of it as the speedometer reading of a car at an exact second.
To estimate instantaneous velocity, we look at positions over very small time intervals. For example, around the time of 0.2 seconds, we compare positions at times immediately before and after 0.2 seconds (such as 0.0 seconds and 0.4 seconds). By calculating the slope of the position-time graph over these small intervals, we get a good approximation of the instantaneous velocity. In our case, we find that the velocity at t=0.2 seconds is approximately 4.25 ft/s.
Change in Position
Change in position, denoted by \( \Delta s\ \), is another term for displacement and is crucial for calculating velocity. It represents how far an object has moved and in which direction over a specific period. The formula used is simple: \ \Delta s = s_{final} - s_{initial}\ \.
For example, from our table, the car's position at time t=0 is 0 ft, and at t=0.2 seconds, it's 0.45 ft. The change in position, \( \Delta s\ \), is 0.45 feet. This change in position divided by the change in time (0.2 seconds) gives us the average velocity. This concept is foundational for understanding how objects move and helps in predicting future positions and velocities.

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Most popular questions from this chapter

The table below shows the number of calories used per minute as a function of an individual's body weight for three sports: $$\begin{array}{|l|l|l|l|l|l|l|} \hline \text { Activity } & 100 \mathrm{lb} & 120 \mathrm{lb} & 150 \mathrm{lb} & 170 \mathrm{lb} & 200 \mathrm{lb} & 220 \mathrm{lb} \\ \hline \text { Walking } & 2.7 & 3.2 & 4 & 4.6 & 5.4 & 5.9 \\ \hline \text { Bicycling } & 5.4 & 6.5 & 8.1 & 9.2 & 10.8 & 11.9 \\ \hline \text { Swimming } & 5.8 & 6.9 & 8.7 & 9.8 & 11.6 & 12.7 \\ \hline \end{array}$$ a) Determine the number of calories that a 200 lb person uses in one half-hour of walking. b) Who uses more calories, a \(120 \mathrm{lb}\) person swimming for one hour, or a \(220 \mathrm{lb}\) person bicycling for a half-hour? c) Does the number of calories of a person swimming increase or decrease as weight increases?

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