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Chapter 2: Problem 6

A Honda Civic travels in a straight line along a road. Its 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 \(\alpha=1.50 \mathrm{m} / \mathrm{s}^{2}\) and \(\beta=\) 0.0500 \(\mathrm{m} / \mathrm{s}^{3} .\) Calculate the average velocity of the car for each time interval: \((\mathrm{a}) t=0\) to \(t=2.00 \mathrm{s} ;\) (b) \(t=0\) to \(t=4.00 \mathrm{s}\) ; (c) \(t=2.00\) s to \(t=4.00 \mathrm{s}.\)

Short Answer

Expert verified
(a) 3 m/s; (b) 4 m/s; (c) 5 m/s.

Step by step solution

01

Understanding Average Velocity Formula

The average velocity \( v_{avg} \) over a time interval \( \Delta t = t_2 - t_1 \) is given by the change in position \( \Delta x \) over that time interval. The formula is \( v_{avg} = \frac{x(t_2) - x(t_1)}{t_2 - t_1} \), where \( x(t) \) is the position function.
02

Calculate Positions at the Given Times

Substitute the given \( \alpha \) and \( \beta \) values into the position function \( x(t) = \alpha t^2 - \beta t^3 \). Calculating: - \( x(0) = 0 \) - \( x(2) = 1.50 \times 2^2 - 0.0500 \times 2^3 = 6 \text{ m} \)- \( x(4) = 1.50 \times 4^2 - 0.0500 \times 4^3 = 16 \text{ m} \)
03

Calculate Average Velocity for t=0 to t=2.00s

Use \( v_{avg} = \frac{x(2) - x(0)}{2 - 0} \). Substituting the calculated positions:\( v_{avg} = \frac{6 - 0}{2} = 3 \text{ m/s} \).
04

Calculate Average Velocity for t=0 to t=4.00s

Use \( v_{avg} = \frac{x(4) - x(0)}{4 - 0} \). Substituting the calculated positions:\( v_{avg} = \frac{16 - 0}{4} = 4 \text{ m/s} \).
05

Calculate Average Velocity for t=2.00s to t=4.00s

Use \( v_{avg} = \frac{x(4) - x(2)}{4 - 2} \). Substituting the calculated positions:\( v_{avg} = \frac{16 - 6}{2} = 5 \text{ m/s} \).

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Key Concepts

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

Understanding the Position Function
In physics, a position function is an equation that denotes the location of an object concerning time. For the Honda Civic example, the position function is given by \( x(t) = \alpha t^2 - \beta t^3 \), where:- \( \alpha = 1.50 \text{ m/s}^2 \) is a constant acceleration factor,- \( \beta = 0.0500 \text{ m/s}^3 \) is a constant deceleration factor. This function tells us how the position \( x \) changes as time \( t \) progresses.
The function is shaped by both accelerating and decelerating terms, affecting how it graphs over time.
Recognizing Position Changes
When you want to find the change in position for an object moving along a road, you must compare its positions at two different timestamps. This change is denoted as \( \Delta x \), or how much the position of the car has shifted.In our exercise, the car's position is evaluated at different time points using the position function. Specifically, we compute positions like:
  • \( x(0) = 0 \)
  • \( x(2) = 6 \text{ m} \)
  • \( x(4) = 16 \text{ m} \)
By using these calculations, we understand how far the car has moved from one time to another.This change is crucial in determining quantities like average velocity. Essentially, it shows us the distance the car travels within specific time intervals.
Exploring Time Intervals
Time intervals are periods that help us analyze how certain variables, like position, change over time. They are critical when looking at motion problems because they tell us the duration over which changes are observed.
For instance, by calculating the average velocity of the Honda Civic, we are observing:
  • From \( t = 0 \) to \( t = 2.00 \text{ s} \)
  • From \( t = 0 \) to \( t = 4.00 \text{ s} \)
  • From \( t = 2.00 \text{ s} \) to \( t = 4.00 \text{ s} \)
The average velocity over each interval is determined by the formula: \( v_{avg} = \frac{\Delta x}{\Delta t}\) Where
\( \Delta x \) is the change in position and
\( \Delta t \) is the length of the time interval. By calculating the average velocity for different intervals, you get distinct insights into how the car behaves over short versus longer periods.

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Most popular questions from this chapter

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

A juggler throws a bowling pin straight up with an initial speed of 8.20 \(\mathrm{m} / \mathrm{s} .\) How much time elapses until the bowling pin returns to the juggler's hand?

A car's velocity as a function of time is given by \(v_{x}(t)=\alpha+\beta t^{2},\) where \(\alpha=3.00 \mathrm{m} / \mathrm{s}\) and \(\beta=0.100 \mathrm{m} / \mathrm{s}^{3} .\) (a) Calculate the average acceleration for the time interval \(t=0\) to \(t=5.00\) s. (b) Calculate the instantaneous acceleration for \(t=0\) and \(t=5.00\) s. (c) Draw \(v_{x^{-}} t\) and \(a_{x^{-}}\) graphs for the car's motion between \(t=0\) and \(t=5.00 \mathrm{s}.\)

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 turtle crawls along a straight line, which we will call the \(x\) -axis with the positive direction to the right. The equation for the turtle's position as a function of time is \(x(t)=50.0 \mathrm{cm}+\) \((2.00 \mathrm{cm} / \mathrm{s}) t-\left(0.0625 \mathrm{cm} / \mathrm{s}^{2}\right) t^{2}\) (a) Find the turtle's initial velocity, initial position, and initial acceleration. (b) At what time \(t\) is the velocity of the turtle zero? (c) How long after starting does it take the turtle to return to its starting point? (d) At what times \(t\) is the turtle a distance of 10.0 \(\mathrm{cm}\) from its starting point? What is the velocity (magnitude and direction) of the turtle at each of these times? (e) Sketch graphs of \(x\) versus \(t, v_{x}\) versus \(t,\) and \(a_{x}\) versus \(t\) for the time interval \(t=0\) to \(t=40 \mathrm{s}.\)
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