/*! 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 5 The position of an object moving... [FREE SOLUTION] | 91Ó°ÊÓ

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The position of an object moving along an \(x\) axis is given by \(x=3 t-4 t^{2}+t^{3}\), where \(x\) is in meters and \(t\) in seconds. Find the position of the object at the following values of \(t\) : (a) \(1 \mathrm{~s}\), (b) \(2 \mathrm{~s}\), (c) \(3 \mathrm{~s}\), and (d) \(4 \mathrm{~s}\). (e) What is the object's displacement between \(t=0\) and \(t=4 \mathrm{~s} ?\) (f) What is its average velocity for the time interval from \(t=2 \mathrm{~s}\) to \(t=4 \mathrm{~s} ?(\mathrm{~g})\) Graph \(x\) versus \(t\) for \(0 \leq t \leq 4 \mathrm{~s}\) and indicate how the answer for (f) can be found on the graph.

Short Answer

Expert verified
Positions are 0 m, -2 m, 0 m, 12 m for t = 1s, 2s, 3s, 4s, respectively. Displacement is 12 m; average velocity is 7 m/s.

Step by step solution

01

Calculate position for t = 1s

Substitute \( t = 1 \) seconds into the position function:\[ x = 3(1) - 4(1)^2 + (1)^3 \]Simplify:\[ x = 3 - 4 + 1 = 0 \]Thus, the position at \( t = 1 \) second is \( 0 \) meters.
02

Calculate position for t = 2s

Substitute \( t = 2 \) seconds into the position function:\[ x = 3(2) - 4(2)^2 + (2)^3 \]Simplify:\[ x = 6 - 16 + 8 = -2 \]Thus, the position at \( t = 2 \) seconds is \( -2 \) meters.
03

Calculate position for t = 3s

Substitute \( t = 3 \) seconds into the position function:\[ x = 3(3) - 4(3)^2 + (3)^3 \]Simplify:\[ x = 9 - 36 + 27 = 0 \]Thus, the position at \( t = 3 \) seconds is \( 0 \) meters.
04

Calculate position for t = 4s

Substitute \( t = 4 \) seconds into the position function:\[ x = 3(4) - 4(4)^2 + (4)^3 \]Simplify:\[ x = 12 - 64 + 64 = 12 \]Thus, the position at \( t = 4 \) seconds is \( 12 \) meters.
05

Calculate displacement from t = 0s to t = 4s

Calculate the initial position at \( t = 0 \):\[ x_0 = 3(0) - 4(0)^2 + (0)^3 = 0 \]The position at \( t = 4 \) seconds is \( 12 \) meters (from the previous step).So, the displacement \( \Delta x = x_4 - x_0 = 12 - 0 = 12 \) meters.
06

Calculate average velocity from t = 2s to t = 4s

The positions at \( t = 2 \) and \( t = 4 \) seconds are \( -2 \) meters and \( 12 \) meters, respectively.Thus, the displacement \( \Delta x = 12 - (-2) = 14 \) meters.The time interval \( \Delta t = 4 - 2 = 2 \) seconds.Average velocity \( v_{avg} = \frac{\Delta x}{\Delta t} = \frac{14}{2} = 7 \text{ m/s} \).
07

Graph the function x versus t

To graph the function, compute key points:At \( t = 0s \), \( x = 0 \); at \( t = 1s \), \( x = 0 \); at \( t = 2s \), \( x = -2 \); at \( t = 3s \), \( x = 0 \); at \( t = 4s \), \( x = 12 \).Plot these points and connect them to see the trajectory.The average velocity from \( t = 2s \) to \( t = 4s \) is represented by the slope of the line connecting these points in the graph.

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Key Concepts

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

Position function
The position function is crucial in understanding motion along a path. In kinematics, it defines the location of an object in space at any given time. For the problem presented, the position function is given as:\[ x = 3t - 4t^2 + t^3 \]where \( x \) denotes the position in meters and \( t \) is the time in seconds. This function helps us pinpoint where exactly the object is located along the x-axis at any specified time.To evaluate the object's position at different times, simply substitute the time value \( t \) into the position function and solve for \( x \). This gives you the exact position of the object at that particular moment.
Displacement
Displacement is a key concept in understanding how far an object has moved over a time period, regardless of its actual path. It is defined as the change in position from the initial point to the final point.In our example, to find the displacement from \( t = 0 \) to \( t = 4 \), calculate the initial position at \( t = 0 \):- Plugging \( t = 0 \) into the position function yields \( x_0 = 0 \)And the final position at \( t = 4 \):- Calculating at \( t = 4 \) gives \( x_4 = 12 \)Thus, the displacement can be found using:\[ \Delta x = x_4 - x_0 = 12 - 0 = 12 \text{ meters} \]The displacement is always a straight-line measure from start to finish and does not concern the path taken.
Average velocity
Average velocity is a measure of how an object's position changes over a specified time interval. It's defined as the total displacement divided by the total time taken.For the interval from \( t = 2 \) to \( t = 4 \):- The positions are \( x_2 = -2 \) meters and \( x_4 = 12 \) meters respectively.- The displacement \( \Delta x \) is \( 12 - (-2) = 14 \) meters- Time interval \( \Delta t \) is \( 4 - 2 = 2 \) secondsHence, the average velocity \( v_{avg} \) can be calculated as:\[ v_{avg} = \frac{\Delta x}{\Delta t} = \frac{14}{2} = 7 \text{ m/s} \]Always remember that average velocity considers only the initial and final positions, not the path or fluctuations between these points.
Graphing motion
Graphing motion helps visualize how an object's position changes over time. The position vs. time graph provides a clear picture of the object's journey.To create the graph for our function:- Calculate key positions at various time points: - At \( t = 0 \), \( x = 0 \) - At \( t = 1 \), \( x = 0 \) - At \( t = 2 \), \( x = -2 \) - At \( t = 3 \), \( x = 0 \) - At \( t = 4 \), \( x = 12 \)Plot these points on a graph and connect them with a smooth curve to represent the object's motion.To determine average velocity from \( t = 2 \) to \( t = 4 \) on a graph, the slope of the line connecting these points represents it. The steeper the slope, the higher the average velocity. This graphical representation enhances understanding by providing a visual link to the calculated values.

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Most popular questions from this chapter

A stone is dropped into a river from a bridge \(53.6 \mathrm{~m}\) above the water. Another stone is thrown vertically down \(1.00 \mathrm{~s}\) after the first is dropped. The stones strike the water at the same time. (a) What is the initial speed of the second stone? (b) Plot velocity versus time on a graph for each stone, taking zero time as the instant the first stone is released.

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