/*! 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 \((\mathrm{d}) 4 \mathrm{~s}\). (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
(a) 0 m, (b) -2 m, (c) 0 m, (d) 12 m, (e) 12 m, (f) 7 m/s. The secant line on the graph from t = 2s to t = 4s shows average velocity.

Step by step solution

01

Evaluate position at t = 1s

To find the position of the object at \( t = 1 \)s, substitute \( t = 1 \) into the equation \( x = 3t - 4t^2 + t^3 \). \( x = 3(1) - 4(1)^2 + (1)^3 = 3 - 4 + 1 = 0 \). Therefore, the position at \( t = 1 \)s is 0 meters.
02

Evaluate position at t = 2s

Substitute \( t = 2 \) into the position equation:\( x = 3(2) - 4(2)^2 + (2)^3 = 6 - 16 + 8 = -2 \).The position at \( t = 2 \)s is -2 meters.
03

Evaluate position at t = 3s

Substitute \( t = 3 \) into the position equation:\( x = 3(3) - 4(3)^2 + (3)^3 = 9 - 36 + 27 = 0 \).The position at \( t = 3 \)s is 0 meters.
04

Evaluate position at t = 4s

Substitute \( t = 4 \) into the position equation:\( x = 3(4) - 4(4)^2 + (4)^3 = 12 - 64 + 64 = 12 \).The position at \( t = 4 \)s is 12 meters.
05

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

Displacement is the change in position from \( t = 0 \) to \( t = 4 \). First, find the position at \( t = 0 \):\( x = 3(0) - 4(0)^2 + (0)^3 = 0 \).Position at \( t = 0 \) is 0 meters, and at \( t = 4 \), it's 12 meters.So, displacement = 12 - 0 = 12 meters.
06

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

Average velocity is given by the total displacement divided by the total time.From \( t = 2 \) to \( t = 4 \), the position changes from -2 meters to 12 meters.Displacement = 12 - (-2) = 14 meters.Time interval = 4s - 2s = 2s.Average velocity = \( \frac{14}{2} = 7 \) m/s.
07

Graph x versus t and indicate average velocity

Plot the graph using the points: \((0,0), (1,0), (2,-2), (3,0), (4,12)\). The curve through these points shows the motion of the object.To find the average velocity from \( t = 2 \) to \( t = 4 \) on the graph, draw a secant line that passes through the points \((2,-2)\) and \((4,12)\). The slope of this line represents the average velocity, calculated as slope = \( \frac{12 - (-2)}{4-2} = 7 \) m/s.

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Key Concepts

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

Average Velocity
When we talk about average velocity, it's about understanding how fast an object is moving over a specified time interval. Unlike instantaneous velocity, which tells you how fast an object is moving at a particular moment, average velocity looks at the whole journey. It is calculated by taking the total displacement of the object and dividing by the total time taken.
This gives us a simple formula: \[\text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Time Interval}}\]For instance, in this exercise, we considered a time interval between 2 seconds and 4 seconds. The displacement was found to be 14 meters (from -2 meters at 2s to 12 meters at 4s). With a time interval of 2 seconds, the average velocity was calculated as:\[\text{Average Velocity} = \frac{14 \text{ meters}}{2 \text{ seconds}} = 7 \text{ m/s}\]This value of 7 m/s represents how fast, on average, the object was moving during the time interval. Average velocity provides a broad overview of an object's motion over a timeframe.
Graphical Analysis
Graphical analysis involves interpreting motion through graphs, which is a powerful tool for visualizing how objects move. This exercise specifically asked us to graph the position, \(x\), of the object over time, \(t\).
Plotting points as given in the exercise: - at \(t = 0\), \(x = 0\)- at \(t = 1\), \(x = 0\)- at \(t = 2\), \(x = -2\)- at \(t = 3\), \(x = 0\)- and at \(t = 4\), \(x = 12\)The curve drawn through these points represents the object's motion over time.
One of the useful aspects of such a graph is the ability to visualize average velocity. In the graph, a secant line can be drawn from the point at \(t = 2\) (where \(x = -2\)) to \(t = 4\) (where \(x = 12\)). The slope of this secant line represents the average velocity during that time span of 7 m/s.
Graphical analysis not only helps in understanding motion but also in verifying calculations done algebraically.
Kinematic Equations
Kinematic equations allow us to describe the motion of objects, providing insights into their positions, velocities, and accelerations over time. For the exercise given, the position of the object as a function of time is described by a polynomial equation:\[x = 3t - 4t^2 + t^3\]This equation distilled the various influences acting on the object's movement, like initial position and acceleration. By substituting different values of \(t\) (time), the position \(x\) is calculated, offering a complete story of motion at any given instance.
Kinematic equations are particularly useful because they allow for the prediction of future motion when the initial conditions are known. In this exercise, substituting \(t = 1, 2, 3, 4\) into the equation showed how the position changed at those instances, painting a dynamic picture of trajectory.
  • At \(t = 1\): position was \(0\)
  • At \(t = 4\): position was \(12\)
  • Displacement can thus be calculated by comparing the initial and final positions
Understanding and using kinematic equations is fundamental in physics to predict how objects behave under various forces.

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Most popular questions from this chapter

A stone is dropped into a river from a bridge \(43.9 \mathrm{~m}\) 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.

When a high-speed passenger train traveling at \(161 \mathrm{~km} / \mathrm{h}\) rounds a bend, the engineer is shocked to see that a locomotive has improperly entered onto the track from a siding and is a distance \(D=676 \mathrm{~m}\) ahead (Fig. 2-32). The locomotive is moving at \(29.0 \mathrm{~km} / \mathrm{h}\). The engineer of the high-speed train immediately applies the brakes. (a) What must be the magnitude of the resulting constant deceleration if a collision is to be just avoided? (b) Assume that the engineer is at \(x=0\) when, at \(t=0\), he first spots the locomotive. Sketch \(x(t)\) curves for the locomotive and high-speed train for the cases in which a collision is just avoided and is not quite avoided.

A steel ball is dropped from a building's roof and passes a window, taking \(0.125 \mathrm{~s}\) to fall from the top to the bottom of the window, a distance of \(1.20 \mathrm{~m}\). It then falls to a sidewalk and bounces back past the window, moving from bottom to top in \(0.125 \mathrm{~s}\). Assume that the upward flight is an exact reverse of the fall. The time the ball spends below the bottom of the window is \(2.00 \mathrm{~s}\). How tall is the building?

An automobile travels on a straight road for \(40 \mathrm{~km}\) at \(30 \mathrm{~km} / \mathrm{h}\). It then continues in the same direction for an- other \(40 \mathrm{~km}\) at \(60 \mathrm{~km} / \mathrm{h}\). (a) What is the average velocity of the car during the full \(80 \mathrm{~km}\) trip? (Assume that it moves in the positive \(x\) direction.) (b) What is the average speed? (c) Graph \(x\) versus \(t\) and indicate how the average velocity is found on the graph.

A ball of moist clay falls \(15.0 \mathrm{~m}\) to the ground. It is in contact with the ground for \(20.0 \mathrm{~ms}\) before stopping. (a) What is the magnitude of the average acceleration of the ball during the time it is in contact with the ground? (Treat the ball as a particle.) (b) Is the average acceleration up or down?

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