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Chapter 2: Problem 66

Starting at \(x=-16 \mathrm{m}\) at time \(t=0 \mathrm{s},\) an object takes \(18 \mathrm{s}\) to travel \(48 \mathrm{m}\) in the \(+x\) direction at a constant velocity. Make a position-time graph of the object's motion and calculate its velocity.

Short Answer

Expert verified
Velocity is approximately 2.67 m/s.

Step by step solution

01

Understand the Problem

The problem is asking for the creation of a position-time graph and calculation of velocity for an object moving from a starting position with a constant velocity.
02

Identify Given Values

The object starts at position \( x = -16 \text{ m} \) and moves in the \( +x \) direction. The time taken for the motion is \( t = 18 \text{ s} \), and the total distance traveled is \( 48 \text{ m} \).
03

Calculate Final Position

The initial position is \(-16 \text{ m}\) and it travels \(48 \text{ m}\) in the positive \(x\)-direction: \(-16 \text{ m} + 48 \text{ m} = 32 \text{ m}\). So, the final position is \( x = 32 \text{ m} \).
04

Calculate Velocity

Velocity \( v \) is given by the formula: \( v = \frac{\text{displacement}}{\text{time}} \). Displacement is \( x_{final} - x_{initial} = 32 \text{ m} - (-16 \text{ m}) = 48 \text{ m} \). Time is \( 18 \text{ s} \). Then, \( v = \frac{48 \text{ m}}{18 \text{ s}} \approx 2.67 \text{ m/s} \).
05

Set Up the Position-Time Graph

The graph will plot position on the vertical axis and time on the horizontal axis. The object starts at \(-16 \text{ m}\) at \( t = 0 \), moves with a constant velocity of \( 2.67 \text{ m/s} \), and reaches \(32 \text{ m}\) at \( t = 18 \text{ s} \).
06

Graph the Motion

Plot a straight line from the point \((-16, 0)\) to \((32, 18)\) on the graph. The slope of this line represents the object's constant velocity of \(2.67 \text{ m/s}\).

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Key Concepts

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

Constant Velocity
In physics, constant velocity is a key concept that describes motion where the speed and direction remain unchanged. When an object moves at constant velocity, it means its speed does not vary, and it travels equal distances in equal intervals of time.
For the exercise:
  • The object moves in the positive direction at a speed that does not change over the entire duration.
  • This simplifies calculations and predictions, as constant velocity means the motion can be graphed as a straight line on a position-time graph.
  • Understanding constant velocity helps in determining future positions and times, knowing the movement won't accelerate or decelerate.
Displacement
Displacement is the change in position of an object. Unlike distance, which considers only how much ground is covered, displacement accounts for the starting and ending points.
For this exercise:
  • The initial position is \(-16 \, \mathrm{m}\).
  • After traveling \(48 \, \mathrm{m}\) in the positive x direction, the final position is \(32 \, \mathrm{m}\).
  • The displacement is calculated as the final position minus the initial position: \(32 \, \mathrm{m} - (-16 \, \mathrm{m}) = 48 \, \mathrm{m}\).

Understanding displacement helps in visualizing how far the object has actually got on its path from its original position.
Velocity Calculation
Velocity is a vector quantity that includes both speed and direction. It's calculated by dividing the displacement by the time taken.
For this problem:
  • The formula used for velocity is \( v = \frac{\mathrm{displacement}}{\mathrm{time}} \).
  • Substituting the values, we get \( v = \frac{48 \, \mathrm{m}}{18 \, \mathrm{s}} \approx 2.67 \, \mathrm{m/s} \).
  • This shows that the object moves approximately \(2.67 \, \mathrm{m}\) each second in the positive x direction.

Understanding how to calculate velocity from displacement and time helps predict future positions and time durations for constant velocity contexts.
Position and Time Relationship
The position-time graph is a tool that visually displays how the position of an object changes over time. On this graph:
  • The vertical axis represents position, while the horizontal axis represents time.
  • The motion at constant velocity is represented by a straight line, indicating uniform motion.
  • The slope of the line on the position-time graph corresponds to the velocity of the object.

Drawing this graph involves plotting the initial position at time \( t = 0 \) and the final position at \( t = 18 \, \mathrm{s} \). By connecting these points with a straight line, you can visualize the object's journey and constant motion over time. This graph not only helps in understanding the current scenario but also projects how future motion would look under the same conditions.

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Most popular questions from this chapter

For each of the three pairs of positions listed in the following table, determine the magnitude and direction (positive or negative) of the displacement. $$\begin{array}{lcc} & \text { Initial position } x_{0} & \text { Final position } x \\\\\hline \text { (a) } & +2.0 \mathrm{m} & +6.0 \mathrm{m} \\\\\hline \text { (b) } & +6.0 \mathrm{m} & +2.0 \mathrm{m} \\\\\hline \text { (c) } & -3.0 \mathrm{m} & +7.0 \mathrm{m} \\\\\hline\end{array}$$

The initial velocity and acceleration of four moving objects at a given instant in time are given in the following table. Determine the final speed of each of the objects, assuming that the time elapsed since \(t=0\) s is 2.0 s. $$ \begin{array}{lcc} & \text { Initial velocity } v_{0} & \text { Acceleration } a \\\\\hline \text { (a) } & +12 \mathrm{m} / \mathrm{s} & +3.0 \mathrm{m} / \mathrm{s}^{2} \\\\\hline \text { (b) } & +12 \mathrm{m} / \mathrm{s} & -3.0 \mathrm{m} / \mathrm{s}^{2} \\\\\hline \text { (c) } & -12 \mathrm{m} / \mathrm{s} & +3.0 \mathrm{m} / \mathrm{s}^{2} \\\\\hline \text { (d) } & -12 \mathrm{m} / \mathrm{s} & -3.0 \mathrm{m} / \mathrm{s}^{2} \\ \hline\end{array}$$

Two arrows are shot vertically upward. The second arrow is shot after the first one, but while the first is still on its way up. The initial speeds are such that both arrows reach their maximum heights at the same instant, although these heights are different. Suppose that the initial speed of the first arrow is \(25.0 \mathrm{m} / \mathrm{s}\) and that the second arrow is fired \(1.20 \mathrm{s}\) after the first. Determine the initial speed of the second arrow.

Electrons move through a certain electric circuit at an average speed of \(1.1 \times 10^{-2} \mathrm{m} / \mathrm{s} .\) How long (in minutes) does it take an electron to traverse \(1.5 \mathrm{m}\) of wire in the filament of a light bulb?

Two motorcycles are traveling due east with different velocities. However, four seconds later, they have the same velocity. During this foursecond interval, cycle A has an average acceleration of \(2.0 \mathrm{m} / \mathrm{s}^{2}\) due east, while cycle B has an average acceleration of \(4.0 \mathrm{m} / \mathrm{s}^{2}\) due east. By how much did the speeds differ at the beginning of the four-second interval, and which motorcycle was moving faster?.
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