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Chapter 2: Problem 66

Starting at \(x=-16 \mathrm{m}\) at time \(t=0 \mathrm{s}\) , an object takes 18 \(\mathrm{s}\) 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 \( \frac{8}{3} \, \text{m/s} \). Draw a graph from \((0, -16)\) to \((18, 32)\).

Step by step solution

01

Understanding the Problem

The problem involves an object moving in the positive x-direction, starting at \(x=-16 \, \text{m}\) at \(t=0 \, \text{s}\), and traveling 48 \(\text{m}\) in 18 \(\text{s}\). We need to graph the object's position over time and find its constant velocity.
02

Calculate the Final Position

The object starts at \(x = -16 \, \text{m}\) and moves 48 \(\text{m}\) in the positive x-direction. Hence, the final position is: \(-16 \, \text{m} + 48 \, \text{m} = 32 \, \text{m}\).
03

Create the Position-Time Table

Create a table of two points for the position-time graph: \((t_0, x_0) = (0 \, \text{s}, -16 \, \text{m})\) and \((t_f, x_f) = (18 \, \text{s}, 32 \, \text{m})\).
04

Plot the Position-Time Graph

Draw the position-time graph with time \(t\) on the x-axis and position \(x\) on the y-axis. Plot the points \((0, -16)\) and \((18, 32)\), and draw a straight line connecting them, which represents constant velocity.
05

Determine the Velocity

The velocity \(v\) is the change in position over time. Compute it using: \[ v = \frac{x_f - x_0}{t_f - t_0} = \frac{32 \, \text{m} - (-16 \, \text{m})}{18 \, \text{s} - 0 \, \text{s}} = \frac{48 \, \text{m}}{18 \, \text{s}} = \frac{8}{3} \, \text{m/s}. \]

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Key Concepts

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

Constant Velocity
When we talk about constant velocity, we refer to an object moving in a straight path without changing its speed or direction. In the exercise above, the object maintains a steady pace as it travels from start to finish within a specific time frame.
  • Constant velocity implies that the object covers equal distances in equal intervals of time.
  • This means there's no speeding up, slowing down, or changing directions; the object moves uniformly.
  • In mathematical terms, a straight line on a position-time graph represents this concept because the slope remains the same throughout.
So, in our exercise, the straight line connecting the initial and final points on the graph tells us directly that the object is moving with constant velocity.
Calculating Velocity
Calculating velocity is crucial when analyzing motion, as it tells us how fast an object is moving and in which direction. To find the object鈥檚 velocity in our scenario, we use the formula:\[ v = \frac{\Delta x}{\Delta t} \]Where:
  • \( \Delta x \) is the change in position (final position minus initial position).
  • \( \Delta t \) is the elapsed time (final time minus initial time).
For our example:
  • The initial position \( x_0 \) is \(-16 \mathrm{m} \), and the final position \( x_f \) is \(32 \mathrm{m}\). Thus, \( \Delta x = 32 \mathrm{m} - (-16 \mathrm{m}) = 48 \mathrm{m} \).
  • The time interval is \( 18 \mathrm{s} \), so \( \Delta t = 18 \mathrm{s} - 0 \mathrm{s} = 18 \mathrm{s} \).
  • So, putting these into our formula: \( v = \frac{48 \mathrm{m}}{18 \mathrm{s}} = \frac{8}{3} \mathrm{m/s} \)
This calculation shows us that the object's velocity remains consistent throughout its motion at \( \frac{8}{3} \mathrm{m/s} \).
Motion in One Dimension
Motion in one dimension refers to movement along a single straight line, which is the simplest form of motion. When discussing such motion, several critical factors come into play:
  • The position of the object, which is usually described by a coordinate, such as the x-coordinate in our exercise.
  • The time it takes for the object to move along this single axis.
  • Velocity, as determined by how far and in which direction the object travels within a set time frame.
In our given task, the object moves along the x-axis from one point to another, providing a straightforward example of one-dimensional motion. There's no change in direction or movement along any other axis, which simplifies both calculation and understanding.
  • Graphically, this is represented as a simple line in a position-time graph, where the steepness of the line indicates speed.
Thus, motion in one dimension is ideal for introductory studies of physics, focusing purely on linear movement without the complexities of additional directional changes.

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Most popular questions from this chapter

Review Conceptual Example 7 as background for this problem. A car is traveling to the left, which is the negative direction. The direction of travel remains the same throughout this problem. The car's initial speed is \(27.0 \mathrm{m} / \mathrm{s},\) and during a 5.0 \(\mathrm{s}\) - interval, it changes to a final speed of (a) 29.0 \(\mathrm{m} / \mathrm{s}\) and \((\mathrm{b}) 23.0 \mathrm{m} / \mathrm{s}\) . In each case, find the acceleration ( magnitude and algebraic sign) and state whether or not the car is decelerating.

In preparation for this problem, review Conceptual Example 7 From the top of a cliff, a person uses a slingshot to fire a pebble straight downward, which is the negative direction. The initial speed of the pebble is 9.0 \(\mathrm{m} / \mathrm{s} .\) (a) What is the acceleration (magnitude and direction) of the pebble during the downward motion? Is the pebble decelerating? Explain. (b) After 0.50 \(\mathrm{s}\) , how far beneath the cliff top is the pebble?

An 18 -year-old runner can complete a \(10.0-\mathrm{km}\) course with an average speed of 4.39 \(\mathrm{m} / \mathrm{s}\) . A 50 -year-old runner can cover the same distance with an average speed of 4.27 \(\mathrm{m} / \mathrm{s}\) . How much later (in seconds) should the younger runer start in order to finish the course at the same time as the older runner?

Two motorcycles are traveling due east with different velocities. However, four seconds later, they have the same velocity. During this four-second interval, cycle \(\mathrm{A}\) has an average acceleration of 2.0 \(\mathrm{m} / \mathrm{s}^{2}\) due east, while cycle \(\mathrm{B}\) 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?

Two soccer players start from rest, 48 \(\mathrm{m}\) apart. They run directly toward each other, both players accelerating. The first player's acceleration has a magnitude of 0.50 \(\mathrm{m} / \mathrm{s}^{2}\) . The second player's acceleration has a magnitude of 0.30 \(\mathrm{m} / \mathrm{s}^{2}\) (a) How much time passes before the players collide? (b) At the instant they collide, how far has the first player run?
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