Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Calculate The position of a ball as a function of time is given by $$ x=3.0 \mathrm{~m}+(-5.0 \mathrm{~m} / \mathrm{s}) t $$ What is the position of the ball at \(1.5 \mathrm{~s}\) ?

Short Answer

Expert verified
The position of the ball at 1.5 seconds is -4.5 meters.

Step by step solution

01

Identify Given Equation

The position function of the ball is given as \( x = 3.0 \, \text{m} + (-5.0 \, \text{m/s}) t \). This equation shows that the ball starts at an initial position of \( 3.0 \text{ m} \) and moves at a velocity of \( -5.0 \text{ m/s} \).
02

Substitute the Time Value

We want to find the position of the ball at \( t = 1.5 \text{ s} \). Substitute \( t = 1.5 \) into the equation: \( x = 3.0 \, \text{m} + (-5.0 \, \text{m/s}) \times 1.5 \, \text{s} \).
03

Calculate the Product

Calculate the product of the velocity and time: \( -5.0 \, \text{m/s} \times 1.5 \, \text{s} = -7.5 \, \text{m} \).
04

Solve for Position

Substitute the calculated product into the equation: \( x = 3.0 \, \text{m} - 7.5 \, \text{m} \).
05

Calculate Final Result

Perform the subtraction to find the position: \( x = 3.0 - 7.5 = -4.5 \, \text{m} \). This means the ball is 4.5 meters in the negative direction from the origin after 1.5 seconds.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Position Function
In kinematics, the position function tells us where an object is located at a specific point in time. It is typically described in the form of an equation that relates an object's position to time. For example, in the exercise, the position of the ball is given by the equation:
  • \( x = 3.0 \, \text{m} + (-5.0 \, \text{m/s}) \cdot t \)
This equation helps us determine the ball's location by substituting different time values into it. The position function commonly includes initial position and velocity, directly showing how the object's movement depends on time progression.
Velocity
Velocity is a key component in understanding motion. It signifies the rate at which the object's position changes over time and includes both magnitude and direction. In our example, the velocity of the ball is:
  • \(-5.0 \, \text{m/s}\)
This negative sign indicates that the ball is moving in the opposite direction, or backwards relative to a specified origin. This information allows us to understand not just how fast the ball is moving, but also in which direction it's headed. Thus, velocity enables us to predict future positions using the position function.
Time
Time plays a crucial role in analyzing kinematic equations. In our problem, we are interested in the ball's position at a specific time, namely 1.5 seconds. By substituting \(t = 1.5 \, \text{s}\) into the position function, we can find the ball's exact position at that moment. Time helps us track how movements unfold and keeps our analysis grounded in real-world scenarios, allowing calculations that translate abstract equations into practical outcomes.
Displacement
Displacement refers to the change in position of an object, providing insight into the object’s relative motion from one point to another. In essence, it is the difference between the initial and final position:
  • Initial position: \(3.0 \, \text{m}\)
  • Final position: \(-4.5 \, \text{m}\)
From the exercise, the displacement would be the change from 3.0 m to -4.5 m, which results in a displacement of -7.5 m. This measurement captures the ball's overall movement over time.
Equation of Motion
The equation of motion is a mathematical expression that describes an object’s position related to time, taking into account constants such as velocity and initial position. In our scenario, the equation is straightforward:
  • \( x = 3.0 \, \text{m} + (-5.0 \, \text{m/s}) \cdot t \)
This particular form of the equation of motion is useful for analyzing linear motion with constant velocity. It allows one to plug different time values into the equation to predict positions efficiently, demonstrating the relationship between time and movement in a clear, mathematical form. By understanding how each variable interplays within the equation of motion, predictions about future positions become straightforward and accurate.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In a well-known novel a person travels around the world in 80 days. (a) What is the person's approximate average speed during the adventure? (b) What is the approximate average velocity for the entire trip? (Note that Earth's circumference at the equator is \(40,075 \mathrm{~km}\).)

Analyze You and your dog go for a walk to a nearby park. On the way your dog takes many short side trips to chase squirrels, examine fire hydrants, and so on. (a) When you arrive at the park, do you and your dog have the same displacement? Explain. (b) Have you and your dog traveled the same distance? Explain.

Predict \& Explain You drive your car in a straight line at \(15 \mathrm{~m} / \mathrm{s}\) for \(10 \mathrm{~km}\), then at \(25 \mathrm{~m} / \mathrm{s}\) for another \(10 \mathrm{~km}\). (a) Is your average speed for the entire trip more than, less than, or equal to \(20 \mathrm{~m} / \mathrm{s}\) ? (b) Choose the best explanation from the following: A. More time is spent driving at \(15 \mathrm{~m} / \mathrm{s}\) than at \(25 \mathrm{~m} / \mathrm{s}\). B. The average of \(15 \mathrm{~m} / \mathrm{s}\) and \(25 \mathrm{~m} / \mathrm{s}\) is \(20 \mathrm{~m} / \mathrm{s}\). C. Less time is spent driving at \(15 \mathrm{~m} / \mathrm{s}\) than at \(25 \mathrm{~m} / \mathrm{s}\).

Challenge A finch rides on the back of a Galapagos tortoise, which walks at the stately pace of \(0.060 \mathrm{~m} / \mathrm{s}\). After \(1.2\) minutes the finch tires of the tortoise's slow pace, and it takes flight, traveling in the same direction for another \(1.2\) minutes at \(13 \mathrm{~m} / \mathrm{s}\). What was the average speed of the finch over the entire \(2.4\)-minute interval?

Is it possible for two different objects to have the same speed but different velocities?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free