### Object's Velocity Vs. TimeAn object moves with the velocity shown on the graph. What is the displacement of the object during the first 3.00 s?#### Graph ExplanationThe graph presented here shows the velocity (\(v\)) of an object over time (\(t\)). The vertical axis represents the velocity in meters per second (m/s), and the horizontal axis represents time in seconds (s). The graph details the velocity of the object from \(t = 0\) to \(t = 5\) seconds.1. From \(t = 0\) to \(t = 1\) second, the velocity increases linearly from 0 m/s to 2 m/s.2. From \(t = 1\) to \(t = 3\) seconds, the velocity decreases linearly from 2 m/s to -1 m/s.3. From \(t = 3\) to \(t = 5\) seconds, the velocity remains constant at -1 m/s.#### Interactive Questions1. **What is the displacement of the object during the first 3.00 s?** - [Input Box] (Submit Answer Tries 0/2)2. **What is the displacement of the object during the first 5 s?** - [Input Box] (Submit Answer Tries 0/2)To calculate the displacement, you need to find the area under the velocity-time graph for the respective time intervals. The areas represent the displacement during those intervals.- First segment (0 to 1 second): Area of the triangle = \(\frac{1}{2} \times base \times height = \frac{1}{2} \times 1 \times 2 = 1 \, \text{m}\)- Second segment (1 to 3 seconds): Area of the trapezoid can be split into that of the rectangle (1 second to 2 seconds, 2 meters) and the triangle (2 seconds to 3 seconds) - Area of the rectangle = \( base \times height = 1 \times 2 = 2 \, \text{m}\) - Area of the triangle = \( \frac{1}{2} \times base \times height = \frac{1}{2} \times 1 \times 2 \, \text{m = 1} \)

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Description: ### Object's Velocity Vs. TimeAn object moves with the velocity shown on the graph. What is the displacement of the object during the first 3.00 s?#### Graph ExplanationThe graph presented here shows the velocity (\(v\)) of an object over time (\(t\

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Object's Velocity Vs. Time An object moves with the velocity shown on the graph, What is the displacement of the object during the first 3.00 s? +2 2 4 -2 Submit Answer Tries 0/2 What is the displacement of the object during the first 5 s? Submit Answer Tries 0/2 (s/w) a t (s)

Object's Velocity Vs. Time An object moves with the velocity shown on the graph, What is the displacement of the object during the first 3.00 s? +2 2 4 -2 Submit Answer Tries 0/2 What is the displacement of the object during the first 5 s? Submit Answer Tries 0/2 (s/w) a t (s)

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Concept explainers

Displacement, Velocity and Acceleration

In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.

Linear Displacement

The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.

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Question

### Object's Velocity Vs. Time

An object moves with the velocity shown on the graph. What is the displacement of the object during the first 3.00 s?

#### Graph Explanation

The graph presented here shows the velocity (\(v\)) of an object over time (\(t\)). The vertical axis represents the velocity in meters per second (m/s), and the horizontal axis represents time in seconds (s). The graph details the velocity of the object from \(t = 0\) to \(t = 5\) seconds.

1. From \(t = 0\) to \(t = 1\) second, the velocity increases linearly from 0 m/s to 2 m/s.
2. From \(t = 1\) to \(t = 3\) seconds, the velocity decreases linearly from 2 m/s to -1 m/s.
3. From \(t = 3\) to \(t = 5\) seconds, the velocity remains constant at -1 m/s.

#### Interactive Questions

1. **What is the displacement of the object during the first 3.00 s?**
- [Input Box] (Submit Answer Tries 0/2)

2. **What is the displacement of the object during the first 5 s?**
- [Input Box] (Submit Answer Tries 0/2)

To calculate the displacement, you need to find the area under the velocity-time graph for the respective time intervals. The areas represent the displacement during those intervals.

- First segment (0 to 1 second): Area of the triangle = \(\frac{1}{2} \times base \times height = \frac{1}{2} \times 1 \times 2 = 1 \, \text{m}\)
- Second segment (1 to 3 seconds): Area of the trapezoid can be split into that of the rectangle (1 second to 2 seconds, 2 meters) and the triangle (2 seconds to 3 seconds)
- Area of the rectangle = \( base \times height = 1 \times 2 = 2 \, \text{m}\)
- Area of the triangle = \( \frac{1}{2} \times base \times height = \frac{1}{2} \times 1 \times 2 \, \text{m = 1} \)

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