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AP Physics 1: Exam Prep13 chapters | 143 lessons | 6 flashcard sets

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Lesson Transcript

Instructor:
*Angela Hartsock*

Angela has taught college Microbiology and has a doctoral degree in Microbiology.

There is a lot of information you can determine by looking at a velocity vs. time graph. In this lesson, we will use a little geometry to calculate the displacement of the object represented by the graph.

I want to start this lesson with a piece of advice. Ready? When doing physics math problems, always show your units. Units like meters, seconds, meters per second, and meters per second squared can give you hints as to what you need to do next, which answers are correct or incorrect, and whether you did the right math. When looking at a basic **velocity vs. time** graph of the straight-line motion of an object, you can see how the units can be helpful.

The units of velocity are meters per second; the units for time are seconds. Remember that you can use the velocity vs. time graph to calculate the acceleration of the object by dividing the change in velocity by the change in time.

Let's look at those units. Velocity divided by time equals acceleration. Meters per second divided by seconds equals meters per second squared. Of course, meters per second squared is the unit of measure for acceleration. Simply by following your units, you can tell if you did the problem right.

But we can also do other calculations with the units on a velocity vs. time graph. What if instead of dividing, we multiply? Velocity times time equals...well, meters per second times seconds equals meters. So, it follows that if we multiply the velocity by the time, we should be able to calculate the meters traveled by the object. In physics, this is called **displacement** and it represents the change in position from the starting to the ending point. The process isn't quite this simple; it's more than just multiplying a couple points. On a velocity vs. time graph, the area between the graph line and the time axis equals the object's displacement over that time period. This is the official definition, but it'll be much clearer if we look at an example.

Take a look at this velocity vs. time graph of a runner on a straight track.

We can see from the graph that the runner accelerates from a dead stop to maximum velocity of 12 meters per second then slows down to a stop. The entire race takes 12 seconds. We can determine the displacement of the runner over the 12-second race by calculating the area between the graph line and the time axis. Hopefully, you'll remember that you need the following formula to calculate the area of the triangle:

**Area = (1/2) x base x height**

For this problem you can see on the graph that the base is 12 seconds. The height is 12 meters per second. Let's plug them into the equation: *Area = (1/2) x (12s) x (12 m/s)*, which equals *72 meters*. Our runner traveled 72 meters in 12 seconds. His displacement is 72 meters.

Remember: displacement is a vector quantity, requiring both magnitude and direction. On a graph with straight-line motion, the direction is either positive - moving forward - or negative - moving backwards. For this problem, the displacement is positive, telling us the runner was moving forward, away from his starting position. This technique also works if the object is moving backwards and if the graph line is below the time axis. Just remember that if the velocity is negative, you need to keep that negative sign. You can also determine displacement if the object is traveling at a constant speed. Instead of calculating the area of the triangle, it would be a rectangle like this.

**Area = base x height**

It's time to briefly review. You can use a velocity vs. time graph to calculate the displacement of an object in straight-line motion. The area between the graph line and the time axis equals the object's displacement over that time period. The displacement can be positive if the line is above the time axis or negative if the line is below the time axis. Always remember to include the sign of the displacement because displacement is a vector quantity.

Review this lesson to learn how to use a velocity vs. time graph to determine an object's displacement.

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AP Physics 1: Exam Prep13 chapters | 143 lessons | 6 flashcard sets

- What is Kinematics? - Studying the Motion of Objects 3:29
- Scalars and Vectors: Definition and Difference 3:23
- What is Position in Physics? - Definition & Examples 4:42
- Distance and Displacement in Physics: Definition and Examples 5:26
- Speed and Velocity: Difference and Examples 7:31
- Acceleration: Definition, Equation and Examples 6:21
- Significant Figures and Scientific Notation 10:12
- Uniformly-Accelerated Motion and the Big Five Kinematics Equations 6:51
- Representing Kinematics with Graphs 3:11
- Ticker Tape Diagrams: Analyzing Motion and Acceleration 4:36
- What are Vector Diagrams? - Definition and Uses 4:20
- Using Position vs. Time Graphs to Describe Motion 4:35
- Determining Slope for Position vs. Time Graphs 6:48
- Using Velocity vs. Time Graphs to Describe Motion 4:52
- Determining Acceleration Using the Slope of a Velocity vs. Time Graph 5:07
- Velocity vs. Time: Determining Displacement of an Object 4:22
- Free Fall Physics Practice Problems 8:16
- Graphing Free Fall Motion: Showing Acceleration 5:24
- The Acceleration of Gravity: Definition & Formula 6:06
- Projectile Motion: Definition and Examples 4:58
- Projectile Motion Practice Problems 9:59
- Kinematic Equations List: Calculating Motion 5:41
- Go to AP Physics 1: Kinematics

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