Representations of Motion - High School Physics

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Question

Q1

Describe the motion of the object.

Answer

This is a position versus time graph. The slope of the line of the graph tells us the velocity of the object.

For the first the object has a constant positive slope indicating that the object is traveling at a constant positive speed.

From to the object remains at the same position which means that the object has stopped.

From to the object has a constant negative slope meaning that the object is traveling at a constant negative speed.

At , the object switches directions and now has a constant positive slope meaning that the object is traveling at a constant positive speed.

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Question

Q2

Describe the motion of the object.

Answer

This is a velocity versus time graph. During the first 5 seconds, the velocity is approaching 0. This means that the object is slowing down.

During the next 5 seconds, the velocity is increasing in the negative direction. This means that the object is speeding up in the negative direction.

During the final 5 seconds, the velocity value does not change. This means that the object is remaining at a constant velocity.

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Question

Q3

Calculate the total displacement of the object.

Answer

This is a velocity versus time graph. The displacement can be determined from the area under the curve of the graph.

Begin with the first . This is a triangle with a base of and height of .

Next look at the next . This is a triangle with a base of and height of . That negative is important as it indicates that the object was traveling in the opposite direction.

For the final , the area is that of a rectangle with a base of and a height of . Again this negative is important as it indicates what direction the object was traveling in.

To find the total displacement, add all of the areas together.

Therefore the final displacement is .

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Question

Q1 6

What is the velocity of the object at 25 s?

Answer

This is a position versus time graph. To find the velocity of an object at a particular time, determine the slope of the graph at that time. Using the points and we can determine the slope of the graph at 25 seconds since it is a constant slope.

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Question

Q1 6

What is the object’s total distance traveled at ? What is the object’s displacement at ?

Answer

This is a position versus time graph. The total distance traveled is the total position that has been traveled during the entire time.

From 0 to 10 seconds, the object traveled .

From 10 seconds to 15 seconds, the object did not move.

From 15 seconds to 30 seconds the object traveled backward.

The distance is the total amount traveled.

The displacement which is the distance traveled from the original position. Since at the object has returned to the , the displacement is .

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Question

Q3

What is the acceleration of the object at ?

Answer

This is a position versus time graph. To find the velocity of an object at a particular time, determine the slope of the graph at that time. Using the points and we can determine the slope of the graph at 5 seconds since it is a constant slope.

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Question

You are given a graph of velocity vs. time. How can you find the change in displacement for a particular time interval?

Answer

Velocity is equal to a change in displacement per unit time. Consider a basic velocity vs. time graph that depicts a constant velocity, given by a straight horizontal line. How can we determine the displacement during a given time interval? Use the kinematics equation:

Acceleration will be zero, allowing us to reduce the equation. We can also assume that there is zero initial displacement.

So, the displacement will be equal to the velocity multiplied by the time. In the graph, velocity will be represented by the vertical location and time will be represented by the horizontal location of any point. Displacement will be equal to the height of the graph times the width of the graph, or the area of the rectangle created by the constant velocity over a unit of time.

Though this is just one example, displacement will always be equal to the area under the curve of a velocity vs. time graph.

In calculus terms, velocity is the derivative of the function for displacement in terms of time. To find the displacement, one must take the integral of the velocity function. The result is the area under the curve of the velocity function.

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Question

You are given a graph of displacement vs. time. Which of the following ways can be used to determine velocity at any given point?

Answer

The important thing to note is that the question asks for the velocity at any given point. The average velocity will be equal to the total displacement divided by the total time, but the question is asking for the instantaneous velocity.

Velocity is calculated by a change in displacement over a change in time. In a displacement versus time graph, this is equal to the slope.

This tells us that the slope at a certain time will be equal to the velocity at that time.

In calculus terms, velocity is the derivative of the function for displacement in terms of time. What this means is that at any point on the graph, the instantaneous slope is the velocity for that given time. To determine velocity, one must find the slope of the line at that particular time interval.

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Question

You are given a graph of velocity vs. time. Which of the following ways can be used to determine acceleration at any given point?

Answer

The important thing to note is that the question asks for the acceleration at any given point. The average acceleration will be equal to the net change in velocity divided by the total time, but the question is asking for the instantaneous acceleration.

Acceleration is calculated by a change in velocity over a change in time. In a velocity versus time graph, this is equal to the slope.

This tells us that the slope at a certain time will be equal to the acceleration at that time.

In calculus terms, acceleration is the derivative of the function for velocity in terms of time. What this means is that at any point on the graph, the instantaneous slope is the acceleration for that given time. To determine acceleration, one must find the slope of the line at that particular time interval.

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