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### Course: Physics library > Unit 1

Lesson 2: Displacement, velocity, and time- Intro to vectors and scalars
- Introduction to reference frames
- What is displacement?
- Calculating average velocity or speed
- Solving for time
- Displacement from time and velocity example
- Instantaneous speed and velocity
- What is velocity?
- Position vs. time graphs
- What are position vs. time graphs?
- Average velocity and average speed from graphs
- Instantaneous velocity and instantaneous speed from graphs

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# What are position vs. time graphs?

See what we can learn from graphs that relate position and time.

## How are position vs. time graphs useful?

Many people feel about graphs the same way they do about going to the dentist: a vague sense of anxiety and a strong desire for the experience to be over with as quickly as possible. But position graphs can be beautiful, and they are an efficient way of visually representing a vast amount of information about the motion of an object in a conveniently small space.

## What does the vertical axis represent on a position graph?

*The vertical axis represents the position of the object.*For example, if you read the value of the graph below at a particular time you will get the position of the object in meters.

Try sliding the dot horizontally on the graph below to choose different times and see how the position changes.

$$

**Concept check: What is the position of the object at time**$t=5$ seconds according to the graph above?

## What does the slope represent on a position graph?

*The slope of a position graph represents the velocity of the object.*So the value of the slope at a particular time represents the velocity of the object at that instant.

To see why, consider the slope of the position vs. time graph shown below:

The slope of this position graph is $\text{slope}={\displaystyle \frac{\text{rise}}{\text{run}}}={\displaystyle \frac{{x}_{2}-{x}_{1}}{{t}_{2}-{t}_{1}}}$ .

This expression for slope is the same as the definition of velocity: $v={\displaystyle \frac{\mathrm{\Delta}x}{\mathrm{\Delta}t}}={\displaystyle \frac{{x}_{2}-{x}_{1}}{{t}_{2}-{t}_{1}}}$ . So the slope of a position graph has to equal the velocity.

This is also true for a position graph where the slope is changing. For the example graph of position vs. time below, the red line shows you the slope at a particular time. Try sliding the dot below horizontally to see what the slope of the graph looks like for particular moments in time.

The slope of the curve between the times $t=0\text{s}$ and $t=3\text{s}$ is positive since the slope is directed upward. This means the velocity is positive and the object is moving in the positive direction.

The slope of the curve is negative between $t=3\text{s}$ and $t=9\text{s}$ since the slope is directed downward. This means that the velocity is negative and the object is moving in the negative direction.

At $t=3\text{s}$ , the slope is zero since the line representing the slope is horizontal. This means the velocity is zero and the object is momentarily at rest.

**Concept check: What is the velocity of the object at**$t=9\text{s}$ according to the graph above?

One more thing to keep in mind is that the slope of a position graph at a given moment in time gives you the instantaneous velocity at that moment in time. The average slope between two points in time will give you the average velocity between those two points in time. The instantaneous velocity does not have to equal the average velocity. However,

*if the slope is constant*for a period of time (i.e., the graph is a straight line segment), then*the instantaneous velocity will equal the average velocity*between any two points on that line segment.## What does the curvature on a position graph mean?

Look at the graph below. It looks curvy since it's not just made out of straight line segments. If a position graph is curved, the slope will be changing, which also means the velocity is changing. Changing velocity implies acceleration. So,

*curvature in a graph means the object is accelerating, changing velocity/slope*.On the graph below, try sliding the dot horizontally to watch the slope change. The first hump between $1\text{s}$ and $5\text{s}$ represents negative acceleration since the slope goes from positive to negative. For the second hump between $7\text{s}$ and $11\text{s}$ , the acceleration is positive since the slope goes from negative to positive.

**Concept check: What is the acceleration of the object at**$t=6\text{s}$ according to the graph above?

To summarize, if the curvature of the position graph looks like an

*upside down bowl, the acceleration will be negative*. If the curvature looks like a*right side up bowl, the acceleration will be positive*. Here's a way to remember it: if your bowl is upside down all your food will fall out and that is*negative*. If your bowl is right side up, all your food will stay in it and that is*positive*.## What do solved examples involving position vs. time graphs look like?

### Example 1: Hungry walrus

The motion of a hungry walrus walking back and forth horizontally looking for food is given by the graph below, which shows the horizontal position $x$ as a function of time $t$ .

**What was the instantaneous velocity of the walrus at the following times:**$2\text{s}$ , $5\text{s}$ , and $8\text{s}$ ?

#### Finding the velocity at $2\text{s}$ :

We can find the velocity of the walrus at $t=2\text{s}$ by finding the slope of the graph at $t=2\text{s}$ :

Now we will pick two points along the line we are considering that conveniently lie at a hashmark so we can determine the value of the graph at those points. We'll choose the points $(0\text{s},1\text{m})$ and $(4\text{s},3\text{m})$ , but we could pick any two points between $0\text{s}$ and $4\text{s}$ . We must plug in the later point in time as point 2, and the earlier point in time as point 1.

So, the

*velocity of the walrus at*$2\text{s}$ was $0.5\text{m/s}$ .#### Finding the velocity at $5\text{s}$ :

To find the velocity at $5\text{s}$ , we just have to note that the graph is horizontal there. Since the graph is horizontal, the slope is equal to zero, which means that the

*velocity of the walrus at*$5\text{s}$ was $0\text{m/s}$ .#### Finding the velocity at $8\text{s}$ :

We'll pick the points at the beginning and end of the final line segment, which are $(6\text{s},3\text{m})$ and $(9\text{s},0\text{m})$ .

So, the

*velocity of the walrus at*$8\text{s}$ was $-1\text{m/s}$ .### Example 2: Happy bird

The motion of an extraordinarily jubilant bird flying straight up and down is given by the graph below, which shows the vertical position $y$ as a function of time $t$ . Answer the following questions about the motion of the bird.

**What was the average velocity of the bird between**$t=0\text{s}$ and $t=10\text{s}$ ?

**What was the average speed of the bird between**$t=0\text{s}$ and $t=10\text{s}$ ?

#### Finding the average velocity of the bird between $t=0\text{s}$ and $t=10\text{s}$ :

To find the average velocity between $t=0\text{s}$ and $t=10\text{s}$ , we can find the average slope between $t=0\text{s}$ and $t=10\text{s}$ . Visually, this would correspond to finding the slope of the line that connects the initial point and the final point on the graph.

The initial point would be $(0\text{s},7\text{m})$ , and the final point would be $(10\text{s},6\text{m})$ .

So, the

*average velocity of the bird between*$t=0\text{s}$ and $t=10\text{s}$ was $-0.1\text{m/s}$ .#### Finding the average speed of the bird between $t=0\text{s}$ and $t=10\text{s}$ :

The definition of average speed is the distance traveled divided by the time. So, to find the distance traveled, we need to add the path lengths of each leg of the trip. Between $t=0\text{s}$ and $t=2.5\text{s}$ , the bird moved $5\text{m}$ down. Then, between $t=2.5\text{s}$ and $t=5\text{s}$ , the bird did not move at all. And finally between $t=5\text{s}$ and $t=10\text{s}$ , the bird flew $4\text{m}$ upward. Adding up all the path lengths gives us a total distance traveled of $\text{distance}=9\text{m}$ .

Now we can divide by the time to get the average speed ${s}_{avg}$ :

So the

*average speed of the bird between*$t=0\text{s}$ and $t=10\text{s}$ was $0.9\text{m/s}$ .## Want to join the conversation?

- are there ways to calculate the slope of a curved graph without using calculus(25 votes)
- It turns out to be possible for the conic sections: circles, parabolas, hyperbolas, and ellipses, but I think that's about it for the functions used by most people today. You could invent or define some curves by what you want their slopes to do, and before Newton came along, people played with these a lot -- osculating curves and evolutes and such. But Newton basically INVENTED CALCULUS
*precisely because*he needed to calculate the slopes of curved graphs of given functions, and there was no way to do it.(44 votes)

- In the average velocity, why don't we simply calculate the instantaneous velocity when t=0 and when t=10, add them and divide them over 2 to get their average? It gave me a different result! what's wrong then?(3 votes)
- it's because "average velocity" in physics is different from the "average of the initial and final velocity". This is admittedly confusing, but the definition of average velocity is displacement over time. The definition of the average of the velocities is the sum of the velocities divided by two. Like Andrew said, if the acceleration was constant then it turns out these two quantities will be equal. But if acceleration was not constant you can't assume they will give the same answer.(23 votes)

- What is the meaning of negative acceleration? Since an object cannot slow down/decelerate beyond zero.

The velocity is negative when object moves in the opposite direction(the negative direction) so is negative acceleration the acceleration when the object is moving in the opposite direction(the negative direction)?(7 votes)- Yes, negative acceleration would be acceleration in the negative direction. Acceleration is a vector quantity.(9 votes)

- I have a few questions:

1. Since velocity is "Speed with given direction", and the acceleration is negative when the slope is going down, why is the velocity constant when the slope is constant? (Refer to graph 4)

2. In example 2, "The motion of an extraordinarily jubilant bird flying straight up and down is given by the graph..." states that the bird flies STRAIGHT UP AND DOWN. But why is the slope of the graph down and then up? Is the sentence just an intro?

3. If the slope is going up, the acceleration remains at a constant rate and will not increase anymore unless the slope goes even higher, the same goes for the velocity, am I right?(6 votes)- 1. Velocity is the slope of position vs. time. If that slope is not changing, the velocity is constant. If the slope is negative and not changing, the velocity is a negative constant.

Acceleration is slope of velocity vs time.

2. Yes, it's an introduction.

3. Can't tell what slope you are referring to, so can't answer.(3 votes)

- For the Hungry Walrus question, what does -1 m/s velocity mean? At first I thought that it meant the walrus was going slower, but actually the walrus went faster, right? What does the -1 imply? He was going back in the direction he came from?(2 votes)
- Yes, the (-) tells us that he is going back in the direction he came from. And yes, he is actually going faster.

At 2 s -> slope = 0.5 m/s.

At 5 s -> slope = 0 m/s.

At 8 s -> slope = -1 m/s.

At 8 s the MAGNITUDE or SIZE (aka number) for the velocity is the greatest from the three (since 1 > 0.5 > 1). Thus, he goes faster at the end.

As for the signs, we only have them to indicate direction, since VELOCITY is speed with direction. For example, if we were just calculating SPEED, which has no direction, we would not put the (-). However, since we were calculating VELOCITY, which has direction, we put the (-) because he went back in the direction he came.

Hope that helps. :)(6 votes)

- Why does the slope relate to the velocity and not the speed?(4 votes)
- As displacement is a vector quantity, the slope of dispacement-time graph should be velocity because velocity is a vector quantity as well.

(Vector quantities have a direction and magnitude)(1 vote)

- In example two, wouldn't the variable x in the formula change to y since it is measuring vertical position?(2 votes)
- How am i supposed to know the slope of a curvy graph(5 votes)

- How do you calculate the Instantaneous Velocity of a Position - Time graph when the acceleration is constant? Is there a way to do it with the graph alone (no calculus)?(2 votes)
- draw a line that is tangent to the curve at that point, and find the slope of that line(5 votes)

- when we are reffering to instantaneous velocity at a particular moment are we considering a little before and a little after it or just after it?this is really important to me because we can consider the velocity to be zero at the highest point in the graph only if we consider the time to be a little after and a little before that point. because in that case we can roughly say our object has turned it to its previous position so(if our object hasn't change it's magnitude of velocity after it has changed it's direction at t that moment).also sal seid the object is momentarily at rest but i think the object was moving at that point but is has just chand it direction.any help is really appreciated because this question has been stuck in my mind for a long time.(3 votes)
- In example two, wouldn't the variable x in the formula change to y since it is measuring vertical position?(3 votes)