How would you calculate the average speed on a velocity/time graph?

The area under the velocity/time curve is the total displacement. If you divide that by the change in time, you will get the average velocity. Velocity is the vector form of speed. If velocity is always non-negative, then average velocity and average speed are the same. In example 2 above, the displacement of the go-kart was found to be 30m. The change in time was 7 seconds, so the average velocity was 30/7 = 4.286 m/s. Since velocity was never negative, this is also the average speed. If velocity is ever negative, you would need to make an adjustment. Speed is always positive, while velocity can be negative. The speed/time graph would be the absolute value of the velocity/time graph. When velocity is negative, the area under the curve would also be negative. To find the average speed, you'll need to change that negative area to a positive.

In the last graph, to find the area of the area under the graph couldn't we use the formula of a trapezoid?

Yeah, you can use the formula of a trapezoid Area of a trapezoid = 1/2 * sum of the parallel sides * the distance between them Area of the trapezoid = displacement = 1/2 * (7 + 3) * 6 =30 thus, the displacement = 30m

If a curve represents the motion(velocity) of a body, then when the curve slopes downwards, is it just slowing down compared to the previous velocity or is it the fact that it immediately decelerates or else is it the case as if it moves too another direction as indicated by the y-axis?

When (on a Position time-graph), the line (which represents velocity) is sloping downwards, it means that the object is moving in the opposite direction and has nothing to do with acceleration or deceleration. Acceleration and deceleration can be determined by the slope, if the velocity is decreasing, the it is decelerating and and if the speed is increasing, there is acceleration.

Is acceleration both for speed and velocity? Cuz' now I'm getting confused

When a position vs time graph is concave up, the acceleration is increasing. When a position vs time graph is concave down, the acceleration is decreasing.

Can speed decrease over time, but acceleration increase over time? If yes, how is it possible?

If you had a ball traveling to the right at a speed and then you applied a force impeding that movement that becomes bigger in magnitude. the speed of the object would decrease because of the force acting opposite its direction of motion however the acceleration of the ball would increase because of the increasing force acting upon it. This is how you can achieve the situation you described.

So you can only use average velocity to find displacement when the acceleration is constant? Which you could still do on this breaking it up where the acceleration is zero and where its 3/2 per second

You can always use average velocity to find displacement. Calculating the average becomes more difficult if acceleration is not constant.

If the triangle is stated to be between t=3s and t=7s, why was t=4 used in the Triangle Formula?

Bcuz you're trying to find the base, the base being 4 (the distance between 3 and 7).

if I have a problem like justin drives west at 20M per S then slows down to stop at a red light and there is a graph that shows his velocity over time where westword is negative and asks for the acceleration what doesthe diretions have to do with this i mean what ireally want is the slope and if it is constant or not and the velocity on the y axis so there is really no use from knowing thedirections ofhis motion

The direction tells you if his change in velocity is negative or positive.

When distance (r) between two objects is zero i.e they are touching each other, the the G force between them must be F = g X m1Xm2/r2 =gXm1Xm2/0 square = infinite That means the force by which they both attract each other is infinite, so they cannot be separated at all. How is this possible?

It's very important to understand that the distance r "between objects" is not the distance between their surfaces, but the *distance between their centers*. So when a person is standing on the earth, and we want to use that formula to find the force between them, we don't say "r = 0 because they are touching", we say "r = 6400 km, the distance between the the center of the earth and the center of the person".

Is negative acceleration still considered "high acceleration" if the graph is curving downwards steeply?

High acceleration usually refers to a large magnitude of acceleration so it doesn't matter if it is positive or negative.

Main content

Course: Physics library > Unit 1

Lesson 3: Acceleration

What are velocity vs. time graphs?

Google Classroom

How to analyze graphs that relate velocity and time to acceleration and displacement.

What does the vertical axis represent on a velocity graph?

The vertical axis represents the velocity of the object. This probably sounds obvious, but be forewarned—velocity graphs are notoriously difficult to interpret. People get so used to finding velocity by determining the slope—as would be done with a position graph—they forget that for velocity graphs the value of the vertical axis is giving the velocity.

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

Concept check: What is the velocity of the object at time $t = 4 seconds$ ‍, according to the graph above?

What does the slope represent on a velocity graph?

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

The slope of a velocity graph will be given by the following formula:

slope = \frac{rise}{run} = \frac{v_{2} - v_{1}}{t_{2} - t_{1}} = \frac{Δ v}{Δ t}

Since

\frac{Δ v}{Δ t}

is the definition of acceleration, the slope of a velocity graph must equal the acceleration of the object.

slope = acceleration

This means that when the slope is steep, the object will be changing velocity rapidly. When the slope is shallow, the object will not be changing its velocity as rapidly. This also means that if the slope is negative—directed downwards—the acceleration will be negative, and if the slope is positive—directed upwards—the acceleration will be positive.

Try sliding the dot horizontally on the example velocity graph below to see what the slope looks like for particular moments in time.

The slope of the curve is positive between the times

t = 0 s

and

t = 2 s

since the slope is directed upward. This means the acceleration is positive.

The slope of the curve is negative between

t = 2 s

and

t = 8 s

since the slope is directed downward. This means the acceleration is negative.

t = 2 s

, the slope is zero since the tangent line is horizontal. This means the acceleration is zero at that moment.

Concept check: Is the object whose motion is described by the graph above speeding up or slowing down at time $t = 4 s$ ‍?

The object is slowing down at

t = 4 s

One way to see this is by dragging the slider to

t = 4 s

. Notice that the slope of the graph is negative, which means the acceleration is negative. The velocity at

t = 4 s

is positive and has a value of

+ 3 m/s

. Since the acceleration and the velocity have opposite signs, the object must be slowing down.

Another way to see this is by recognizing that at

t = 4 s

the curve is approaching the horizontal axis. Since the horizontal axis represents

v = 0 m/s

, a curve that is heading toward the horizontal axis represents an object that is slowing down. Similarly, a curve that is heading away from the horizontal axis represents an object that is speeding up. So at

t = 7 s

in the graph above, the curve is heading away from the horizontal axis and the object's speed is increasing as the object heads in the negative direction.

What does the area under a velocity graph represent?

The area under a velocity graph represents the displacement of the object. To see why, consider the following graph of motion that shows an object maintaining a constant velocity of 6 meters per second for a time of 5 seconds.

To find the displacement during this time interval, we could use this formula

Δ x = v Δ t = (6 m/s) (5 s) = 30 m

which gives a displacement of

30 m

Now we're going to show that this was equivalent to finding the area under the curve. Consider the rectangle of area made by the graph as seen below.

The area of this rectangle can be found by multiplying height of the rectangle, 6 m/s, times its width, 5 s, which would give

area = height \times width = 6 m/s \times 5 s = 30 m

This is the same answer we got before for the displacement. The area under a velocity curve, regardless of the shape, will equal the displacement during that time interval.

For an arbitrary shape, you can consider breaking the area into rectangles of very small width as seen in the graph below. The displacement during each small time interval would be given by

Δ x = v Δ t

since the velocity would be roughly constant if the time intervals are small enough. If you were to make the rectangles infinitesimally small, they would exactly represent the area under the curve. Adding up all the small areas would equal the total area and also the displacement of the object. When doing an actual problem, you never have to actually break a curve into an infinite amount of rectangles. Once you have proven to yourself that the area represents the displacement, you are free to determine the area in whichever way is most convenient.

area under curve = displacement

You would still find the area from the curve to the time axis. But since the area lies beneath the time axis, the displacement would be equal to negative the area. This makes sense since the velocity would be negative during this time which would lead to a negative displacement.

For this reason, many people think of areas that lie underneath the time axis as "negative areas under the curve" so that they remember to add the negative sign. Strictly speaking this is nonsense, since area is zero or positive by definition, but that doesn't stop people from using "negative area under the curve" as a useful mnemonic device.

What do solved examples involving velocity vs. time graphs look like?

Example 1: Windsurfing speed change

A windsurfer is traveling along a straight line, and her motion is given by the velocity graph below.

Select all of the following statements that are true about the speed and acceleration of the windsurfer.

(A) Speed is increasing.
(B) Acceleration is increasing.
(C) Speed is decreasing.
(D) Acceleration is decreasing.

Options A, speed increasing, and D, acceleration decreasing, are both true.

The slope of a velocity graph is the acceleration. Since the slope of the curve is decreasing and becoming less steep this means that the acceleration is also decreasing.

It might seem counterintuitive, but the windsurfer is speeding up for this entire graph. The value of the graph, which represents the velocity, is increasing for the entire motion shown, but the amount of increase per second is getting smaller. For the first 4.5 seconds, the speed increased from 0 m/s to about 5 m/s, but for the second 4.5 seconds, the speed increased from 5 m/s to only about 7 m/s.

Example 2: Go-kart acceleration

The motion of a go-kart is shown by the velocity vs. time graph below.

A. What was the acceleration of the go-kart at time $t = 4 s$ ‍?
B. What was the displacement of the go-kart between $t = 0 s$ ‍ and $t = 7 s$ ‍?

A. Finding the acceleration of the go-kart at $t = 4 s$ ‍

We can find the acceleration at

t = 4 s

by finding the slope of the velocity graph at

t = 4 s

slope = \frac{rise}{run}

For our two points, we'll choose the start—

3 s, 6 m/s

—and end—

7 s, 0 m/s

—of the diagonal line as points one and two respectively. Plugging these points into the formula for slope we get

slope = \frac{v_{2} - v_{1}}{t_{2} - t_{1}} = \frac{0 m/s - 6 m/s}{7 s - 3 s} = \frac{- 6 m/s}{4 s} = - 1.5 \frac{m}{s^{2}}

acceleration = - 1.5 \frac{m}{s^{2}}

B. Finding the displacement of the go-kart between $t = 0 s$ ‍ and $t = 7 s$ ‍

We can find the displacement of the go-kart by finding the area under the velocity graph. The graph can be thought of as being a rectangle (between

t = 0 s

and

t = 3 s

) and a triangle (between

t = 3 s

and

t = 7 s

). Once we find the area of these shapes and add them, we will get the total displacement.

The area of the rectangle is found by

area = h \times w = 6 m/s \times 3 s = 18 m

The area of the triangle is found by

area = \frac{1}{2} b h = \frac{1}{2} (4 s) (6 m/s) = 12 m

Adding these two areas together gives the total displacement.

total area = 18 m + 12 m = 30 m

total displacement = 30 m

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ayalaalyssa42
Posted 9 years ago. Direct link to ayalaalyssa42's post “How would you calculate t...”
How would you calculate the average speed on a velocity/time graph?
Button navigates to signup pageButton navigates to signup page
(30 votes)
Answer
- Delaine Hare
  Posted 9 years ago. Direct link to Delaine Hare's post “The area under the veloci...”
  The area under the velocity/time curve is the total displacement. If you divide that by the change in time, you will get the average velocity. Velocity is the vector form of speed. If velocity is always non-negative, then average velocity and average speed are the same.
  
  In example 2 above, the displacement of the go-kart was found to be 30m. The change in time was 7 seconds, so the average velocity was 30/7 = 4.286 m/s. Since velocity was never negative, this is also the average speed.
  
  If velocity is ever negative, you would need to make an adjustment. Speed is always positive, while velocity can be negative. The speed/time graph would be the absolute value of the velocity/time graph. When velocity is negative, the area under the curve would also be negative. To find the average speed, you'll need to change that negative area to a positive.
  Button navigates to signup page
  (47 votes)
Zhong Ze
Posted 6 years ago. Direct link to Zhong Ze's post “In the last graph, to fin...”
In the last graph, to find the area of the area under the graph couldn't we use the formula of a trapezoid?
Button navigates to signup pageComment on Zhong Ze's post “In the last graph, to fin...”
(13 votes)
Answer
- Divy Shah
  Posted 4 years ago. Direct link to Divy Shah's post “Yeah, you can use the for...”
  Yeah, you can use the formula of a trapezoid
  Area of a trapezoid = 1/2 * sum of the parallel sides * the distance between them
  Area of the trapezoid = displacement
  = 1/2 * (7 + 3) * 6
  =30
  thus, the displacement = 30m
  Button navigates to signup page
  (9 votes)
Pratham Gupta
Posted 9 years ago. Direct link to Pratham Gupta's post “When distance (r) between...”
When distance (r) between two objects is zero i.e they are touching each other, the the G force between them must be
F = g X m1Xm2/r2
=gXm1Xm2/0 square
= infinite
That means the force by which they both attract each other is infinite, so they cannot be separated at all. How is this possible?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- robshowsides
  Posted 9 years ago. Direct link to robshowsides's post “It's very important to un...”
  It's very important to understand that the distance r "between objects" is not the distance between their surfaces, but the distance between their centers. So when a person is standing on the earth, and we want to use that formula to find the force between them, we don't say "r = 0 because they are touching", we say "r = 6400 km, the distance between the the center of the earth and the center of the person".
  Comment on robshowsides's post “It's very important to un...”
  (19 votes)
Paul
Posted 4 years ago. Direct link to Paul's post “Is acceleration both for ...”
Is acceleration both for speed and velocity? Cuz' now I'm getting confused
Button navigates to signup pageComment on Paul's post “Is acceleration both for ...”
(5 votes)
Answer
- Jaden Cutinha
  Posted 3 years ago. Direct link to Jaden Cutinha's post “When a position vs time g...”
  When a position vs time graph is concave up, the acceleration is increasing. When a position vs time graph is concave down, the acceleration is decreasing.
  Button navigates to signup page
  (4 votes)
Navneet
Posted 6 years ago. Direct link to Navneet's post “If a curve represents the...”
If a curve represents the motion(velocity) of a body, then when the curve slopes downwards, is it just slowing down compared to the previous velocity or is it the fact that it immediately decelerates or else is it the case as if it moves too another direction as indicated by the y-axis?
Button navigates to signup pageButton navigates to signup page
(6 votes)
Answer
- Shreyas
  Posted 6 years ago. Direct link to Shreyas's post “When (on a Position time-...”
  When (on a Position time-graph), the line (which represents velocity) is sloping downwards, it means that the object is moving in the opposite direction and has nothing to do with acceleration or deceleration.
  
  Acceleration and deceleration can be determined by the slope, if the velocity is decreasing, the it is decelerating and and if the speed is increasing, there is acceleration.
  Button navigates to signup page
  (3 votes)
Stinky
Posted 4 years ago. Direct link to Stinky's post “Can speed decrease over t...”
Can speed decrease over time, but acceleration increase over time? If yes, how is it possible?
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- obiwan kenobi
  Posted 4 years ago. Direct link to obiwan kenobi's post “If you had a ball traveli...”
  If you had a ball traveling to the right at a speed and then you applied a force impeding that movement that becomes bigger in magnitude. the speed of the object would decrease because of the force acting opposite its direction of motion however the acceleration of the ball would increase because of the increasing force acting upon it. This is how you can achieve the situation you described.
  Button navigates to signup page
  (5 votes)
jakelumia
Posted 8 years ago. Direct link to jakelumia's post “So you can only use avera...”
So you can only use average velocity to find displacement when the acceleration is constant? Which you could still do on this breaking it up where the acceleration is zero and where its 3/2 per second
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- Andrew M
  Posted 8 years ago. Direct link to Andrew M's post “You can always use averag...”
  You can always use average velocity to find displacement.
  Calculating the average becomes more difficult if acceleration is not constant.
  Button navigates to signup page
  (5 votes)
Mariem Bakr
Posted 6 years ago. Direct link to Mariem Bakr's post “if I have a problem like ...”
if I have a problem like justin drives west at 20M per S then slows down to stop at a red light and there is a graph that shows his velocity over time where westword is negative and asks for the acceleration what doesthe diretions have to do with this i mean what ireally want is the slope and if it is constant or not and the velocity on the y axis so there is really no use from knowing thedirections ofhis motion
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Andrew M
  Posted 6 years ago. Direct link to Andrew M's post “The direction tells you i...”
  The direction tells you if his change in velocity is negative or positive.
  Comment on Andrew M's post “The direction tells you i...”
  (6 votes)
Ellen.JS.WANG
Posted a year ago. Direct link to Ellen.JS.WANG's post “Is negative acceleration ...”
Is negative acceleration still considered "high acceleration" if the graph is curving downwards steeply?
Button navigates to signup pageComment on Ellen.JS.WANG's post “Is negative acceleration ...”
(0 votes)
Answer
- Charles LaCour
  Posted a year ago. Direct link to Charles LaCour's post “High acceleration usually...”
  High acceleration usually refers to a large magnitude of acceleration so it doesn't matter if it is positive or negative.
  Comment on Charles LaCour's post “High acceleration usually...”
  (11 votes)
Gabriel Vazquez
Posted 4 years ago. Direct link to Gabriel Vazquez's post “If the triangle is stated...”
If the triangle is stated to be between t=3s and t=7s, why was t=4 used in the Triangle Formula?
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(4 votes)
Answer
- John Cena
  Posted 4 years ago. Direct link to John Cena's post “Bcuz you're trying to fin...”
  Bcuz you're trying to find the base, the base being 4 (the distance between 3 and 7).
  Button navigates to signup page
  (3 votes)

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