I don't get why the slope is shown to be negative in the two examples, while both objects are gaining speed. Seems like the slope should be positive... What's the reasoning behind this?

These are acceleration vs time graphs. Any line ABOVE the time axis (a=0) indicates positive acceleration. and a line below the time axis indicates negative acceleration (Slowing down) BUT The slope of the lines says NOTHING about the amount of acceleration. The slope is only a measure of 'jerkiness' (or rate of change) of acceleration.

Are there quantities like acceleration of jerk, jerk of jerk and so on?

Change in jerk is called jounce or snap. Change in jounce is called crackle. Change in crackle is called pop. (Yes, the names are hilarious!) More information can be found here: https://en.wikipedia.org/wiki/Jounce https://en.wikipedia.org/wiki/Crackle_(physics) https://en.wikipedia.org/wiki/Pop_(physics)

ok, but how do you go about making a acceleration time graph?

Or you could measure acceleration at each point in time with an accelerometer.

The area under the acceleration-time graph is velocity. That under the velocity-time graph is displacement (or may be distance). What about that under the displacement-time graph, what would it be?

I believe it represents a negative derivative of displacement. This is called Absement and is essentially the "total" displacement. Essentially, the derivative of this is displacement, the "change" in Absement, and velocity would the derivative of displacement, the "change" in displacement, the acceleration being the second-order derivative, and so on. The area under the curve is the anti-derivative, and in lay terms moving upwards. For instance, the area under acceleration-time graph is the velocity, moving upwards. For reference, I located a list of the derivatives of displacement. -1. Absement 0. Displacement 1. Velocity 2. Acceleration 3. Jerk 4. Jounce (snap) 5. Crackle 6. Pop 7. Lock 8. Drop 9. Shot 10. Put

in Example 1: Race car acceleration it is said that the driver had a constant velocity of 20 m/s and again she had it at 0s. but in the graph it is shown that at 0s she had a acceleration of 6 m/s2. so my question is if the graph is mistaken or it is right. since its clear that constant velocity means 0 acceleration and she had a constant velocity of 20 m/s at 0s, so it means the acceleration at 0s would also be 0m/s2 . am i right?

It states that the constant velocity was until the driver began accelerating at the end. At time = 0, velocity was no longer constant.

How can we calculate the jerk using only the information given by a velocity-time graph ?

Take the second derivative of the v(t) graph to get J(t). J(t) = da(t)/dt = d²v(t)/dt²

last graph how did you find Vi = 10 m/s ?

It is written in the text of Example 2:"A sailboat is sailing in a straight line with a velocity of 10 m/s." Otherwise you can not know exact velocity from acceleration vs. time graph. You only know change of velocity (Δv).

In the example 1 , retardation is taking place instead of acceleration (according to the graph because the line is pointing downwards). So how can the velocity of the car after 8 seconds be more than 20m/s, similar confusion in example 2 . I don't understand at all!!

I think the slope should be rising because acceleration at time t=0 is 0 since velocity is constant............ but yes velocity can be more than 20m/s even if acceleration is decreasing ..... see the top question's answer......... but still it should be a rising slope..

Main content

Course: Physics library > Unit 1

Lesson 3: Acceleration

What are acceleration vs. time graphs?

Google Classroom

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

What does the vertical axis represent on an acceleration graph?

The vertical axis represents the acceleration of the object.

For example, if you read the value of the graph shown below at a particular time, you will get the acceleration of the object in meters per second squared for that moment.

Try sliding the dot horizontally on the graph below to choose different times, and see how the acceleration—abbreviated Acc—changes.

Concept check: According to the graph above, what is the acceleration at time

t = 4 s

What does the slope represent on an acceleration graph?

The slope of an acceleration graph represents a quantity called the jerk. The jerk is the rate of change of the acceleration.

For an acceleration graph, the slope can be found from

slope = \frac{rise}{run} = \frac{a_{2} - a_{1}}{t_{2} - t_{1}} = \frac{Δ a}{Δ t}

, as can be seen in the diagram below.

This slope, which represents the rate of change of acceleration, is defined to be the jerk.

jerk = \frac{Δ a}{Δ t}

As strange as the name jerk sounds, it fits well with what we would call jerky motion. If you were in a ride where the acceleration was increasing and decreasing significantly over short periods of time, the motion would feel jerky, and you would have to keep applying different amounts of force from your muscles to stabilize your body.

To finish up this section, let's visualize the jerk with the example graph shown below. Try moving the dot horizontally to see what the slope—i.e., jerk—looks like at different points in time.

Concept check: For the acceleration graph shown above, is the jerk positive, negative, or zero at

t = 6 s

What does the area represent on an acceleration graph?

The area under an acceleration graph represents the change in velocity. In other words, the area under the acceleration graph for a certain time interval is equal to the change in velocity during that time interval.

area = Δ v

It might be easiest to see why this is the case by considering the example graph below which shows a constant acceleration of 4

\frac{m}{s^{2}}

for a time of 9 s.

If we multiply both sides of the definition of acceleration,

a = \frac{Δ v}{Δ t}

, by the change in time,

Δ t

, we get

Δ v = a Δ t

Plugging in the acceleration 4

\frac{m}{s^{2}}

and the time interval 9 s we can find the change in velocity:

Δ v = a Δ t = (4 \frac{m}{s^{2}}) (9 s) = 36 \frac{m}{s}

Multiplying the acceleration by the time interval is equivalent to finding the area under the curve. The area under the curve is a rectangle, as seen in the diagram below.

The area can be found by multiplying height times width. The height of this rectangle is 4

\frac{m}{s^{2}}

, and the width is 9 s. So, finding the area also gives you the change in velocity.

area = 4 \frac{m}{s^{2}} \times 9 s = 36 \frac{m}{s}

The area under any acceleration graph for a certain time interval gives the change in velocity for that time interval.

Good question. We only proved that the area under an acceleration graph for a certain time interval gives the change in velocity for that time interval for the case of constant acceleration, but the area under any acceleration graph will give the change in velocity, even if the acceleration is not constant.

The reason is that the area under an acceleration graph that is changing can be broken up into small rectangles of very thin width, as seen in the diagram below. If the width of the rectangles is small enough, the acceleration will roughly be constant. Multiplying height times width will give the area and the change in velocity,

Δ v = a Δ t

, during each small interval of time. Adding up the areas of the rectangles gives roughly the total area under the curve and the total change in velocity for a given time interval.

If the width of the rectangles is taken to be infinitesimally small, the sum of the areas of all the rectangles will give exactly the area under the curve and the exact change in velocity. So the area under any acceleration graph gives the change in velocity. If this discussion sounds like mathematical witchcraft it's probably because the proper derivation of this proof requires the ideas of calculus—the integral in calculus is an operation that allows you to find areas under curves.

Note: When solving a problem or answering a question, we don't have to actually break a curve into rectangles when finding the area. Since we know that the area under the curve is equal to the change in velocity, we are free to find the area under the curve in whichever way is most convenient. In other words, for the graph shown above, we could just use the formula for the area of a triangle,

\frac{1}{2} b h

, to find the area and change in velocity.

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

Example 1: Race car acceleration

A confident race car driver is cruising at a constant velocity of 20 m/s. As she nears the finish line, the race car driver starts to accelerate. The graph shown below gives the acceleration of the race car as it starts to speed up. Assume the race car had a velocity of 20 m/s at time

t = 0 s

What is the velocity of the race car after the 8 seconds of acceleration shown in the graph?

We can find the change in velocity by finding the area under the acceleration graph.

Δ v = area = \frac{1}{2} b h = \frac{1}{2} (8 s) (6 \frac{m}{s^{2}}) = 24 m/s (Use the formula for area of triangle: \frac{1}{2} b h .)

Δ v = 24 m/s (Calculate the change in velocity.)

But this is just the change in velocity during the time interval. We need to find the final velocity. We can use the definition of the change in velocity,

Δ v = v_{f} - v_{i}

, to find that

Δ v = 24 m/s

v_{f} - v_{i} = 24 m/s (Plug in v_{f} - v_{i} for Δ v .)

v_{f} - 20 m/s = 24 m/s (Plug in 20 m/s for the initial velocity v_{i} .)

v_{f} = 24 m/s + 20 m/s (Solve for v_{f} .)

v_{f} = 44 m/s (Calculate and celebrate!)

The final velocity of the race car was 44 m/s.

Example 2: Sailboat windy ride

A sailboat is sailing in a straight line with a velocity of 10 m/s. Then at time

t = 0 s

, a stiff wind blows causing the sailboat to accelerate as seen in the diagram below.

What is the velocity of the sailboat after the wind has blown for 9 seconds?

The area under the graph will give the change in velocity. The area of the graph can be broken into a rectangle, a triangle, and a triangle, as seen in the diagram below.

The blue rectangle between

t = 0 s

and

t = 3 s

is considered positive area since it is above the horizontal axis. The green triangle between

t = 3 s

and

t = 7 s

is also considered positive area since it is above the horizontal axis. The red triangle between

t = 7 s

and

t = 9 s

, however, is considered negative area since it is below the horizontal axis.

We'll add these areas together—using

h w

for the rectangle and

\frac{1}{2} b h

for the triangles—to get the total area between

t = 0 s

and

t = 9 s

Δ v = area = (4 \frac{m}{s^{2}}) (3 s) + \frac{1}{2} (4 s) (4 \frac{m}{s^{2}}) + \frac{1}{2} (2 s) (- 2 \frac{m}{s^{2}}) (Add areas of rectangle and two triangles.)

Δ v = 18 m/s (Calculate to get total change in velocity.)

But this is the change in velocity, so to find the final velocity, we'll use the definition of change in velocity.

v_{f} - v_{i} = 18 m/s (Use definition of change in velocity.)

v_{f} = 18 m/s + v_{i} (Solve for the final velocity.)

v_{f} = 18 m/s + 10 m/s (Plug in initial velocity.)

v_{f} = 28 m/s (Calculate and celebrate!)

The final velocity of the sailboat is

v_{f} = 28 m/s

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Frodochips
Posted 9 years ago. Direct link to Frodochips's post “I don't get why the slope...”
I don't get why the slope is shown to be negative in the two examples, while both objects are gaining speed. Seems like the slope should be positive... What's the reasoning behind this?
Button navigates to signup pageButton navigates to signup page
(14 votes)
Answer
- Teacher Mackenzie (UK)
  Posted 9 years ago. Direct link to Teacher Mackenzie (UK)'s post “These are acceleration vs...”
  These are acceleration vs time graphs. Any line ABOVE the time axis (a=0) indicates positive acceleration. and a line below the time axis indicates negative acceleration (Slowing down)
  BUT The slope of the lines says NOTHING about the amount of acceleration.
  The slope is only a measure of 'jerkiness' (or rate of change) of acceleration.
  Comment on Teacher Mackenzie (UK)'s post “These are acceleration vs...”
  (38 votes)
RobinZhangTheGreat
Posted 8 years ago. Direct link to RobinZhangTheGreat's post “Are there quantities like...”
Are there quantities like acceleration of jerk, jerk of jerk and so on?
Button navigates to signup pageButton navigates to signup page
(8 votes)
Answer
- Mahi
  Posted 8 years ago. Direct link to Mahi's post “Change in jerk is called ...”
  Change in jerk is called jounce or snap. Change in jounce is called crackle. Change in crackle is called pop. (Yes, the names are hilarious!)
  More information can be found here:
  https://en.wikipedia.org/wiki/Jounce
  https://en.wikipedia.org/wiki/Crackle_(physics)
  https://en.wikipedia.org/wiki/Pop_(physics)
  Comment on Mahi's post “Change in jerk is called ...”
  (38 votes)
Scarlet
Posted 8 years ago. Direct link to Scarlet's post “They keep referring to st...”
They keep referring to straight lines as curves. (i.e. "But multiplying the acceleration by the time interval is equivalent to finding the area under the curve. The area under the curve is a rectangle, as seen in the diagram below.") Why is this? Is there a reason?
Button navigates to signup pageButton navigates to signup page
(8 votes)
Answer
- Niranjana
  Posted 4 years ago. Direct link to Niranjana's post “Area under a curve genera...”
  Area under a curve generally talks about the area under a specific geometric shape, be it a line or a curvature(sometimes known as concavity). Some teachers also teach it as 'area under the graph', so as long as you know what you're calculating, you should be fine.
  Button navigates to signup page
  (5 votes)
Haitham Alhad Hyder
Posted 6 years ago. Direct link to Haitham Alhad Hyder's post “The area under the accele...”
The area under the acceleration-time graph is velocity.
That under the velocity-time graph is displacement (or may be distance).

What about that under the displacement-time graph, what would it be?
Button navigates to signup pageButton navigates to signup page
(6 votes)
Answer
- ForrestZengMusic
  Posted 4 years ago. Direct link to ForrestZengMusic's post “I believe it represents a...”
  I believe it represents a negative derivative of displacement. This is called Absement and is essentially the "total" displacement. Essentially, the derivative of this is displacement, the "change" in Absement, and velocity would the derivative of displacement, the "change" in displacement, the acceleration being the second-order derivative, and so on.
  
  The area under the curve is the anti-derivative, and in lay terms moving upwards. For instance, the area under acceleration-time graph is the velocity, moving upwards.
  
  For reference, I located a list of the derivatives of displacement.
  
  -1. Absement
  0. Displacement
  1. Velocity
  2. Acceleration
  3. Jerk
  4. Jounce (snap)
  5. Crackle
  6. Pop
  7. Lock
  8. Drop
  9. Shot
  10. Put
  Button navigates to signup page
  (8 votes)
Bridget Healy
Posted 9 years ago. Direct link to Bridget Healy's post “ok, but how do you go abo...”
ok, but how do you go about making a acceleration time graph?
Button navigates to signup pageButton navigates to signup page
(7 votes)
Answer
- Delaine Hare
  Posted 9 years ago. Direct link to Delaine Hare's post “Or you could measure acce...”
  Or you could measure acceleration at each point in time with an accelerometer.
  Button navigates to signup page
  (2 votes)
loaymohamedabd
Posted 8 years ago. Direct link to loaymohamedabd's post “How can we calculate the ...”
How can we calculate the jerk using only the information given by a velocity-time graph ?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Mark Zwald
  Posted 8 years ago. Direct link to Mark Zwald's post “Take the second derivativ...”
  Take the second derivative of the v(t) graph to get J(t).
  J(t) = da(t)/dt = d²v(t)/dt²
  Button navigates to signup page
  (7 votes)
Zahidul Islam Bayzid
Posted 5 years ago. Direct link to Zahidul Islam Bayzid's post “in Example 1: Race car ac...”
in Example 1: Race car acceleration it is said that the driver had a constant velocity of 20 m/s and again she had it at 0s. but in the graph it is shown that at 0s she had a acceleration of 6 m/s2. so my question is if the graph is mistaken or it is right. since its clear that constant velocity means 0 acceleration and she had a constant velocity of 20 m/s at 0s, so it means the acceleration at 0s would also be 0m/s2 . am i right?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Davin V Jones
  Posted 5 years ago. Direct link to Davin V Jones's post “It states that the consta...”
  It states that the constant velocity was until the driver began accelerating at the end. At time = 0, velocity was no longer constant.
  Comment on Davin V Jones's post “It states that the consta...”
  (3 votes)
thiongane86
Posted 9 years ago. Direct link to thiongane86's post “last graph how did you fi...”
last graph how did you find Vi = 10 m/s ?
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(2 votes)
Answer
- Lovre Kardum
  Posted 9 years ago. Direct link to Lovre Kardum's post “It is written in the text...”
  It is written in the text of Example 2:"A sailboat is sailing in a straight line with a velocity of 10 m/s." Otherwise you can not know exact velocity from acceleration vs. time graph. You only know change of velocity (Δv).
  Button navigates to signup page
  (3 votes)
Shivansh Kamboj
Posted 6 years ago. Direct link to Shivansh Kamboj's post “Sir, why the velocity - t...”
Sir, why the velocity - time and Distance - itme graph not start from 0 for constant accelartion
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(3 votes)
Answer
sumeshg0591
Posted 8 years ago. Direct link to sumeshg0591's post “In the example 1 , retard...”
In the example 1 , retardation is taking place instead of acceleration (according to the graph because the line is pointing downwards). So how can the velocity of the car after 8 seconds be more than 20m/s, similar confusion in example 2 . I don't understand at all!!
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Muhammad Nawal
  Posted 8 years ago. Direct link to Muhammad Nawal's post “I think the slope should ...”
  I think the slope should be rising because acceleration at time t=0 is 0 since velocity is constant............ but yes velocity can be more than 20m/s even if acceleration is decreasing ..... see the top question's answer......... but still it should be a rising slope..
  Button navigates to signup page
  (2 votes)