If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

What are acceleration vs. time graphs?

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, equals, 4, start text, space, s, end text?

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 start text, s, l, o, p, e, end text, equals, start fraction, start text, r, i, s, e, end text, divided by, start text, r, u, n, end text, end fraction, equals, start fraction, a, start subscript, 2, end subscript, minus, a, start subscript, 1, end subscript, divided by, t, start subscript, 2, end subscript, minus, t, start subscript, 1, end subscript, end fraction, equals, start fraction, delta, a, divided by, delta, t, end fraction, as can be seen in the diagram below.
This slope, which represents the rate of change of acceleration, is defined to be the jerk.
start text, j, e, r, k, end text, equals, start fraction, delta, a, divided by, delta, t, end fraction
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, equals, 6, start text, space, s, end text?

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.
start text, a, r, e, a, end text, equals, delta, v
It might be easiest to see why this is the case by considering the example graph below which shows a constant acceleration of 4space, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction for a time of 9 s.
If we multiply both sides of the definition of acceleration, a, equals, start fraction, delta, v, divided by, delta, t, end fraction, by the change in time, delta, t, we get delta, v, equals, a, delta, t.
Plugging in the acceleration 4space, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction and the time interval 9 s we can find the change in velocity:
delta, v, equals, a, delta, t, equals, left parenthesis, 4, space, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction, right parenthesis, left parenthesis, 9, start text, space, s, end text, right parenthesis, equals, 36, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction
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 4space, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction, and the width is 9 s. So, finding the area also gives you the change in velocity.
start text, a, r, e, a, end text, equals, 4, space, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction, times, 9, start text, space, s, end text, equals, 36, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction
The area under any acceleration graph for a certain time interval gives the change in velocity for that time interval.

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, equals, 0, start text, space, s, end text.
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.
delta, v, equals, start text, a, r, e, a, end text, equals, start fraction, 1, divided by, 2, end fraction, b, h, equals, start fraction, 1, divided by, 2, end fraction, left parenthesis, 8, start text, space, s, end text, right parenthesis, left parenthesis, 6, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction, right parenthesis, equals, 24, start text, space, m, slash, s, end text, start text, left parenthesis, U, s, e, space, t, h, e, space, f, o, r, m, u, l, a, space, f, o, r, space, a, r, e, a, space, o, f, space, t, r, i, a, n, g, l, e, colon, space, start fraction, 1, divided by, 2, end fraction, b, h, point, right parenthesis, end text
delta, v, equals, 24, start text, space, m, slash, s, end text, start text, left parenthesis, C, a, l, c, u, l, a, t, e, space, t, h, e, space, c, h, a, n, g, e, space, i, n, space, v, e, l, o, c, i, t, y, point, right parenthesis, end text
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, delta, v, equals, v, start subscript, f, end subscript, minus, v, start subscript, i, end subscript, to find that
delta, v, equals, 24, start text, space, m, slash, s, end text
v, start subscript, f, end subscript, minus, v, start subscript, i, end subscript, equals, 24, start text, space, m, slash, s, end text, start text, left parenthesis, P, l, u, g, space, i, n, space, v, start subscript, f, end subscript, minus, v, start subscript, i, end subscript, space, f, o, r, space, delta, v, point, right parenthesis, end text
v, start subscript, f, end subscript, minus, 20, start text, space, m, slash, s, end text, equals, 24, start text, space, m, slash, s, end text, start text, left parenthesis, P, l, u, g, space, i, n, space, 20, space, m, slash, s, space, f, o, r, space, t, h, e, space, i, n, i, t, i, a, l, space, v, e, l, o, c, i, t, y, space, v, start subscript, i, end subscript, point, right parenthesis, end text
v, start subscript, f, end subscript, equals, 24, start text, space, m, slash, s, end text, plus, 20, start text, space, m, slash, s, end text, start text, left parenthesis, S, o, l, v, e, space, f, o, r, space, v, start subscript, f, end subscript, point, right parenthesis, end text
v, start subscript, f, end subscript, equals, 44, start text, space, m, slash, s, end text, start text, left parenthesis, C, a, l, c, u, l, a, t, e, space, a, n, d, space, c, e, l, e, b, r, a, t, e, !, right parenthesis, end text
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, equals, 0, start text, space, s, end text, 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, equals, 0, start text, space, s, end text and t, equals, 3, start text, space, s, end text is considered positive area since it is above the horizontal axis. The green triangle between t, equals, 3, start text, space, s, end text and t, equals, 7, start text, space, s, end text is also considered positive area since it is above the horizontal axis. The red triangle between t, equals, 7, start text, space, s, end text and t, equals, 9, start text, space, s, end text, 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 start fraction, 1, divided by, 2, end fraction, b, h for the triangles—to get the total area between t, equals, 0, start text, space, s, end text and t, equals, 9, start text, space, s, end text.
delta, v, equals, start text, a, r, e, a, end text, equals, left parenthesis, 4, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction, right parenthesis, left parenthesis, 3, start text, space, s, end text, right parenthesis, plus, start fraction, 1, divided by, 2, end fraction, left parenthesis, 4, start text, space, s, end text, right parenthesis, left parenthesis, 4, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction, right parenthesis, plus, start fraction, 1, divided by, 2, end fraction, left parenthesis, 2, start text, space, s, end text, right parenthesis, left parenthesis, minus, 2, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction, right parenthesis, start text, left parenthesis, A, d, d, space, a, r, e, a, s, space, o, f, space, r, e, c, t, a, n, g, l, e, space, a, n, d, space, t, w, o, space, t, r, i, a, n, g, l, e, s, point, right parenthesis, end text
delta, v, equals, 18, start text, space, m, slash, s, end text, start text, left parenthesis, C, a, l, c, u, l, a, t, e, space, t, o, space, g, e, t, space, t, o, t, a, l, space, c, h, a, n, g, e, space, i, n, space, v, e, l, o, c, i, t, y, point, right parenthesis, end text
But this is the change in velocity, so to find the final velocity, we'll use the definition of change in velocity.
v, start subscript, f, end subscript, minus, v, start subscript, i, end subscript, equals, 18, start text, space, m, slash, s, end text, start text, left parenthesis, U, s, e, space, d, e, f, i, n, i, t, i, o, n, space, o, f, space, c, h, a, n, g, e, space, i, n, space, v, e, l, o, c, i, t, y, point, right parenthesis, end text
v, start subscript, f, end subscript, equals, 18, start text, space, m, slash, s, end text, plus, v, start subscript, i, end subscript, start text, left parenthesis, S, o, l, v, e, space, f, o, r, space, t, h, e, space, f, i, n, a, l, space, v, e, l, o, c, i, t, y, point, right parenthesis, end text
v, start subscript, f, end subscript, equals, 18, start text, space, m, slash, s, end text, plus, 10, start text, space, m, slash, s, end text, start text, left parenthesis, P, l, u, g, space, i, n, space, i, n, i, t, i, a, l, space, v, e, l, o, c, i, t, y, point, right parenthesis, end text
v, start subscript, f, end subscript, equals, 28, start text, space, m, slash, s, end text, start text, left parenthesis, C, a, l, c, u, l, a, t, e, space, a, n, d, space, c, e, l, e, b, r, a, t, e, !, right parenthesis, end text
The final velocity of the sailboat is v, start subscript, f, end subscript, equals, 28, start text, space, m, slash, s, end text.

Want to join the conversation?

  • old spice man green style avatar for user Frodochips
    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?
    (13 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Teacher Mackenzie (UK)
      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.
      (31 votes)
  • male robot hal style avatar for user RobinZhangTheGreat
    Are there quantities like acceleration of jerk, jerk of jerk and so on?
    (6 votes)
    Default Khan Academy avatar avatar for user
  • aqualine ultimate style avatar for user Scarlet
    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?
    (8 votes)
    Default Khan Academy avatar avatar for user
    • aqualine tree style avatar for user Niranjana
      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.
      (5 votes)
  • blobby green style avatar for user Bridget Healy
    ok, but how do you go about making a acceleration time graph?
    (7 votes)
    Default Khan Academy avatar avatar for user
  • leaf blue style avatar for user Shivansh Kamboj
    Sir, why the velocity - time and Distance - itme graph not start from 0 for constant accelartion
    (3 votes)
    Default Khan Academy avatar avatar for user
  • old spice man green style avatar for user Haitham Alhad Hyder
    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?
    (4 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user ForrestZengMusic
      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
      (5 votes)
  • starky ultimate style avatar for user loaymohamedabd
    How can we calculate the jerk using only the information given by a velocity-time graph ?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Zahidul Islam Bayzid
    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?
    (3 votes)
    Default Khan Academy avatar avatar for user
  • duskpin ultimate style avatar for user sumeshg0591
    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!!
    (2 votes)
    Default Khan Academy avatar avatar for user
    • spunky sam green style avatar for user Muhammad Nawal
      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..
      (2 votes)
  • blobby green style avatar for user vinayahalu14
    What is the definition of "v𝒇"? What is the definition of "vᵢ"?
    (1 vote)
    Default Khan Academy avatar avatar for user