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Homework Help: Find maximum velocity in Acceleration-Time Graph

  1. May 29, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the maximum velocity of the particle using the graph given.(See attachment)

    2. Relevant equations

    Initial velocity was not given. I assumed it to be 0.

    In book, the solution was 55m/s^2

    3. The attempt at a solution

    Initial acceleration was ##10m/s^2##. Final acceleration after 11 seconds was ##0m/s^2##.

    I figured out that ##a_n=\frac{110-10(n-1)}{11}##, where n is the second and ##a_n## is the acceleration in ##n##th second.

    I hope I am right at this step.

    I think that velocity will be maximum at 11th second.

    So I calculate the velocity at 1st second by

    I derived this using identities of motion and my already derived ##a_n## formula.


    I should get 55m/s.

    Please help me out.

    Attached Files:

  2. jcsd
  3. May 29, 2014 #2


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    Gold Member

    Does this equation recover the values of a at n=0 and n=11 that you know from looking at the graph?
    Last edited: May 29, 2014
  4. May 29, 2014 #3
    No, you can't do that. Don't use n. Use t and find an expression for the acceleration as a function of time a(t) where t is any number - not just a whole number. Than integrate the acceleration to find the velocity. If you don't know integrals than you probably know some theorem relating the change in velocity to the area in a graph of acceleration vs time. In this case the area is a triangle which can be solved without integrals.
  5. May 29, 2014 #4
    Wait a minute!

    I got a brilliant idea.

    I know that the area below velocity-time graph gives the total displacement of the body.

    I thought about it a while.

    Velocity is the change of displacement over time.

    Acceleration is the change of velocity over time.

    Then , presto! Instead of velocity in velocity-time graph, we have acceleration.

    Instead of distance(factor of velocity), we have velocity(factor of acceleration).

    Hence the area under the graph is the max. velocity i.e. 55m/s.

    After solving it, I came to this thread and also saw dauto saying the same.

    Thank you CAF123.
  6. May 29, 2014 #5

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