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-   -   Solving for velocity with changing acceleration (https://www.askmehelpdesk.com/showthread.php?t=334578)

  • Mar 27, 2009, 10:17 AM
    cdedmundson
    solving for velocity with changing acceleration
    I know I will need to use some integration but I am not yet familiar enough with it to use it successfully. It would be greatly appreciated if someone could walk me through this problem...

    The acceleration of a bus is given by the equation a(t)=βt where β=1.2m/s^2.

    a.) if the bus's velocity at time t=1.0s is 5m/s, what is the bus's velocity at 2s?
    b.) 8.7s?
  • Mar 28, 2009, 02:54 PM
    stingRay18
    Quote:

    Originally Posted by cdedmundson View Post
    I know I will need to use some integration but I am not yet familiar enough with it to use it successfully. It would be greatly appreciated if someone could walk me through this problem...

    The acceleration of a bus is given by the equation a(t)=βt where β=1.2m/s^2.

    a.) if the bus's velocity at time t=1.0s is 5m/s, what is the bus's velocity at 2s?
    b.) 8.7s?

    Given V(t) = Vi + a*t where Vi = Initial Velocity.

    If V(1) at 1 sec = 5 Then solving for Vi
    5 = Vi + (1.2*t)*t
    5 = Vi + 1.2*(t^2)
    5 = Vi + 1.2(1^2)
    5 = Vi + 1.2
    Vi = 3.8

    Then solving for V at 2 and 8.7 sec
    V(2) = 3.8 + 1.2*(2^2) = 8.6
    V(8.7) = 3.8 + 1.2*(8.7^2) = 94.628

    That's how I see it.
  • Mar 29, 2009, 10:41 AM
    cdedmundson
    that would work if the acceleration was CONSTANT, but since the acceleration is changing with relation to time you need to use integration...

    I figured this out about 20 minutes after I posted this problem and anyone is welcome to use this example to learn basic integration...

    Integration is also known in some other countries as "anti-derivatives", which is exactly what they are. The integral equation to solve for the acceleration is ∫(0→t) β/2*t^2 + c, you solve for c at t=1 and you get c=4.4, plug c back into the equation and solve for the total equation at 2 and 8.7 (6.8 and 49.8 respectively)
  • Mar 29, 2009, 01:32 PM
    stingRay18

    That makes sense - thanks.

    I need to take the rust off my calculus :)

    Be well.

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