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Finding the Average Acceleration

  1. Aug 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Starting from rest, a subway train first accelerates to 25m/s and then begins to brake. Forty-eight seconds after starting, it is moving at 17 m/s. What is its average acceleration in this 48-s interval?


    2. Relevant equations
    1. v=v0+at
    2. x=x0+(1/2)(v0+v)t
    3. x=x0+v0t+(1/2)at^2
    4. v^2=v0^2+2a(x-x0)


    3. The attempt at a solution
    I am unsure of this question here - I tried drawing the situation in a velocity vs time graph with velocity increasing from rest to 25ms^-1 and then decreasing to 17ms^-1 over 48s. However the question didn't specify the time at which it reached 25ms^-1, I attempted the question without this information but couldn't solve this question. How do you know at what point in this 48s it reaches 25ms^-1 and subsequently it's displacement?

    I tried using the equation of motion for constant acceleration but could not find a suitable one.

    Thus tried using the average acceleration=[itex]\Delta[/itex]X/[itex]\Delta[/itex]t in hope but I was unsuccessful there.

    The answer is 0.354m/s^2
     
  2. jcsd
  3. Aug 10, 2012 #2

    CWatters

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    Science Advisor
    Homework Helper

    Is it simply 17/48 = 0.354 ? That sounds too easy.
     
  4. Aug 10, 2012 #3
    With an average it doesn't matter what happens in between - just what the state is at the two ends. The point of the question is to get you to see that the maximum speed reached is irrelevant, all that matters it the initial and final speed and the time taken.
     
  5. Aug 10, 2012 #4
    Thanks - that makes sense now ΔV=(Vf-V0). I thought too deeply into this question and didn't see the simplicity of it.
     
  6. Aug 10, 2012 #5
    I'm not sure differences between constant and average.

    If we have a constant acceleration then we use this formula.
    x=x0+ut+(1/2)at2
    or
    x=(1/2)at2


    The displacement of the OP question depends on how fast it attains the 25m/s velocity.
    The shorter time to attain that velocity means greater distance travelled within 48sec.
    Thus greater constant acceleration.
     
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