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Average acceleration of an object

  1. Sep 6, 2006 #1
    Here's my question:
    An object has a speed of 18 m/s at a certain time, and 2.4 seconds later, it's speed is 30 m/s in the opposite direction.
    So my question is, what were the magnitude and direction of the average acceleration of the object during this 2.4 second time period? And, if you give me an answer, how did you figure it out? Thanks!
     
  2. jcsd
  3. Sep 6, 2006 #2

    berkeman

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    Staff: Mentor

    We don't give answers here, but we do try to help you figure out how to get to the answer. So you can simplify the question a bit to assume a constant acceleration the whole time period. That would give the same results as having a varying acceleration with an average equal to the constant acceleration.

    Now, what equations have you used to relate velocity and acceleration? What are the units of these quantities in the equation? How does time come into the equation?
     
  4. Sep 6, 2006 #3
    I'm sure this is wrong, but would the constant acceleration be 12 m/s?

    Here is an equation I've used for average velocity; (can't make it look the same on the monitor though, but I'll try)
    v = D x/ D t = (x^2 - x^1)/D t - The D's being delta, or the little triangle.
    And acceleration:
    v = D v/ D t = (v^2 - v^1)/D t
     
  5. Sep 7, 2006 #4
    I still have no idea what to do next. I'm not looking for the answer so much, just the next step.
     
  6. Sep 7, 2006 #5

    Doc Al

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    Staff: Mentor

    Rewrite this correctly:
    [tex]a_{ave} = \Delta v / \Delta t = (v_2 - v_1)/ \Delta t[/tex]

    Hint: Direction (and thus sign) matters!
     
  7. Sep 7, 2006 #6
    So I'm guessing the direction of the acceleration is - (negative), sense after the 2.4 second time interval it's moving in the opposite direction faster than it was moving in the positive direction. I'm still lost on this one; any extra help would be great!
     
  8. Sep 7, 2006 #7
    The number one thing in mechanics is drawing an image.

    Draw an image, write the velocity vectors and define a positive direction and use the values in the equation outlined above.
     
  9. Sep 7, 2006 #8

    berkeman

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    Staff: Mentor

    Whether the acceleration is negative or positive depends on how you define the initial and final velocities. From the problem statement you posted, it looks like it is left up to you.

    So let's try this -- let's call the initial motion in the + x direction, so the final motion ("in the opposite direction) will be in the - x direction. That would give you the following for initial and final velocities:

    initial --> [tex]v_i = +18 \frac{m}{s}[/tex]

    final --> [tex]v_f = -30 \frac{m}{s}[/tex]

    Now, using the equation that Doc Al suggested, and being careful with the signs, what do you get for the change in velocity divided by the time interval?
     
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