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Calculate the average acceleration of 2 or more objects?

  1. Jan 9, 2009 #1
    can i take calc. the average acceleration (m/s2) of 3 objects by adding them and dividing the answer by 3 if the measuretime is more or less equal for the 3 objects? or what should i do different?
    do you not have to take into account that you are adding and dividing logarithms? and can you do it with any number of objects?
     
  2. jcsd
  3. Jan 9, 2009 #2

    berkeman

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    Staff: Mentor

    What do you mean by logarithms? Accelerations can be averaged as you say. You will want to do a 3-dimensional average, in general, so calculate the 3 averages (one in each of the x, y and z dimensions, for example).
     
  4. Jan 9, 2009 #3
    Thank you Berkeman, I made a mistake and meant exponentials. I took several vibration measurements of persons driving cars. Due to the amount of unavoidable variables I wanted to average the x,y,z directions (separately for each of the axis of course) for each of the multiple measurements taken per person. Because acceleration means exponential growth, I thought I could not add them and divide them by the number of measurements, can I?
     
  5. Jan 9, 2009 #4

    berkeman

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    A constant acceleration results in a linearly increasing velocity, and an exponentially growing displacement. The accelerations are not exponential in themselves.

    I'm not understanding what your experiment was, so it's hard for me to give you a firm answer about how to do the math. You also have to be careful about acceleratsions of the drivers and accelerations of the vehicle, and be consistent about what coordinate system you are using. You probably need to use an inertial frame of reference, which would not be the car if it is accelerating or going in a circle.

    I guess if you could post a diagram or more clearly state where the sensors were and what the vehicle and driver were doing, we would have a better chance at guiding your calculations.
     
  6. Jan 9, 2009 #5
    many thanks for your assistance, it is already very helpfull. if i run into problems with the calc. I will add them to this threat later. thank you
     
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