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G-Force homework problem

  1. Jul 5, 2007 #1
    1. If I measure the acceleration of an object every 1 second, I will have a g-force which I can use formula 1 to convert into meters per second per second. Therefore I know from the result of this formula how much the object has accelerated/de-accelerated in the last second.

    If however I wish to sample the acceleration of an object every 0.5 of a second, I can get a more accurate reading as the object may not accelerate/deaccelerate at the same rate for an entire second. My problem is converting the output to suit the 0.5 second interval.

    2. Relevant equations

    formula 1 = meters per second per second = (gForce - 1) * 9.8

    gForce is subtracted by 1 because at 1g an object is stationery.

    3. The attempt at a solution
    I again use formula 1 to convert g-force into meters per second per second - but this time I think I should divide the result of the answer by 2, as only half a second has passed, so if we have an acceleration of 2G at 0.5 seconds, then we've actually only accelerated by (9.8 / 2 = 4.7 meters per second per second).

    I sample the acceleration again at exactly 1 second and again divide the result by 2 as I'm only interested in the acceleration from 0.5 seconds to 1 second. If i add this result to the previous result, i should have the acceleration for the object in 1 second.

    Does this sound right to you?

    Many thanks in advance
    Last edited by a moderator: Mar 7, 2013
  2. jcsd
  3. Jul 5, 2007 #2


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    I think you are making things overly complicated. Acceleration is just (change in velocity)/(change in time). So if you have velocity samples at 1 sec intervals then you divide the difference by 1 sec. If you have samples at 1/2 sec intervals then you divide by 1/2 sec. Your formula 1 is strange. What makes you think that 'at 1g objects are stationary'???
  4. Jul 5, 2007 #3

    Thanks for your reply, as far as I am aware all objects at rest are subject to a 1g static acceleration due to the Earth’s gravitation attraction, because I am only interested in the dynamic acceleration, the above calculation of acceleration is off by 1g due to the aforementioned gravitational attraction of Earth. The actual dynamic acceleration should be the above value minus 1g.

    I guess what its confusing me is the units "meters per second per second".

  5. Jul 5, 2007 #4


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    Objects at rest aren't subject to any acceleration at all. They are subject to a FORCE (mg) which would cause an acceleration of 1g, except that it is balanced by equal and opposite normal force from the floor which gives a net acceleration of 0g. Don't confuse forces and acceleration. They are not the same thing. Referring to a 'force' as a 'static acceleration' is going to cause nothing but confusion. And since a=v/t, v is measured in m/sec, t is measured in sec, hence 'a' is measured in (m/sec)/(sec).
  6. Jul 5, 2007 #5
    Thanks Dick

    Think you've cleared up most of my confusion.

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