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Accelleration and dispacement of planes and babiess

  1. Nov 18, 2007 #1
    one type of aeroplane has a maximum acceleration on the ground of 3.5ms-2
    a)for how many seconds must it accellerare along a runway in order to reach its take-off speed of 115ms-1?
    b) what is the minimum length of runway needed for it to reach this length?

    ummm, not entirely sure
    av acceleration=total change in velocity/time taken


    for a) i thought that it might be 115/3.5 which gives 82.86 (4 sig fig)
    but then b) i thought maybe v=d/t
    so 3.5=115/x
    but that obviosly comes out to 115/3.5 and so the same answer, i don't really know which i have write, if either.

    also i have a question about a baby carriage on a ramp
    it accellerates at an average rate and takes 15s to reach the bottom of a 5m ramp.

    i got that the average velocity is 15/5= 3m/s
    but i need to work out the velocity at the bottom of the ramp and then the average acceleration. I can quite happily get the acceleration if i can find the final velocity, but i don't really know. i thought that as the acceleration was uniform the final velocity might be 6m/s, but this doesn't seem right
  2. jcsd
  3. Nov 18, 2007 #2


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    Science Advisor

    You considered two different ways and they come out to the same answer! Why does that bother you? In any case you shouldn't use a formula without thinking about it. Yes, "av acceleration=total change in velocity/time taken". Here you are told that the maximum acceleration is 3.5 m/s2 and that the change in velocity was 115 m/s. You are asked for the time: putting those numbers into the appropriate variables, 3.5= 115/t. Solve for t: t= 115(m/s)/3.5(m/s2)= 82.86 s. You can also check the units: to "divide" m/s by m/s2, invert the "fraction" and multiply: (m/s)(s2/m) which cancels to s (seconds) exactly what you want.

    I think that is intended to say it accelerates at a constant rate. "it accelerates at an average rate" doesn't make any sense.

    Presumably the baby carriage starts, at the top of the ramp, at 0 m/s. Now, HOW did you get "that the average velocity is 15/5= 3 m/s"? I hope you DIDN'T divide 15 s by 3 m! That would give you 15s/3m= 5 "s/m" which is NOT a unit of velocity! velocity is "distance divided by time" and you have it the wrong way around! Once you have correct average velocity you can find the velocity at the end of the ramp by using a very nice fact: as long as the acceleration is a constant, the average velocity is just the average of the velocity at the beginning and the final velocity. Since the velocity at the start is 0, the velocity at the bottom is just twice the average velocity.
  4. Nov 18, 2007 #3
    thank you (also i've just noticed i wrote "write" meaning "right" which is more than mildly embarraing) only for b) of the first question it was looking for a distance (i realise i may have caused confusion with my bit about the answers i got as i clearly went momentarily insane) so i don't really know what the answer would be for that as i cant now figure out what to do
  5. Nov 18, 2007 #4
    oh wait...v=d/t and now i have t, yet another blunder, woops!!!! thanks
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