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Acceleration and distance :S ?

  1. Sep 16, 2007 #1
    ok so heres the question

    She picks up a large stone and drops it into a pit, listening for the sound of it hitting the bottom. The pit has a uniform temperature of 8 degrees celcius. From the time the rock leaves her hadn to the time the sound of it striking bottom reachers her ears, 24 seconds pass. What is the depth of this whole?



    ...... okay so i know that acceleration is change in velocity over time.
    this gives you the time , and also the temperature which by using v=332+0.6(T) will give me a velocity

    i'm assuming that the initial velocity would be 0 ? .... and then using the formula above the other velocity would be 336.8 m/s

    thennnnnn do 336.8 divided by time ( 24 s ) so get acceleration ... but then how do i get the distanceee?

    d=t*v butttt wouldnt acceleration effect that somehow?
     
  2. jcsd
  3. Sep 16, 2007 #2
    you have two phases in the system:

    phase 1. the stone is falling in free fall state (if we neglect the air friction), takes it some time

    -transition- the stone hits the bottom, this transition does not take time (assume it is instantaneous)

    phase 2. the sound propagates from the bottom to the ears of the girl

    now the total time (24s) is the sum of durations of phase 1 & phase 2

    - I hope this helps you

    BTW, i dont see what does the temperature do here, unless you considering the velocity propagation of sound in air as funtion of temperature (which i doubt in such situation)
     
    Last edited: Sep 16, 2007
  4. Sep 16, 2007 #3
    thanksss , i dont seee how that gets me the depth of the whole though ?
     
  5. Sep 16, 2007 #4

    learningphysics

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    Write an equation for the depth.
     
  6. Sep 16, 2007 #5
    like jsut d=v*t
    doesnt the acceleration effect that somehow though?
     
  7. Sep 16, 2007 #6
    ok, let's see more in details

    phase 2- speed of sound in air Vs = 1200 m/s = d / t2 , so t2 = Vs / d, also d = Vs t2

    phase 1- gravity is the only force applied,
    F=m a = m G, so G = a = dv/dt =d²y/dt², so v(t) = G t + V(0), so y(t) = G t² / 2 + v(0)t + y(0)
    if you consider v(0) = 0 & y(0)=0, then y(t) = G t² / 2
    then at the time of the impact (which is t1), y(t1) = d

    so G t1² / 2 = d
    or d = Vs t2

    and also t1 + t2 = 24 seconds

    the rest is clear
     
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