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Stumped on 2 motion problems

  1. Jan 17, 2009 #1
    Problem #1.
    A particle is moving along a line such that when it is at the origin it has a velocity of 4m/s. If it begins to decelerate at the rate of a = (-1.5v1/2) m/s2, where v is in m/s, determine the particles position and celocity when t = 2s.

    Therefore the givens are: v=4 m/s when t=0s when s=0m and v=? when t=2s when s=?
    as the particle decelerates during this period by a factor of a =(-1.5v1/2) m/s2




    All Available Equations are: a=(dv/dt), v=(ds/dt), ads=vdv. The other equations are for constant acceleration which doesn't apply here because it's changing.



    In my thinking, the only equation which would be available is a = (dv/dt) since you know "a" and can integrate "dv" to find the final velocity since you know both "t" values.
    Therefore: (-1.5v1/2)=dv/dt
    2/3 = v1/2
    v=0.816m/s

    This value is wrong though because the answer in the book is 0.25m/s and s = 3.5m.
    Ill be able to find the position once I get the velocity but I've thought about this too much for how simple it is and I've gotten nowhere.





    Problem #2
    Ball A is released from rest at a height of 40ft at the same time a second ball B is thrown upward starting at a height of 5ft from the ground. If the balls pass one another at a height of 20ft determine the speed at which ball B was thrown upward.




    All Available Equations are: a=(dv/dt), v=(ds/dt), ads=vdv. The other equations are for constant acceleration which can be applied in this case since a = g.



    Since I knew the initial velocity for ball A I used it to determine the time at which it would pass 20ft.
    Using: S=S0+V0t + 1/2at2 therefore
    I changed the heights so that it was simpler by viewing it as if Ball be were at the origin (i.e. all distances are -5)
    15ft=35ft+0t+1/2(+9.8)t2,
    I used +9.8 because the ball is falling therefore its helping to speed the ball up.
    I got t = 2.02s.

    I then used this answer to solve for the velocity of Ball B, the velocity it would have as it passed by 20ft. S=S0 + Vt + 1/2at2
    V0=17.32
    This velocity would be the velocity at t=2.02 when the ball is at 20ft not when it was thrown, therefore I used:
    V2=V02+2a(S-S0)
    V0 2.45mft/s or 24.36ft/s depending on + or - gravity just
    to check.

    These are not correct because the answer is 31.4ft/s I think that my problem is in the last step that the ball is going up and coming down therefore it could pass 20ft with the other ball on the way down or the way up, depending, but I havn't been able to set it up to work out correctly.

    I appreciate any help anyone could provide with either of these.
     
  2. jcsd
  3. Jan 17, 2009 #2

    LowlyPion

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    Welcome to PF.

    You are on the right track in 1).

    You have correctly noted that you need to integrate.

    Unfortunately you have integrated incorrectly.
     
  4. Jan 17, 2009 #3
    The equation you had in the first problem, -1.5v^(1/2)=dv/dt, is a differential equation. You need to separate the variables, to write it as
    (v)^(-1/2)*dv = (2/3)dt
    Now integrate both sides.

    In the second problem, take a second look at your value for g. You seem to have your unts mixed up.
     
  5. Jan 17, 2009 #4
    I understand the v^(-1/2)*dv but I'm confused about the other side. Shouldn't it be:
    v^(-1/2)*dv = (-3/2)*dt?

    ...Even with that though when I integrate it would be (2/1)*(v^(-1/2)+v^(2/2)) from 0 to
    vf = -3/2t from 0 to 2. Therefore 17/4=vf from that solution...I'm definitely doing something wrong and making this overly complicated.

    I was completely oblivious to my mistake on the second problem..I even made the mistake of typing m/s when I was typing in the questions and didn't even realize when I was correcting then.

    Thanks.
     
    Last edited: Jan 17, 2009
  6. Jan 17, 2009 #5

    Dick

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    Yes, it should be v^(-1/2)*dv=(-3/2)*dt. But your integration of v^(-1/2)*dv is completely wrong.
     
  7. Jan 18, 2009 #6
    Can you clarify how it is wrong? v^(-1/2)dv >>> 2*v^(1/2) when integrated, correct? ...I see how I just typed it wrong part way through before but the answer I typed is the correct one for this integration which is wrong.
     
  8. Jan 18, 2009 #7

    Dick

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    2*v^(1/2) is correct. "(2/1)*(v^(-1/2)+v^(2/2))" wasn't. That's what I was looking at.
     
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