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Homework Help: Acceleration, Velocity and Position (calculus)

  1. Mar 16, 2010 #1
    1. The problem statement, all variables and given/known data

    The acceleration of a particle is directly proportional to time, t. At t = 0, the velocity of that particle, v, = 16 in./sec. Knowing that v = 15 in/sec and x = 20in when t = 1, determine the velocity, acceleration and position at x = 7.

    2. Relevant equations

    3. The attempt at a solution

    I took
    [tex]\frac{dv}{dt} = kt, so \: v = \frac{k}{2}t^2 + C[/tex]

    Solving for C:
    [tex]16 = \frac{k}{2}(0)^2 + C[/tex]
    C = 16

    Solve for k

    [tex]1 = \frac{k}{2}(1)^2 + 16[/tex]

    k = -30

    Therefore:

    [tex]v(t) = \-15*(t)^2 + 16[/tex]

    This is already wrong; v(7) = -719 in/sec but the book states the answer is -33 in/sec

    Any advice on what's gone wrong would be greatly appreciated; this is a revision question too...

    Cheers,
    Adrian
     
  2. jcsd
  3. Mar 16, 2010 #2
    k is -2
     
  4. Mar 16, 2010 #3

    ehild

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    You need to find the velocity when x = 7. Find x(t) first.

    ehild
     
  5. Mar 16, 2010 #4
    Ah, thankyou ehild; I failed to read the question once again... I will give it a shot. How is k = -2?
     
  6. Mar 16, 2010 #5

    ehild

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    Looks OK.
     
  7. Mar 16, 2010 #6
    I found x(t) by integrating and finding C. I obtained this:

    [tex]x(t)=16t-5t^3+9[/tex]

    Then solving for t when x = 7

    [tex]16t-5t^3+2 = 0[/tex]

    Sorry, how does one solve this?

    Cheers,
    Adrian
     
    Last edited: Mar 16, 2010
  8. Mar 16, 2010 #7

    ehild

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    Don't forget that k=-2, so v(t)=16-t^2.

    Looking at the solution for v(t) at x=7 , it should be rather t=7 instead of x. Otherwise why would they ask the position at x=7?

    ehild
     
    Last edited: Mar 16, 2010
  9. Mar 16, 2010 #8
    Sorry, this is probably really stupid, but where/how does k = -2?
     
  10. Mar 16, 2010 #9

    ehild

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    "v = 15 in/sec and x = 20in when t = 1"

    15=k/2*1^2+16

    ehild
     
  11. Mar 16, 2010 #10
    Ah! I put in a bad number...dodgy mistake indeed; Cheers, I will retry!
     
  12. Mar 16, 2010 #11
    Got it!! Thanks so much, both of you! ;)
     
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