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S>V>T help with velocity

  1. Apr 26, 2009 #1
    Hi all,
    This is my first post so I hope this doesnt break any rules, to the best of my knowledge it doesnt, but let me know.

    i am curently revising for a maths exam and i am practising questions on velocity.

    how should i go about calculating a questoin in the form

    v= e^t

    at what value of t does the particle have a velocity of x ?


    I have removed the numbers so that you guys arent answering the actual question for me. I am unsure if the question can still be answered as i have written it , if not i will post the actual figures.


    also just to check an answer I have calculated ( I think):

    find the acceleration of a particle whose velocity v=e^3t-2 at t=1

    My answer acceleration T= 1 m/s^2

    thanks
     
  2. jcsd
  3. Apr 26, 2009 #2
    For the first question, if v denotes velocity, then you have v = e^t. You want to know when v, or e^t, is equal to x. So you have x = e^t. Use the natural logarithm to solve for t.

    For your second question, your answer is wrong. If v = e^(3t) - 2, then you know that the acceleration = dv/dt. Calculate dv/dt at t = 1.
     
  4. Apr 26, 2009 #3
    okay so dv/dt at t=1

    v=e3t-2 t=1 so v= e1

    differentiation = 1e0= 1

    I'm afraid I am a maths dunce so i cant see where i went wrong

    I am unsure how to use ln here?

    apologies for my simpleness of mind :P
     
  5. Apr 26, 2009 #4
    v = e^(3t - 2)

    Differentiate using the chain rule. Then just plug in 1 wherever you see t.

    For the second question if x = e^t, then ln(x) = ln(e^t) = tln(e)?

    What's ln(e) equal to?
     
  6. Apr 26, 2009 #5
    i am going to say, tentatively, 1 ?
     
  7. Apr 27, 2009 #6
    v = e^(3t-2)
    First take the derivative, then plug in t=1.
    dv/dt = 3e^(3t-2) (chain rule)
    t=1
    dv/dt = 3e

    v=e^t
    ln v = lne^t = tlne = t
    So: t = lnv
     
  8. Apr 28, 2009 #7
    Evening all,

    another day, another revision question in the endless battle against my brain to make it understand chain rule.

    this time I think I have worked through it and would much apreciate it if someone would notify me of the myrriad mistakes with which it is ridden i am sure.

    okay question :
    The velocity v in m s-1 of a moving body at time tseconds is given by v=e2t-1 when t=0.5 displacement equals 10m. Find the displacement when t=1


    My workings :

    Chain rule v= eu where u= 2t-1


    intergrate 2t-1 = t2-1t+c


    when t = 0.5 displacement (a) = 10

    Then using this to find out value for c:

    ((0.5)2-0.5+c) *e 2(0.5)-1=10

    (-0.25+c)*e0=10

    (-0.25+c)* 1 = 10

    -0.25= 10 -c

    -10.25= c

    substituiting this information back into the question



    a=((1)2 - 1(1) -10.25)* e 2(1)-1

    -0.25 * e = -27.9



    the first thing that makes me think i am wrong is the negative value for displacement .



    Im not sure if any of this even makes sense. please let me know
     
  9. Apr 28, 2009 #8
    Looks like you have to calculate the definite integral!

    The anti-derivative of e2t-1 is (1/2)e2t-1.
    You probably don't understand why, so that's why I'll explain. If you take the derivative of, say: eax, the derivative is aeax (chain rule- a is the derivative of ax). When you take the anti-derivative, you have to 'undo' this extra factor a that comes up front, so you divide by it. Suppose we have again eax, the anti-derivative will be (1/a)eax, because if we differentiate it again we get back to eax.

    So what you have to calculate is:

    int01[(1/2)e2t-1] dt

    Remember that you don't have to find the constant when taking a definite integral (this makes physical sense as well; it only matters how much distance has been traveled in the time interval from 0 to 1, not how much had been travelled before or after, which would basically be the meaning of having a constant in this case. Hope this doesn't confuse you. Forget about it if you don't get it )
     
    Last edited: Apr 28, 2009
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