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Velocity differential equations

  1. Apr 27, 2008 #1
    The 1997 Dodge Viper has mass 1547 kg and an engine that can produce a maximum driving force of 12.36 kN. Suppose drag is proportional to speed such that k = 164 Ns/m.
    (a) Determine the initial acceleration of the Viper from rest. (b) Determine the maximum speed (i.e. terminal velocity). (c) Calculate the distance and speed for t = 12.2 s. (d) Find time and distance for the car to reach 99% of its top speed.

    So basically I found the answers for a and b they are as follows.
    A)F=ma 12360N=1547kg(a) a=7.99N/kg
    B)F=ma Fcar-Fdrag= 0 Fcar=Fdrag F=kv 12360N=164(v) v=75.37

    So those two are checked and fine but c im havin trouble still
    c) F-Fd=ma F-kv=ma F-kv=m dv/dt
    integral 1/m dt= integral dv/(f-k(v))
    t/m= integral dv/(f-kv)

    so im stuck here. o and i think the integral limit is 0 and V. my teacher said to do
    u substitution and make u=F-kv and du=-kdv
    thereby making t/m= -1/k integral du/u with limit f and f-kv
    i have no idea what to do next with the integration. Basically i need someone to help integrate it and solve for v. then with that equation i can solve for part d. also i have no idea how to calculate the distance. o i need this for tomorow morning so any quick help would be very helpful thanks.
     
  2. jcsd
  3. Apr 27, 2008 #2

    alphysicist

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    Hi prezmoneymike,

    From a table of integrals, you can find:

    [tex]
    \int \frac{du}{u} = \ln |u| + C
    [/tex]

    where [tex]C[/tex] is the normal additive integration constant. What do you get with this?
     
  4. Apr 27, 2008 #3
    so it would be t/m= -1/k ln (f-kv) + c?
     
  5. Apr 27, 2008 #4

    alphysicist

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    You have to evaluate [itex]\ln u [/itex] at the beginning and ending points, and then subtract them in the usual way to get the result for the definite integral. What do you get?
     
  6. Apr 27, 2008 #5
    umm im not sure how to do that because i took calculus last year by a teacher who taught it horribly. so do u think u could show me how to accomplish this?
     
  7. Apr 27, 2008 #6

    alphysicist

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    Let me show you on a different integral (since the forum rules state I'm not supposed to do work for you).

    If your integral is

    [tex]
    \int\limits_a^b x\ dx
    [/tex]

    The formula from a table would be:

    [tex]
    \int\limits x\ dx= \frac{x^2}{2}
    [/tex]

    up to an additive constant. Then since the limits on your integral are from a to b, you would do:

    [tex]
    \left. \frac{x^2}{2} \right|_a^b = \frac{b^2}{2} - \frac{a^2}{2}
    [/tex]

    So you evaluate the expression from the integral at the ending point, then the beginning point, and then subtract (ending point expression - beginning point expression).
     
  8. Apr 27, 2008 #7
    so a= f and b= f-kv. so (f-kv)-f? making just -kv?
     
  9. Apr 27, 2008 #8
    to get rid of the ln dont i have to make something go to the e^??
     
  10. Apr 27, 2008 #9

    alphysicist

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    You have to keep the natural log there for now. So the result of your integral is:

    [tex]
    \ln (f-kv) - \ln (f)
    [/tex]

    Then you can use this in your full expression. You will probably want to combine these natural logs (which is possible since they are being subtracted).

    Later on, like you say you'll need to get rid of the ln by using the fact that:

    [tex]
    e^{\ln x} = x
    [/tex]
     
  11. Apr 27, 2008 #10
    so ln(-kv). then take the whole expression to the e. Making it e^t/m= e^(-1/k) * -kv. would this be right?
     
  12. Apr 28, 2008 #11

    alphysicist

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    The way that these natural logs combine is by using the property:

    [tex]
    \ln a - \ln b = \ln \left(\frac{a}{b}\right)
    [/tex]

    so here you will get:

    [tex]
    \ln \left(\frac{ f-kv}{f}\right)
    [/tex]
     
  13. Apr 28, 2008 #12
    ok then you would get e^t/m= e^(-1/k) * ((f-kv)/f)?
    then would you solve for v? how would accomplish this?
     
  14. Apr 28, 2008 #13

    alphysicist

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    Before this step you had:

    [tex]
    \frac{t}{m} = -\frac{1}{k} \ln \left(\frac{f-kv}{f}\right)
    [/tex]

    Then move the k to the other side and take the natural log:

    [tex]
    \begin{align}
    \frac{-kt}{m} &= \ln \left(\frac{f-kv}{f}\right)\nonumber\\
    e^{-kt/m} &= \frac{f-kv}{f}\nonumber
    \end{align}
    [/tex]

    and solve this for v.
     
  15. Apr 28, 2008 #14
    making it v= (f*e^-kt/m + f)/-k
    v= (12360N*e^((-164)(12.25)/1547) + 12360N)/164
    v= 95.93 m/s

    but this doesnt make sense because in part b the max speed was found to be 75.37. so did i do something wrong? and how would i calculate distance for this and part d?
     
  16. Apr 28, 2008 #15

    alphysicist

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    I think your equation should be:

    v = ( f e^{-k t /m} - f) / (-k)

    because the f that was originally in the numerator on the right side was positive, and so gained a negative sign when you moved it to the left.
     
  17. Apr 28, 2008 #16
    ok now it is 55.34 That makes more sense. ok so how do i figure out the distance?
     
  18. Apr 28, 2008 #17

    alphysicist

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    Remember how we found the velocity: we had an equation with the acceleration (which is dv/dt), the velocity and the time, and then we integrated it.

    So now, find an equation for velocity (which is dx/dt) at an arbitrary time t. (I think you can find one in the work you did in finding the velocity.) Then follow the same type of procedure. What do you get?
     
  19. Apr 28, 2008 #18
    yea i know what u mean i got it. thanks you so much for all your help. I think ive learned more from you than my calculus teacher lol. thanks again.
     
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