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Homework Help: Acceleration and Velocity differential equations

  1. Apr 21, 2008 #1
    [SOLVED] Acceleration and Velocity differential equations

    Hey I am having trouble solving this problem.

    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.

    I know you have to do something with an integral but I really dont know how. I have to explain this problem to the class tomorrow. so help on this would be much appreciated.
  2. jcsd
  3. Apr 21, 2008 #2
    for a and b you just need to use newtons law of acceleration.
    A) whats the net force on the car?
    B) what happens to the acceleration when the car hits terminal velocity? what does that tell you about the forces?
    C) you have to write a formula expressing the acceleration as a function of time, then you can integrate to find velocity, and again to find distance. (same for part d)
    Does that help?
  4. Apr 21, 2008 #3
    so when u say use n's law of accleration you mean what equation? sorry if I sound really dumb but lets just say im in physics AP and im a senior. so i kinda have had seniorittis this year so i really dont know all that much about physics. so if u could tell me a little more simply it would help more.
  5. Apr 21, 2008 #4
    If you don't know what 'Newton's law of acceleration' is (technically, Newton's second law) than I suggest you read up on basic physics. You are not going to get very far without knowing Newton's laws. (This is not meant to flame you are anything, just a bit of advice).

    Anyway, newton's second law is: [tex]\sum{\vec F} = m \vec a[/tex] or without vector notition just [tex]F_r = ma[/tex] where [tex]F_r[/tex] is the total resulting force.
  6. Apr 21, 2008 #5
    well nick89 thanks for the equation but school is over in another week for me so im just trying to finish up this little bit more.
    so it would be f=ma 12.36=1547*a a=0.0079??
    what would I do then? if someone could give me a step by step guide that be great.
  7. Apr 21, 2008 #6
    Fnet = MA = Facc + Fdrag
    Facc = the force the viper provides, Fdrag is the force of drag from air resistance (which should be negative - i.e. in the opposite direction as the accelerating force)?
    Does that help?
  8. Apr 21, 2008 #7
    so your saying just add 12.36kN and 164 Ns/m? and that equals Fnet?
  9. Apr 21, 2008 #8
    almost, the 12.36 is constant (the viper acceleration), but the drag force is dependent on velocity (linearly), i.e. Fdrag = - 164 * V (the velocity - and we see that Ns/m * m/s = N a force, as we would expect. just like 2x+3y, you can't add numbers in different units).
  10. Apr 22, 2008 #9
    I've looked at the question more thoroughly.

    For the first one, a, it asks you to find the initial acceleration from rest. This means there is no velocity yet which in turn means there is no drag (wind resistance) yet.
    So you can use F = ma to find a.

    The maximum speed is a little bit more difficult.
    Consider it like this. The formula F = ma can still be used here, however, since the car is at a maximum speed, the acceleration a will be 0. Thus, from the formula, the total net force F will also be zero.
    That means all forces acting on the car will cancel eachother out exactly.
    The only forces (in this example) are the car's driving force and the drag.
    This means:
    [tex]F_{car} - F_{drag} = 0[/tex] (drag is negative since it pulls the car back! The force points in the opposite direction than the driving force.)

    And thus: [tex]F_{car} = F_{drag}[/tex].

    You know both these forces (F_car is the given maximum car driving force, and F_drag is the drag force which was proportional to the speed.)

    [tex]F_{drag} = kv[/tex]

    The equation becomes:
    [tex]12.36*10^3 = 164v[/tex] , solve for v, which is the maximum speed.
    Last edited: Apr 22, 2008
  11. Apr 22, 2008 #10
    For question c it becomes yet again a bit harder.

    Now, we need to find the speed and distance for t = 12.2 s.
    This time, the net force does not equal 0 since there is still some acceleration at t = 12.2 s.
    Let's call F the driving force and [tex]F_d[/tex] the drag force. Now:
    [tex]F - F_d = ma[/tex] or substiting [tex]F_d[/tex]:
    [tex]F - kv = ma[/tex].

    Now you may know that the acceleration a is just the derivative of the speed v.
    So we can write: [tex]a = \frac{dv}{dt}[/tex]

    [tex]F - kv = m \frac{dv}{dt}[/tex]

    As you may see, this is a differential equation for v.

    If you know how to solve this equation than you will find a formula for v depending on time. Substitute t = 12.2 s in that equation and you will find the speed at that time.
  12. Apr 22, 2008 #11
    Thanks i get a and b and c now. anyone know d?
  13. Apr 22, 2008 #12
    Do you know the relation between distance and speed? (Watch out, the speed is not constant so you cannot simply use x = vt !)
  14. Apr 22, 2008 #13
    i dont know what equation to use for it. would i use 99% of the answer for b? or ..??
  15. Apr 23, 2008 #14
    Ok so you got a formule for the speed v that depends on time t right?
    You also got a value for the maximum speed, right?

    Now you can relate the two and solve for t ! something like this:
    [tex]v(t) = 0.99v_{\text{max}}[/tex]

    You now found the time to reach 99% of it's max speed, and you can use that time to determine the distance again just like you did in part c.
  16. Apr 23, 2008 #15
    thanks so much for the help nick89
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