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Variable acceleration and terminal velocity

  1. Oct 14, 2013 #1
    1. The problem statement, all variables and given/known data

    A cube is falling from a height of 200 m (initial velocity =0) , it reaches it's terminal velocity . Given that :its mass is 100 kg , cross sectional area 40 cm2 and coefficient of drag 2. Putting only weight and drag in consideration , calculate its terminal velocity and time of the fall (g=10)

    2. Relevant equations

    W=mg , Fd= Cd*A*1/2*ρ*v2 ( ρ
    =1)
    3. The attempt at a solution
    it tried to solve it using calculus , but all the formulas i had was integration or differentiation in respect to time , and i didnt have a function of time . i figured out a relation between velocity and acceleration , Fr=W-FD, after working it out , it was something like this , a=(100)/(1000-(0.08 v2) , i don't know what to do next , i dont know if i can substitute v with some something to get a function time
     
    Last edited: Oct 14, 2013
  2. jcsd
  3. Oct 14, 2013 #2
    $$a=\frac{dv}{dt}=\frac{dv}{dt}.\frac{dx}{dx}=\frac{dv}{dx}.\frac{dx}{dt}=\frac{dx}{dt}.\frac{dv}{dx}=v.\frac{dv}{dx}$$

    Now, you have relation with which you don't need a function of time. And acceleration is a function of velocity.

    Edit: check your acceleration again, it should be a little different.
     
  4. Oct 14, 2013 #3
    Recheck value of Fd.

    Your expression of net acceleration also looks incorrect .

    mdv/dt = Fd-mg .

    Substituting value of Fd ,you will get mdv/(Fd-mg) =dt .

    Integrate with proper limits and you will get velocity as a function of time.What do you get ?
     
    Last edited: Oct 14, 2013
  5. Oct 14, 2013 #4
    He'll need to calculate velocity as a function of distance first, otherwise "velocity as a function of time" information would be useless.

    But yes, he does need to calculate "velocity as a function of time".
     
  6. Oct 14, 2013 #5
    okay , this is how i got a
    F= m/a
    100/a = (100*10)-(1/2 * 1 * 2 * (0.42) v2)
    100/a= 1000- (0.16 v2)
    a=100/(1000-0.16v2)
    if it's wrong please correct me
     
  7. Oct 14, 2013 #6
    I believe distance doesn't play a role in calculating terminal velocity ,unless you want to determine if the object achieves terminal velocity before reaching the ground.
     
  8. Oct 14, 2013 #7
    Hey, ##F=m*a##.

    Tanya already posted about this!!!.....understanding the situation is more important. But here is the result:

    taking +ve direction to be downwards, we have:

    $$mg-F_{d}=ma$$
    $$a=g-\frac{F_{d}}{m}$$
    Substitute the correct values, you'll get acceleration as a function of velocity.
     
  9. Oct 14, 2013 #8
    terminal velocity is just the condition when the object is in equilibrium. ##F_{d}=mg## for the terminal velocity in this question. But the question asks the time of fall, not the time to reach terminal velocity. Time of fall is surely enough, dependent on the distance.

    I hope you get the point!!!

    Edit: Yes, distance doesn't play important role here to calculate terminal velocity. But I was talking about velocity in my previous posts. And its time of fall depends on velocity and not terminal velocity. But to calculate final velocity and hence calculate time, we need "velocity as a function of distance"...................to reach 100% of terminal velocity you need infinite/very large time, because velocity reaches terminal velocity asymptotically with time.
     
    Last edited: Oct 14, 2013
  10. Oct 14, 2013 #9
    worst mistake i've ever made , sorry , i drank like 5 cups of coffee , didnt sleep , got a test in 2 days
     
  11. Oct 14, 2013 #10
    get some sleep, sleepiness and physics don't go hand in hand!!!!:tongue:
     
  12. Oct 14, 2013 #11
    Will do, thanks
     
  13. Oct 14, 2013 #12
    :thumbs:
     
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