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How far does it take for an object to reach maximum velocity

  1. Jun 23, 2015 #1
    1. The problem statement, all variables and given/known data
    At what distance (OR at what time) does it take for an object falling at 32 ft per sec per sec. to reach terminal velocity in its descent??

    2. Relevant equations
    The only equation I can think of to use here would be distance = (rate of descent) times time.

    3. The attempt at a solution
    ?? I need the time for this equation: distance = rate of descent times time
     
  2. jcsd
  3. Jun 23, 2015 #2

    tony873004

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    You need an air drag formula. Set it equal to the weight. At what point does air drag equal weight?
     
  4. Jun 23, 2015 #3
    What if I don't know the weight?
     
  5. Jun 23, 2015 #4
    How does setting the equation equal to the object's weight tell me at what point air drag is equal to the weight??
     
  6. Jun 23, 2015 #5

    haruspex

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    I'm unsure how to respond to this. It depends whether the question means literally achieving terminal velocity or getting within some margin.
    If it means the second, you need to know what margin, and it won't be something you can solve analytically. And you will need to make some assumption about how the drag force relates to speed.
    If it means the first, well, what would you guess?
     
  7. Jun 23, 2015 #6

    CWatters

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    Tonia - Is that the whole question you were set word for word? If so then it's possibly a trick question...

    Firstly.. An object reaches terminal velocity (eg it no longer accelerates) when the objects weight is exactly balanced by air drag. When that happens there is no net force acting on the object so it stops accelerating and continues falling at a constant velocity. To work out what the terminal velocity is you need to know the objects weight, and drag coefficient (which aren't stated in the question it seems).

    But never mind there is a worse problem that will actually make it easier to answer the question! The problem is that air drag is also proportional to velocity. So as the object accelerates drag increases and that reduces the rate of acceleration.

    To understand this consider this problem... Imagine you are standing 16 meters away from a wall. Each time your teacher rings a bell you are to walk forwards reducing the distance between you and the wall by a half. How many rings of the bell until you reach the wall?

    Ring , Distance
    0 , 16
    1 , 8
    2 , 4
    3 , 2
    4 , 1
    5 , 0.5
    6 , 0.25
    7 , 0.125
    8 , 0.0625
    etc

    In theory you get closer and closer but you never actually reach the wall.

    The terminal velocity question is a similar problem. Unless there is more information in the question a valid answer is that the object never quite reaches terminal velocity (before it hits the ground), although it might come exceptionally close to terminal velocity somewhere along the way.

    This is also what harpspex is hinting at. To calculate the distance you not only need to know the weight and drag coefficient but you also need to know "how close is close enough"? Within 5%? Within 1%?
     
  8. Jun 23, 2015 #7
    A relative asked the question, he didn't give the weight or the air drag. Apparently, the question can't be answered unless I have those two pieces of information.
     
  9. Jun 23, 2015 #8

    tony873004

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    In addition to the drag coefficient CD of the object, which is based on it shape, and the object's mass m, you also need to know the cross-sectional area A of the falling object, and the density of air, ρ. Air's density decreases as you gain altitude. Wikipedia gives you the air drag formula. Just set it equal to weight
    [tex]F_D = \frac{1}{2}\rho v^2 C_D A = mg[/tex]
    [tex]v = \sqrt{\frac{{2mg}}{{\rho C_D A}}}[/tex][/I]
    A coffee filter will hit terminal velocity in a fraction of a second, after falling only a few centimeters. A person will take a few seconds after falling hundreds of feet.
     
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