Time taken to reach 90% of terminal velocity

In summary, the conversation discusses finding the time it takes for a 4.5 kg metal sphere to reach 90% of its terminal velocity in a fluid with a drag coefficient of 10.5 N s^2/m^s. The equations used include the force of drag and the sum of forces equations, and the kinematics equations for constant acceleration are not applicable due to the changing acceleration. There is a method to solve this without graphing, but the exact method is unknown. The conversation also mentions solving the problem numerically using a spreadsheet and solving it using calculus, although the latter is a more difficult approach.
  • #1
newageanubis
15
0

Homework Statement


A 4.5 kg metal sphere is released in a fluid where k = 10.5 N s^2/m^s. How long does it take to reach 90% of its terminal velocity?

Homework Equations


Force of drag = kv^2, where k is the drag coefficient
(I believe we're not considering buoyancy.)

The Attempt at a Solution


ƩF = ma
a = (mg-kv^2)/m
a = [(4.5)((9.8) - (10.5)v^2]/4.5
a = 9.8 - 2.3v^2

At terminal velocity,
ƩF = 0
mg = kv^2
v = √(mg/k)
v = √[(4.5)(9.8)/10.5]
v = 2.0 m/s [down]
90% of this is 1.8 m/s [down].

So I know I have to find the time taken for the ball to achieve a velocity of 1.8 m/s^2 [down], and a have an equation with both acceleration and velocity. However, acceleration is not constant, so all of my kinematics knowledge (the constant acceleration equations) are useless, so I don't know how to proceed. We're supposed to create a graph to get the answer, but my teacher said there's a way to do this without graphing. I'm trying to find out what this method is. Thanks!
 
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  • #2
Disregard what I had here before, misread what time you were trying to find. Not sure what other way to do it.
 
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  • #3
I don't see how to get it with a graph! I would be very interested in seeing how you do it.

I did solve it numerically on a spreadsheet, with headings for t, d, v and a.
You CAN use the constant accelerated motion formulas for short time intervals (the smaller the interval the more accurate they are, and you can make the time interval between rows on the spreadsheet as small as you like; keep decreasing it until the answer no longer changes significantly).

It should be possible to solve it with calculus, Replace a with dv/dt and it becomes a differential equation. Not an easy one, though. I don't know how to solve it.
 

1. What factors affect the time taken to reach 90% of terminal velocity?

The time taken to reach 90% of terminal velocity is affected by several factors, including the mass and shape of the object, the density of the fluid it is falling through, and the gravitational pull of the Earth.

2. How does air resistance impact the time taken to reach 90% of terminal velocity?

Air resistance plays a significant role in determining the time taken to reach 90% of terminal velocity. As an object falls, air resistance increases, slowing down its acceleration and ultimately causing it to reach a constant velocity known as terminal velocity. The greater the air resistance, the longer it will take for an object to reach 90% of this final velocity.

3. What is the formula for calculating the time taken to reach 90% of terminal velocity?

The formula for calculating the time taken to reach 90% of terminal velocity is t = (1/9.8) * ln(10/(10-9vt)), where t is the time, vt is the terminal velocity, and ln is the natural logarithm. This formula takes into account the effects of air resistance on the object's acceleration.

4. Can the time taken to reach 90% of terminal velocity be manipulated?

The time taken to reach 90% of terminal velocity can be manipulated by changing the factors that affect it, such as the object's mass, shape, and the density of the fluid it is falling through. For example, a lighter and more aerodynamic object will reach 90% of terminal velocity faster than a heavier and less streamlined object.

5. How does altitude affect the time taken to reach 90% of terminal velocity?

Altitude can affect the time taken to reach 90% of terminal velocity due to changes in air density. At higher altitudes, the air is less dense, resulting in less air resistance. This means that an object falling at a high altitude will reach 90% of terminal velocity faster than the same object falling at a lower altitude.

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