Calculating Resistive Force on Styrofoam Dropping from 2.00m

[Gear300]

I come back once more with a question:
So far, when it comes to resistive forces, I've only done resistance for slow moving objects in liquids and small particles through the air, along with larger objects going against heavier air resistance. This question refers to a smaller particle:

A small piece of styrofoam packing material is dropped from a height of 2.00m. Until it reaches terminal velocity (velocity when the net force is 0N), the acceleration is given as a = g - bv. After falling .500m, the styrofoam reaches terminal speed and takes 5.00s to reach the ground. I need to find the value of the constant b.

Thus far, resistance R = -bv and that mg (weight) is pulling the styrofoam down. That would imply ma = mg - bv, in which a = g - (b/m)v. Terminal velocity is when Vt = mg/b and is only approached, not reached, so the equation for v would be
v = Vt(1 - e^(-bt/m)). The question said that a = g - bv...so I would assume m = 1kg (a bit big for styrofoam)...or do I have to come up with another equation for this? and how would I incorporate displacement into the equations if the acceleration is not constant?

 

[PhanthomJay]

I come back once more with a question:
So far, when it comes to resistive forces, I've only done resistance for slow moving objects in liquids and small particles through the air, along with larger objects going against heavier air resistance. This question refers to a smaller particle:

A small piece of styrofoam packing material is dropped from a height of 2.00m. Until it reaches terminal velocity (velocity when the net force is 0N), the acceleration is given as a = g - bv. After falling .500m, the styrofoam reaches terminal speed and takes 5.00s to reach the ground. I need to find the value of the constant b.

Thus far, resistance R = -bv and that mg (weight) is pulling the styrofoam down. That would imply ma = mg - bv, in which a = g - (b/m)v. Terminal velocity is when Vt = mg/b and is only approached, not reached, so the equation for v would be
v = Vt(1 - e^(-bt/m)). The question said that a = g - bv...so I would assume m = 1kg (a bit big for styrofoam)...or do I have to come up with another equation for this? and how would I incorporate displacement into the equations if the acceleration is not constant?

You are making the problem complex. It is given that a = g-bv. Terminal velocity can be calculated knowing that it takes 5 seconds to travel 1.5m. What is the acceleration at terminal velocity? Solve for b.

 

[Gear300]

I see...I get what you're saying; with the time, I can find b knowing the at some point v is approximately Vt. I have just one more question. How did they find that
v = Vt(1 - e^(-bt/m)).

 

[Shooting Star]

I told you the other day -- by integration of the eqn of motion. You'll soon learn it.