A small piece of Styrofoam packing material is dropped from a height of 2.00 m a

Question

A small piece of Styrofoam packing material is dropped from a height of 2.00 m above the ground. Until it reaches terminal speed, the magnitude of its acceleration is given by a = g - bv. After falling 0.500 m, the styrofoam effectively reaches its terminal speed, and then takes 5.10 s more to reach the ground.

(a) What is the value of the constant b?

______________________s-1

(b) What is the acceleration at t = 0?

_____________________m/s2 (down)

(c) What is the acceleration when the speed is 0.150 m/s?

_____________________m/s2 (down)

Explanation / Answer

First we solve for vterm, then for b.
vterm = (2-0.5) m / 5.1 s = 0.294 m/s
1. At vterm, accel a = 0, so g = bv ==> b = g/v = 9.8/0.294= 33.33 s^-1
2. When t = 0, v = 0, so acceleration a = g = 9.8 m/s^2.
3. Acceleration a when v = 0.15 m/s = g-bv = 9.8-33.33*0.15 = 4.8005 m/s^2 (downward)

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