Question

A small piece of Styrofoam packing material is dropped from a height of 2.60 m above the ground. Until it reaches terminal speed, the magnitude of its acceleration is given by a = g − Bv. After falling 0.600 m, the Styrofoam effectively reaches terminal speed, and then takes 4.50 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)].

Answer 1

Answer:

Explanation:

a ) After the attainment of terminal speed , object takes 4.5 s to cover a distance of 2 m

So terminal speed V = 2 / 4.5

= .444 m /s

When it attains terminal speed , acceleration becomes zero

0 = g - B x .444

B = 22.25 s⁻¹

b ) At t = 0 , v = 0

a = g - B v

a = g at t = 0

c ) When v = .15

a = g - 22.25 x .15

= 9.8 - 3.31

= 6.5  m /s²