Question

A small piece of Styrofoam packing material is dropped from a height of 1.90 m above the ground. Until it reaches terminal speed, the magnitude of its acceleration is given by a = g − Bv. After falling 0.400 m, the Styrofoam effectively reaches terminal speed, and then takes 5.40 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

a

`v  =  0.28 \  m/s`

b

`a =  9.8  m/s^2`

c

`a = 4.55 \ m/s^2`

Step-by-step explanation:

From the question we are told that

The fall height is
`h = 1.90 \ m`

The magnitude of the acceleration is
`a = g -Bv`

The height at which terminal speed is attained is
`h_t = 0.400 \ m`

The time taken to reach the ground from the terminal velocity height is t = 5.40 s

Generally the height which object traveled with terminal velocity is

`H = h- h_t`

=>
`H = 1.90 - 0.400`

=>
`H = 1.50 \ m`

Generally the terminal velocity is mathematically represented as

`v = (H)/(t)`

=>
`v  =  (1.50)/(5.40)`

=>
`v  =  0.28 \  m/s`

Generally at terminal velocity , acceleration is zero so

`g -Bv = 0`

substituting

`9.8 \ m/s^2`

=>
`9.8 -B (0.28) = 0`

=>
`v = 35 \  m/s`

Generally at t = 0 s the velocity v = 0 (That no motion at time zero)

So from acceleration equation

`a =  9.8  -35(0)`

=>
`a =  9.8  m/s^2`

Generally when the speed v = 0.150 m/s

The acceleration is

`a =  9.8  -35(0.150)`

=>
`a =  4.55 \  m/s^2`

Answer 2

Step-by-step explanation:

If a small piece of Styrofoam packing material is dropped from a height of 1.90 m above the ground and reaches a terminal speed after falling 0.400m, the Change in distance will be 1.90m - 0.400 = 1.50m

If it takes 5.4secs fo r the Styrofoam to reach the ground, the terminal velocity will be expressed as;

Vt = change in distance/time

Vt = 1.5m/5.4s

Vt = 0.28m/s

Note that the Styrofoam reaches its final velocity when the acceleration is zero.

To get the constant value B from the equation a = g-Bv

a = 0m/s²

g = 9.81m/s²

v = 0.28m/s

Substituting the parameters into the formula.

0 = 9.81-0.28B

-9.81 = -0.28B

Divide both sides by -0.28

B = -9.81/-0.28

B = 35.04

b) at t = 0sec, the initial terminal velocity is also zero.

Substituting v = 0 into the equation to get the acceleration.

a = g-Bv

a = g-B(0)

a = g

Hence the acceleration at t =0s is equal to the acceleration due to gravity which is 9.81m/s²

c) Given speed v = 0.150m/s

g = 9.81m/s²

B = 35.04

Substituting the given data into the equation a = g-Bv

a = 9.81-35.04(0.15)

a = 9.81 - 5.26

a = 4.55m/s²