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

Answer:

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

Answer 2

Answer:

a

    $v = 0.28 \ m/s$

b

    $a = 9.8 m/s^2$

c

$a = 4.55 \ m/s^2$

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

$v = \frac{H}{t}$

=>        $v = \frac{1.50}{5.40}$

=>       $v = 0.28 \ m/s$

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

      $a = 9.8 -35(0.150)$

=>    $a = 4.55 \ m/s^2$