(1 point) A hollow steel ball weighing 4 pounds is suspended from a spring. This stretches the spring 12 feet. The ball is started in motion from the equilibrium position with a downward velocity of 2 feet per second. The air resistance (in pounds) of the moving ball numerically equals 4 times its velocity (in feet per second) . Suppose that after t seconds the ball is y feet below its rest position. Find y in terms of t. (Note that the positive direction is down.) Take as the gravitational acceleration 32 feet per second per second.

(-0.377 - 1.117i)texp(-0.5 + 1.55i)t

Step-by-step explanation:

When the steel ball of mass m = 4 pounds is first suspended, the spring extends a distance of y₀ = 12 ft. From ky₀ = mg , k = mg/x₀ where k is the spring constant, and g = 32 ft/s² So, k = mg/y₀ = 4 × 32/12 = 10.67  

The force of air resistance F = -4v

The spring force after it has been extended y ft is F = -ky

The net force on the steel ball is thus F = ma = -4v - ky

-4v - ky = ma (v = dy/dt = y' and a =dy/dt = y")

-4v - ky - ma = 0 ⇒ 4v + ky + ma = 0 (v = dy/dt = y' and a =dy/dt = y")

4y' + ky + my" = 0 ⇒ my" + 4y' + ky = 0

y" + 4y'/m + ky/m = 0

y" + 4y'/4 + 32y/12 = 0

y" + y' + 8y/3 = 0

3y" + 3y' + 8y = 0. Let y" = D², y' = D, y = 1

3D² + 3D + 8 = 0

Using the quadratic formula,

D = [-3 ±√(3² - 4 × 3 × 8)]/(2 × 3) = [-3 ±√(9 - 96)]/(6) = [-3 ±√( - 87)]/(6) = [-3 ±√( - 9.33)]/(6) = -0.5 ±√-1.55 = -0.5 ± 1.55i

y(t) = C₁exp(-0.5 + 1.55i)t + C₂texp(-0.5 + 1.55i)t

y'(t) = (-0.5 + 1.55i)C₁exp(-0.5 + 1.55i)t + (-0.5 + 1.55i)C₂texp(-0.5 + 1.55i)t + (-0.5 + 1.55i)C₂exp(-0.5 + 1.55i)t

y(0) = 0, y'(0) = 2

y(0) = C₁exp(-0.5 + 1.55i) × 0 + C₂× 0 ×exp(-0.5 + 1.55i) × 0

y(0) = C₁ exp(0)  + 0 = C₁

0 = C₁

C₁ = 0

y'(0) = (-0.5 + 1.55i)C₁exp(-0.5 + 1.55i) × 0 + (-0.5 + 1.55i)C₂ × 0 × exp(-0.5 + 1.55i) × 0 + (-0.5 + 1.55i)C₂exp(-0.5 + 1.55i) × 0

2 =  (-0.5 + 1.55i) × 0 × exp(0) + 0 + (-0.5 + 1.55i)C₂exp(0)

2 =  0 + 0 + (-0.5 + 1.55i)C₂

2 = (-0.5 + 1.55i)C₂

C₂ = 2/(-0.5 + 1.55i) = 2(-0.5 - 1.55i)/2.6525 = (-1 - 3.1i)/2.6525 = -0.377 - 1.117i

So,

y(t) = C₁exp(-0.5 + 1.55i)t + C₂texp(-0.5 + 1.55i)t

= 0 × exp(-0.5 + 1.55i)t + (-0.377 - 1.117i)texp(-0.5 + 1.55i)t

= 0 + (-0.377 - 1.117i)texp(-0.5 + 1.55i)t

= (-0.377 - 1.117i)texp(-0.5 + 1.55i)t

a.   we have two

lines:   y = 4-x   and  

y = 8-x^-1

given two simultaneous equations that are both to be

true, then the solution is the points where the lines cross. the intersection is

where the two equations are equal. therefore the solution that works for both

equations is when

4-x = 8-x^-1

this is where the two lines will cross and that is the

common point that satisfies both equations.

b.   4-x = 8-x^-1

  x           4-x       8-x^-1

-3           7        

8.33

-2           6         8.5

-1           5         9

  0           4        

-

  1           3        

7

  2           2        

7.5

  3           1         7.67

the table shows that none of the x values from -3 to 3 is

the solution because in no case does

4-x = 8-x^-1

to find the solution we need to rearrange the equation to

find for x:

4-x = 8-x^-1

multiply both sides with x:

4x-x^2 = 8x-1

x^2+4x-1=0

x= -4.236, 0.236

therefore there are two points that satisfies the

equation.

find y:  

x=-4.236

y = 4-x   = 4 – (-4.236)

= 8.236

y = 8-x^-1 =   .236)^-1

= 8.236

x=0.236

  y = 4-x   = 4 – (0.236) = 3.764

y = 8-x^-1 =   8-(0.236)^-1

= 3.764

thus the two lines cross at 2 points:

(-4.236, 8.236) & (0.236, 3.764)

c.   to solve

graphically the equation 4-x = 8-x^-1

we would graph both lines: y = 4-x   and   y

= 8-x^-1

the point on the graph where the lines cross is the

solution to the system of equations.

just graph the points on part b on a cartesian coordinate

system and extend the two lines.   the

solution is, as stated, the point where the two lines cross on the graph.

step-by-step explanation:

mark brainliest!

lol.

it's ok we all do this sometimes.

-aparri