A steel ball weighing 128 pounds is suspended from a spring. This stretches the spring 12865 feet. The ball is started in motion from the equilibrium position with a downward velocity of 8 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 this means that the positive direction for y is down.)

Answer and Step-by-step explanation::

Downward force mg = upward force kδg

Spring stiffness k = Load/extension = 128/(128/257) = 257 lb/ft

At instant t when spring extension from equilibrium position is y, downward force on the ball is weight mg and upward forces are (k(y+δ))g and air resistance.

But air resistance = (4v)g. Hence, by Newton's second law, my'' = mg - k(y+δ)g - 4vg

Further, v = dy/dt = y' and mg = kδg

So, my'' = -kyg - 4y'g

or, (m/g)y'' + 4y' + ky = 0,

Boundary conditions: y(0) = 0, y'(0) = 8 ft/s

Characteristic equation is (m/g)r^2 + 4r + k = 0

r = (g/2m)[-4 ± √(16-4mk/g)]

r = (32.2/2*128)[-4 ± √(16-4*128*257/32.2)] = (-0.503125 ± 8.024 i)

Roots are imaginary

So, y = e^(-0.503125*t) [C1*Cos (8.024*t) + C2*Sin (8.024*t)]

Putting y(0) = 0 we get, C1 = 0.

So, y' = C2*e^(-0.503125*t)* [8.024*Cos (8.024*t) + Sin(8.024*t)*(-0.503125)]

Putting y'(0) = 8 ft/s, we get C2 = 0.9969

Thus, y = 0.9969*e^(-0.503125*t) Sin (8.024*t)

the first one

step-by-step explanation:

just id the test.

45.36 feet per second.

step-by-step explanation:

to find the most accurate determination for the number of feet per second tyrone traveled on his way to work let us convert miles to feet and minutes to seconds.

1 mile = 5280 feet.

1 minute = 60 seconds.

now we will divide 158400 by 3492 to find the most accurate determination for the number of feet per second tyrone traveled on his way to work.

therefore, tyrone traveled on his way to work at the rate of 45.36 feet per second.