In an experiment to determine the viscosity of some oil, a ball is dropped into

Question

In an experiment to determine the viscosity of some oil, a ball is dropped into the oil. The position of the ball is given by the formula y(t)=A+Bt+CeDt, where y is given in m, and t is the time in s. At t=0, y(0)= 0.1 m, the y component of the velocity is zero and the y component of the acceleration is ay(0)= 0.02 m/s2. As t approaches infinity, the y component of the velocity approaches vy(infinity) = 0.55 m/s. What are the values for A, B, C, D.

I understand that I need to take the derivative of the position equation to get the velocity equation, and then from there take the derivative of the velocity equation to get the acceleration equation. I know I then have! enough information to shuffle some equations around to solve for the constants (A,B,C,D) but i am having a hard time deriving. Any explanations/step by step help would be greatly appreciated. Thank you!

Explanation / Answer

y(t) = A + Bt + C e^(-Dt)

for t = 0 , y = 0.1 m

0.1 = A + C ....... (i)

v(t) = dy/dt = B - CD e^(-Dt)

t = 0 , v = 0

B - C D = 0

B = C D ...... (ii)

a = dv/dt = C D^2 e^(-Dt)

0.01 = C D^2 ...... (iii)

at t = infinity, v = 0.55

v (infinity) = B = 0.55 m/s

B = 0.55 m/s ..........Ans

0.01 = CD (D) = B D

D = 0.01 / B = 0.01 / 0.55 = 0.018 ........Ans

putting in (iii), C = 0.01 / 0.018^2 = 30.25 ......Ans

A = 0.1 - C = - 30.15 ........Ans

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.