Question

"Suppose an object falling in the atmosphere has mass m=15kg and the drag coefficient is γ=9kg/s. Recall that the differential equation governing the motion of the particle is given by mv′=mg−γv(∗).If the velocity of the object at time t=0s is v(0)=9.55m/s, find the solution of the differential equation given in (∗) corresponding to this initial value (use g=9.8 and do not worry about units),Answer: v(t)= ?find the velocity of the object at time t=0.5s (be sure to enter units). Answer: ?"

Answer 1

Answer:

a. v(t)= -6.78$e^{-16.33t}$ + 16.33 b. 16.33 m/s

Step-by-step explanation:

The differential equation for the motion is given by mv' = mg - γv. We re-write as mv' + γv = mg ⇒ v' + γv/m = g. ⇒ v' + kv = g. where k = γ/m.Since this is a linear first order differential equation, We find the integrating factor μ(t)=$e^{\int\limits^ {}k \, dt }$ =$e^{kt}$. We now multiply both sides of the equation by the integrating factor.

μv' + μkv = μg ⇒ $e^{kt}$v' + k$e^{kt}$v = g$e^{kt}$ ⇒ [v$e^{kt}$]' = g$e^{kt}$. Integrating, we have

∫ [v$e^{kt}$]' = ∫g$e^{kt}$

    v$e^{kt}$ = $\frac{g}{k}$$e^{kt}$ + c

    v(t)=   $\frac{g}{k}$ + c$e^{-kt}$.

From our initial conditions, v(0) = 9.55 m/s, t = 0 , g = 9.8 m/s², γ = 9 kg/s , m = 15 kg. k = y/m. Substituting these values, we have

9.55 = 9.8 × 15/9 + c$e^{-16.33 * 0}$ = 16.33 + c

       c = 9.55 -16.33 = -6.78.

So, v(t)=   16.33 - 6.78$e^{-16.33t}$. m/s = - 6.78$e^{-16.33t}$ + 16.33 m/s

b. Velocity of object at time t = 0.5

At t = 0.5, v = - 6.78$e^{-16.33 x 0.5}$ + 16.33 m/s = 16.328 m/s ≅ 16.33 m/s