3. a plane drops a crate of medical supplies by parachute. at the moment the parachute opens, the velocity…

3. a plane drops a crate of medical supplies by parachute. at the moment the parachute opens, the velocity of the crate satisfies the differential equation $\frac{dv}{dt}=-4(v + 8)$, where $v$ is measured in feet per second, $t$ is measured in seconds, and $v(0)=-40$. the equation for the velocity at any time $t$ is (a) $v=-8 - 32e^{-4t}$ (b) $v=-40e^{-4t}$ (c) $v=8 + 32e^{-4t}$ (d) $v=8 - 48e^{-4t}$

3. a plane drops a crate of medical supplies by parachute. at the moment the parachute opens, the velocity of the crate satisfies the differential equation $\frac{dv}{dt}=-4(v + 8)$, where $v$ is measured in feet per second, $t$ is measured in seconds, and $v(0)=-40$. the equation for the velocity at any time $t$ is (a) $v=-8 - 32e^{-4t}$ (b) $v=-40e^{-4t}$ (c) $v=8 + 32e^{-4t}$ (d) $v=8 - 48e^{-4t}$

Answer

Explanation:

Step1: Separate variables

We have the differential equation $\frac{dv}{dt}=-4(v + 8)$. Separate variables: $\frac{dv}{v + 8}=-4dt$.

Step2: Integrate both sides

Integrate $\int\frac{dv}{v + 8}=\int-4dt$. The left - hand side integral is $\ln|v + 8|=-4t + C$.

Step3: Solve for $v$

Exponentiate both sides: $v+8 = Ce^{-4t}$, so $v=Ce^{-4t}-8$.

Step4: Use the initial condition

Given $v(0)=-40$, substitute $t = 0$ and $v=-40$ into $v=Ce^{-4t}-8$. We get $-40=C - 8$, so $C=-32$.

Step5: Find the velocity equation

Substitute $C=-32$ into $v=Ce^{-4t}-8$, we have $v=-8-32e^{-4t}$.

Answer:

A. $v=-8 - 32e^{-4t}$