Question

) A motor boat weighs 32,000 lb and its motor provides a force of 5000 lb. Assume that the water resistance is 100 pounds per foot per second of the velocity of the boat. Then 1000 dv dt = 5000 − 100v. If the boat starts from rest, what is the maximum velocity of the boat?

Answer 1

Answer:

Maximum velocity, v = 50 ft/s

Step-by-step explanation:

Given

1000$\frac{dv}{dt}=5000-100v$    -----------(1)

Dividing (1) by 1000, we get

$\frac{dv}{dt}=5-\frac{v}{10}$

$\frac{dv}{dt}+\frac{v}{10}=5$     -----------------(2)

Now we can solve the above equation using method of integrating factors

$u(t)=e^{\int \frac{1}{10}dt}$

$u(t)=e^{\frac{1}{10}t}$

Now multiplying each side of (2) by integrating factor,

$e^{\frac{1}{10}t}(\frac{dv}{dt})+\frac{v}{10}e^{\frac{1}{10}t}=5e^{\frac{1}{10}t}$

Combining the LHS into one differential we get,

$\frac{d}{dt}\left ( e^{\frac{1}{10}t}v \right ) = \int 5e^{\frac{1}{10}t}.dt$

$e^{\frac{1}{10}t}v = 50e^{\frac{1}{10}t}$ + c

v(t)=50+ce

Appltying the initial condition v(0)=0, we get

$0=50+ce^{-\frac{1}{10}(0)}$

0=50+c

c=-50

So the particular solution is

$v(t)=50-50e^{-\frac{1}{10}t}$

$v(t)=50\left (1-e^{-\frac{1}{10}t}\right)$

Therefore, the maximum velocity is 50 ft/s