Question

A boat is traveling at speed v0 on a lake when it turns off its engine. The boat is subject to a drag force that is proportional to the cube of the boat’s speed. In other words Fd = −bv3 Find the time it takes for the boat’s speed to decrease to half its initial value.

Answer 1

`t = 1.5\cdot (m)/(b\cdot v_(o)^(2))`

Step-by-step explanation:

The equation of equilibrium for the boat is:

`\Sigma F = - F_(D) = m\cdot a`

`-b\cdot v^(3) = m \cdot a`

`m\cdot (dv)/(dt) = - b \cdot v^(3)`

This is a first-order linear differential equation with separable variables:

`-(m)/(b)\cdot (dv)/(v^(3)) = dt`

`-(m)/(b) \int\limits^{v_(f)}_{v_(o)} {v^(-3)} \, dv = t`

`-(m)/(b)\cdot \left(-0.5\cdot v^(-2) \right)|_{v_(o)}^{v_(f)} = t`

`0.5\cdot (m)/(b)\cdot \left(v_(f)^(-2) - v_(o)^(-2) \right) = t`

The time taken to decrease the speed of the boat to half of its initial value is:

`t = 0.5\cdot (m)/(b)\cdot \left[\left((1)/(2) \right)^(-2) - 1 \right]\cdot v_(o)^(-2)`

`t = 1.5\cdot (m)/(b\cdot v_(o)^(2))`