Question

A 500,000 kg cargo ship is traveling at 75 km/h when its engine is shut off. The magnitude of the frictional force between boat and water is proportional to the speed v of the boat: f k = 85v, where v is in meters per second and f k is in newtons. Find the time in hours for the ship to slow to one-third of its original speed because of the friction with the water.

Answer 1

Answer:1.8 hr

Explanation:

Given

mass of cargo=500,000 kg

initial speed$=75 km/h\approx 20.833 m/s$

Frictional force=85 v

at t=0 engine is shut off so acquired speed is slowly decreasing

$85v=-500,000\frac{\mathrm{d} v}{\mathrm{d} t}$

$85dt=-\frac{dv}{v}$

Integrating both sides

$85\int_{0}^{t}dt=-500,000\int_{20.83}^{6.944}\frac{dv}{v}$

$85t=500,000\ln \frac{1}{3}$

t=6462.425 s

t=1.8 hr