Question

A 40000-Mg ocean liner has an initial velocity of 4 km/h. Neglecting the frictional resistance of the water, determine the time required to bring the liner to rest by using a single tugboat which exerts a constant force of 225 kN. (Round the final answer to the nearest whole number.)

Answer 1

197.2 s

Step-by-step explanation:

First of all, let's convert the mass of the ocean liner into kilograms:

`m=40,000 Mg = 40,000,000 kg = 4\cdot 10^7 kg`

and the initial velocity into m/s:

`u=4 km/h =1.11 m/s`

The force applied is

`F=-225 kN = -2.25\cdot 10^5 N`

So we can find first the deceleration of the liner:

`a=(F)/(m)=(-2.25\cdot 10^5 N)/(4\cdot 10^7 kg)=-5.63\cdot 10^(-3) m/s^2`

And now we can use the following equation:

`a=(v-u)/(t)`

with v = 0 being the final velocity, to find t, the time it takes to bring the liner to rest:

`t=(v-u)/(a)=(0-1.11 m/s)/(-5.63\cdot 10^(-3) m/s^2)=197.2 s`