Question

Modern oil tankers weigh over a half-million tons and have lengths of up to one-fourth mile. Such massive ships require a distance of 5.1 km (about 3.2 mi) and a time of 18 min to come to a stop from a top speed of 34 km/h.
(a) What is the magnitude of such a ship's average acceleration in m/s2 in coming to a stop?



(b) What is the magnitude of the ship's average velocity in m/s?

Answer 1

(a) $-1.46\cdot 10^{-4} m/s^2$

The average acceleration of the ship is given by

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time elapsed

Here we have:

$u=34 km/h =9.44 m/s$ is the initial velocity

v = 0 is the final velocity

$t=18 min =64800 s$ is the time elapsed

Substituting, we find

$a=\frac{0-9.44 m/s}{64800 s}=-1.46\cdot 10^{-4} m/s^2$

(b) 4.72 m/s

Assuming the acceleration is uniform, the average velocity of the ship is given by:

$v_{avg} = \frac{v+u}{2}$

where

v is the final velocity

u is the initial velocity

Here we have:

v = 0

u = 9.44 m/s

So the average velocity of the ship is

$v_{avg} = \frac{0+9.44 m/s}{2}=4.72 m/s$