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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 4.3 km (about 2.7 mi) and a time of 16 min
to come to a stop from a top speed of 32 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?
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(a) This motion is uniform deceleration with constant deceleration.
Let "a" be the value of the constant deceleration.
From Physics (Mechanics and Kinematics), having given the initial speed of 32 km/h
and time 16 minutes to stop, we can calculate the average deceleration
a = = = 2*60 = 120 km/h^2.
Here the ratio is the time of 16 minutes converted to hours.
Converted to m/s^2, this value of 120 km/h^2 is
= = = m/s^2 = 0.00926 m/s^2 (rounded).
This value 0.00926 m/s^2 is the answer (average deceleration in m/s^2) for part (a).
(b) The speed is linear function of time in this motion.
The maximum value of the speed is 32 km/h; the minimum value is 0 km/h, when the ship stops.
The magnitude of the ship's average velocity in km/h is = 16 km/h,
or m/s = m/s = = m/s = 4 m/s = 4.44444 m/s (rounded).
This value 4 = 4.44444 m/s is the answer (ship's average speed in m/s) for part (b).
Solved.