Question 1205775
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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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<pre>
(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 = {{{32/((16/60))}}} = {{{(32*60)/16}}} = 2*60 = 120 km/h^2.


     Here the ratio {{{16/60}}} is the time of 16 minutes converted to hours.

     Converted to m/s^2, this value of 120 km/h^2 is  

         {{{(120*1000)/(3600*3600)}}} = {{{12/(36*36)}}} = {{{1/(3*36)}}} = {{{1/108}}} 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  {{{32/2}}} = 16 km/h,

     or  {{{(16*1000)/3600}}} m/s = {{{16000/3600}}} m/s = {{{160/36}}} = {{{40/9}}} m/s = 4{{{4/9}}} m/s = 4.44444 m/s (rounded).

     This value  4{{{4/9}}} = 4.44444 m/s is the answer (ship's average speed in m/s) for part (b).
</pre>

Solved.