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Modern oil tankers weigh over a half-million tons and have lengths of up to a quarter of a mile. Such massive ships require a distance of 5.0 km (about 3.0 mi) and a time of 22 min to come to a stop from a top speed of 26 km/h. What is the magnitude of such a ship's average acceleration in m/s2 in coming to a stop? What is the magnitude of the ship's average velocity in m/s?Comment on the potential of a tanker running aground

Sagot :

AL2006

Acceleration = (change in speed) / (time for the change)

Change in speed= (0 - 26 km/hr) = -26 km/hr

(-26 km/hr) x (1,000 m/km) x (1 hr / 3,600 sec) = -7.222 m/sec

Average acceleration = (-7.222 m/s) / (22 min x 60sec/min) = -0.00547 m/sec²

Average speed during the stopping maneuver =

               (1/2) (start speed + end speed) = 13 km/hr = 3.6111 m/sec

Chances of running aground:

       If he stays in the designated shipping lanes, he ought to be OK.

Answer :

Average acceleration,  [tex]a=-7.71\times 10^{-5}\ m/s^2[/tex]

Average velocity, v = 13 km/h

Explanation:

It is given that,

Initial velocity of ship, v = 26 Km/h

Final velocity of ship, u = 0

Distance covered by ship, d = 5 km

Time taken, t = 22 min = 0.36 h

(a) Average acceleration, [tex]a=\dfrac{v-u}{t}[/tex]

[tex]a=\dfrac{0-26\ km/h}{0.36\ h}[/tex]

[tex]a=-72.2\ km/h^2=-7.71\times 10^{-5}\ m/s^2[/tex]

So, the magnitude of ship's acceleration is [tex]a=-7.71\times 10^{-5}\ m/s^2[/tex].

(b) Average velocity, [tex]v_a=\dfrac{v+u}{2}[/tex]

[tex]v_a=\dfrac{26\ m/s+0}{2}[/tex]

[tex]v_a=13\ km/h[/tex]

So, ship's average velocity is 13 km/h.

Hence, this is the required solution.  

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