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
Answers
Answer :
Average acceleration,
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,
So, the magnitude of ship's acceleration is .
(b) Average velocity,
So, ship's average velocity is 13 km/h.
Hence, this is the required solution.
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.
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In this physics problem, by using the equation of motion, it was found that the time taken for a free falling object (under gravity and ignoring air resistance) to fall the second half of its total distance (t2) is sqrt(3) times greater than the time taken to fall the first half (t1).
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