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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 4.5 km (about 2.8 mi) and a time of 17 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?

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 4.5 km (about 2.8 mi) and a time of 17 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?

check_circleAnswer
Step 1

Given information:

Distance required to come to rest (s) = 4.5 km = 4500 m.

Time required to stop (t) = 17 min = 17 × 60 s = 1020 s.

Top speed of the ship (u) = 32 kmph = 32 × (5/18) m/s = 8.8 m/s

Step 2

From 3rd equation of motion we can find the acceleration (which is retardation) as follows:

v2 u2
2as
Since the final velocity of the ship is "Zero". And the rest of the values are given we get the
acceleration of the ship as:
a 0.0087 m/s2
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v2 u2 2as Since the final velocity of the ship is "Zero". And the rest of the values are given we get the acceleration of the ship as: a 0.0087 m/s2

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Step 3

The magnitude of average velocity (Vavg) is total dista...

4500
4.4 m/s
Vavg
1020
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Image Transcriptionclose

4500 4.4 m/s Vavg 1020

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