document.write( "Question 1181762: Modern Oil tankers weigh more than a half - million tons and have a length of up to one fourth - mile. Such massive ships require a distance of 5.0 km ( about 3.0 mi ) And a time of 20 minutes to come to a stop from top speed of 30 km/h
\n" ); document.write( "A. What is the magnitude of such a ship's average acceleration in m/s²
\n" ); document.write( " and coming to a stop ?
\n" ); document.write( "B. What is the magnitude of the ship's average velocity in m/s ?
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Algebra.Com's Answer #849966 by CPhill(1959)\"\" \"About 
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Here's how to calculate the average acceleration and velocity:\r
\n" ); document.write( "\n" ); document.write( "**A. Average Acceleration**\r
\n" ); document.write( "\n" ); document.write( "1. **Convert units:**
\n" ); document.write( " * Initial speed (vᵢ) = 30 km/h * (1000 m/km) * (1 h/3600 s) = 8.33 m/s
\n" ); document.write( " * Final speed (v_f) = 0 m/s (since it comes to a stop)
\n" ); document.write( " * Time (t) = 20 minutes * (60 s/minute) = 1200 s\r
\n" ); document.write( "\n" ); document.write( "2. **Calculate average acceleration:**
\n" ); document.write( " Average acceleration (a) = (v_f - vᵢ) / t = (0 m/s - 8.33 m/s) / 1200 s ≈ -0.00694 m/s²\r
\n" ); document.write( "\n" ); document.write( "3. **Magnitude of acceleration:**
\n" ); document.write( " The magnitude is the absolute value of the acceleration. |a| ≈ 0.00694 m/s²\r
\n" ); document.write( "\n" ); document.write( "**B. Average Velocity**\r
\n" ); document.write( "\n" ); document.write( "1. **Calculate average velocity:**
\n" ); document.write( " Average velocity (v_avg) = (vᵢ + v_f) / 2 = (8.33 m/s + 0 m/s) / 2 ≈ 4.17 m/s\r
\n" ); document.write( "\n" ); document.write( "**Answers:**\r
\n" ); document.write( "\n" ); document.write( "* **A. Magnitude of average acceleration:** Approximately 0.00694 m/s²
\n" ); document.write( "* **B. Magnitude of average velocity:** Approximately 4.17 m/s
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