Multiple choice

$A$ ship is moving in the westward direction with a speed of $10\ kmph$ and a ship $B\ 100km$ south of $A$, is moving northward with a speed oh $10\ kmph$. The time after which the distance between them becomes shortest, is

  1. $15\ h$
  2. $5\surd {2}\ h$
  3. $10\surd {2}\ h$
  4. $5\ h$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

After t hours, the horizontal separation is 10t and the vertical separation is 100 - 10t. The squared distance is (10t)^2 + (100 - 10t)^2, which is minimized when t = 5 hours.

AI explanation

Let ship A be at the origin, so ship B is 100 km north and moving south at 10 kmph while ship A moves west at 10 kmph. The square of the distance between them at time t is given by the expression 100 minus 10t squared plus 10t squared. To find the minimum distance, take the derivative of this expression with respect to t and set it to zero, which gives 400t minus 2000 equals 0. Solving this equation results in t equals 5, meaning the shortest distance occurs at 5 hours.

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