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December 21, 2014

December 21, 2014

Posted by **Anon** on Saturday, November 23, 2013 at 10:52pm.

- Calc -
**Reiny**, Sunday, November 24, 2013 at 12:29amAt a time of t hrs,

let the position of ship A be P and let the position of ship B be Q

Join PQ, and complete the large righ-angled triangle

having a base of 100 and a height of 15t + 30t or 45t

(the horizontal distance between them is always 100 km

PQ^2 = 100^2 + (45t)^2

2 PQ d(PQ)/dt = 0 + 2(45t)(45)

d(PQ)/dt = 2025t/D

at 4:00 , t = 4

PQ = √(100^2 + 180^2) = appr 205.913

d(PQ)/dt = 2025(4)/205.913

= 39.34 km/h

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