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December 18, 2014

December 18, 2014

Posted by **John** on Thursday, October 13, 2011 at 5:18pm.

- calculus -
**Reiny**, Thursday, October 13, 2011 at 5:37pmlet the time passed since noon be t hours.

So their paths form a right-angled triangle, with sides

(110 + 25t) and 15t

let d be the distance between them, then

d^2 = (110+25t)^2 + (15t)^2

2d dd/dt = 2(110+25t)(25) + 2(15t)(15)

when t = 4 , (4:00 pm)

d^2 = 210^2 + 60^2 = 47700

d = √47700

dd/dt = ( 50(210) + 30(60) )/(2√47700) = 28.16 mph

- calculus -
**John**, Thursday, October 13, 2011 at 5:48pmThat's not right...

- calculus -
**Reiny**, Thursday, October 13, 2011 at 5:54pmYou are right, I see my error.

I have ship A going west instead of east.

let the time passed since noon be t hours.

So their paths form a right-angled triangle, with sides

(110 - 25t) and 15t

let d be the distance between them, then

d^2 = (110-25t)^2 + (15t)^2

2d dd/dt = 2(110-25t)(25) + 2(15t)(15)

when t = 4 , (4:00 pm)

d^2 = 10^2 + 60^2 = 3700

d = √3700

dd/dt = ( 50(10) + 30(60) )/(2√3700) = 18.9 mph

- calculus -
**John**, Thursday, October 13, 2011 at 6:00pmYeah, I originally got 18.9 km/h, but the website is not accepting my answer. Is there something else?

- calculus -
**Steve**, Friday, October 14, 2011 at 10:11am2d dd/dt = 2(110-25t)

**(-25)**+ 2(15t)(15)

- calculus -
**some kid**, Friday, October 24, 2014 at 2:57amyou guys are retards

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