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February 1, 2015

February 1, 2015

Posted by **Jo** on Tuesday, October 25, 2011 at 12:10am.

- Calculus -
**Reiny**, Tuesday, October 25, 2011 at 8:37amlet the time passed since the biker passed directly underneath be t seconds

Make a diagram showing the distance between them as d ft

I see it as

d^2 = (17t)^2 + (65+t)^2

when t=3

d^2 = 51^2 + 68^2

d = √7225 = 85

2d dd/dt = 2(17)(17t) + 2(65+t)

divide by 2 and solve for dd/dt

dd/dt = (17(17t) + 65+t)/d

when t=3 and d = 85

dd/dt = (17(51) + 68)/85 = 11 ft/s

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