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A boat leaves a dock at 6:00 PM and travels due south at a speed of 20 km/h. Another boat has been heading due east at 15 km/h and reaches the same dock at 7:00 PM. How many minutes after 6:00 PM were the two boats closest together?

User ABu
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1 Answer

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The two boats are closest together 0.36 hours after 6:00 PM, which is 21.6 minutes after 6:00 PM.

The two boats are closest together when the distance between them is minimized.

The boat traveling due south at 20 km/h is at position (0, -20t) at time t, and the boat traveling due east at 15 km/h is at position (-15 + 15t, 0) at time t.

The distance between the two boats at time t is given by the distance formula:

D(t) = sqrt((0 - (-15 + 15t))^2 + (-20t - 0)^2)

= sqrt((15 - 15t)^2 + 400t^2)

= sqrt(225 - 450t + 225t^2 + 400t^2)

= sqrt(625t^2 - 450t + 225)

To find the time t at which the distance between the two boats is minimized, we can take the derivative of D(t) with respect to t, set it equal to 0, and solve for t:

dD/dt = (1/2)(625t^2 - 450t + 225)^(-1/2)(1250t - 450) = 0

1250t - 450 = 0

t = 450/1250

t = 0.36 hours

Therefore, the two boats are closest together 0.36 hours after 6:00 PM, which is 21.6 minutes after 6:00 PM.

User Yanet
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