A boat leaves a dock at 1:00pm and travels due south at a speed of 10 mi/hr. Another boat has been heading due

let's call dock point (0,0).
@ 1 pm, first boat @ (0,0) heading vertically away.
@ time, second boat @ (10,0) heading horizontally toward (0,0).

we'll call 1 pm "time 0".

position of boat #1 given by:
x = 0, y = 10 t (where t hours after 1 pm)
position of boat #2 given by:
x = 10 - 5 t, y = 0.

use pythagorean theorem compute distance between them , minimize value.

distance = sqrt( (0 - 10 + 5t)^2 + (10 t - 0) ^2 )
since distance positive can minimize distance-squared well.

(0 - 10 + 5t) ^2 + (10 t)^2
(-10 + 5t) ^2 + (10t) ^2
100 - 100 t + 25 t^2 + 100 t^2
125 t^2 - 100 t + 100
derivative of
250 t - 100
minimum distance occurs when 0
250 t - 100 = 0
250 t = 100
t = 100 /250 = 10 / 25 = 2 / 5

2/5 hours after 1 pm or 1:24 pm.

let's check result:

@ 1:15, distance sqrt ( 8.75 ^2 + 2.5^2 ) = 9.1
@ 1:30, distance sqrt ( 7.5 ^ 2 + 5 ^ 2) = 9.01
@ 1:24, distance sqrt (8^2 + 4^2) = 8.94

not proof, support answer.

a boat leaves dock @ 1:00pm , travels due south @ speed of 10 mi/hr. boat has been heading due east @ 5mi/hr , reaches same dock @ 3:00 pm .at time 2 boats closest together?

Science & Mathematics Mathematics Next