Answer:
22 minutes
Explanation:
Let t be the time after 6 PM.
D be the distance
During time t, the First boat travelling south = 20t km.
Meanwhile, the second boat travelling east is= 15t km,
Then the second boat started 15 km from the dock, so the distance from the dock is= (15 - 15t) km.
The distance D between the two boats is is expressed as = (15 - 15t)^2 + (20t)^2 = D*2
Let differentiate with respect to time and making dD/dt to 0 (D minimization is important),
then 2(15 - 15t)(-15) + 2(20t)(20) = 0
Therefore t = 9/25 h or 0.36 h or 22 minutes
The boats were closest together for approximately ; 22 minutes
Given data :
Boat 1 leaves at 6:00 pm
Boat 1 travels due south at 20 km/h
Boat 2 travels due east at 15 km/h
Time of arrival for both boats = 7:00pm
Determine the number of minutes both boats were closest after 6 pm
Time After 6 pm = t
Speed of boat 1 travelling south wards after 6 pm ( t ) = 20 t
Speed of boat 2 traveling east after 6 pm = 15 t
Distance of Boat 2 from Dock = 15 - 15 t
Distance of Boat 1 = 20t
∴ The distance between the boats
= ( 15 - 15 t )² + ( 20t )² = D * 2 ----- ( 1 )
Differentiating equation ( 1 ) ( note : make dD/dt = 0 )
2( 15 - 15 t )* (-15) + 2 (20t)*(20)
make t subject of the expression
t = 9 /25
= 0.36 h ≈ 22 minutes
Hence we can conclude that The boats were closest together for approximately ; 22 minutes
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