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Physics question. Help!?
A fishing boat in the ocean is moving at a speed of 20.0 km/h and heading in a direction of 40.0° east of north. A lighthouse spots the fishing boat at a distance of 19.0 km from the lighthouse and in a direction of 15.0° east of north. At the moment the fishing boat is spotted, a speedboat launches from a dock adjacent to the lighthouse. The speedboat travels at a speed of 52.0 km/h and heads in a straight line such that it will intercept the fishing boat.
a) How much time does the speedboat take to travel from the dock to the point where it intercepts the fishing boat?
b) In what direction does the speedboat travel? (Express the direction as a compass bearing with respect to due north.)
2 Answers
- ?Lv 76 years agoFavorite Answer
This is easiest for me to solve if I envision the original line of sight from the lighthouse to be due north.
That means the fishing boat is moving 40.0 - 15.0 = 25.0° east of north.
The eastward velocity of the fishing boat is 20.0sin25.0 = 8.452 km/hr
The northward velocity of the fishing boat is 20.0cos25.0 = 18.126 km/hr
The speed boat will make the shortest fastest trip if he matches his eastward velocity to that of the fishing boat so that the two boats remain on a line directly north/south of each other.
The speed boat makes northward progress at
√(52.0² - 8.452²) = 51.308 km/hr
The net northward speed difference is
51.308 - 18.126 = 33.182 km/hr
so the gap between them closes in
19 km / 33.182 km/hr = 0.5726 hr (or about 34 min 21 s)
the speed boat will be traveling in the direction θ to the right of our artificial north
sinθ = 8.452 / 52.0
θ = 9.35°
so the pilot should aim the speedboat 15.0 + 9.35 = 24.35° east of true north from the lighthouse to meet the fishing boat in the shortest time and distance.
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