Asked by lei
A ship leaves a port at 1:00 PM and sails in the direction N 24° W at the rate of 10 km/p for 30 minutes. It then sails N 66°E for 1.2 kilometers at the same speed. Find its distance and bearing from the port at 2:42 PM.
Answers
Answered by
oobleck
If the port P is at (0,0)
It turns at Q and ends up at R, then
PQ = 10 km/hr * 1/2 hr= 5km
QR = 1.2 km at 10 km/hr = 0.12 hr = 7:20
total time is 0:30 + 7:20 = 37:20
You don't say what it does between 1:37:20 and 2:42 PM
In any case, use the law of cosines to find distances, and then find bearings as usual.
It turns at Q and ends up at R, then
PQ = 10 km/hr * 1/2 hr= 5km
QR = 1.2 km at 10 km/hr = 0.12 hr = 7:20
total time is 0:30 + 7:20 = 37:20
You don't say what it does between 1:37:20 and 2:42 PM
In any case, use the law of cosines to find distances, and then find bearings as usual.
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