10am.-7am.= 3 hours due east from port..
at 10 am it changes direction to the vector specified above..so
2pm-10am= 4 hours from the point where it shifts direction
east of north means vehicle pointing north changes direction 57 deg.. to east..
so..3hrs. X 34 knots = 102naut. miles from port and due east..
and 4hrs X 34 knots= 136 naut miles. from the point of shifting direction..
if you draw it. it would form an oblique triangle..
with the angle for the unknown longer side x to be 147 degrees..
to solve for the longer side use cosine law..c^2=a^2+b^2-2abCosC
where c=147 degrees, a=102 mi. b=136 mi.
so the distance (inclined distance) of the ship from port is= 228.40 naut miles..
if you are asking the total distance travelled due east till it shifts then it would be 238 naut miles..
A ship leaves port at 7 am and heads due east at 34 knots. At 10 am, to avoid a storm the ship changes course to N 57° east of north). Find the ships distance from port at 2 pm. Round to the nearest tenth.
To determine nautical miles multiply the speed in knots by the number of hours.
(ex. 10knots in 2hrs would be;
10knots x 2hrs = 20 nautical miles.)
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