1. A boat sails at a constant speed in a straight line. its position at time S is (30S- 300,10S +500). in the water there are two buoys, A and B. at positions A- (7100,2800) and B - (125700,42500).

a) write an expression in terms of S that is the square of the distance between the boat and buoy A at time S, simplify the answer.

b) use answer a and completing the square to find the closest distance as the boat passes buoy A (answer to nearest ten meters). what time did the closest distance occur?

c) show that the boat passes over buoy b and tell us in minutes how long it takes to get to buoy b. assuming the boat starts at S=0.

d) what is the speed of the boat in metres per second (to 1 decimal place) and it direction as a bearing (to nearest degree).

2. on a second boat trip the boat starts in the same position. (-300,500) and it heading N70^oE. its speed is 30 ms ^-1 but it sails in a current of 12 ms^-1 from direction S10^o E.

Va is velocity of boat in still water.
Vw is velocity of the current
V resultant velocity of the boat.

a) draw a diagram illustrating these three velocities as a triangle, what is the angle between Va and Vw.

b) use the triangle and trigonometry to calculate the overall speed of the boat in m/s^-1 to 1dp. and its direction (bearing) correct to nearest degree.

(without using component form for vectors)

Solution Summary

This solution is comprised of a detailed explanation to write an expression in terms of S that is the square of the distance between the boat and buoy A at time S, simplify the answer.

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