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)
Please see the attachment.
(a) Let be the distance between the boat and ...
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.