physics
posted by jhon on .
A motorboat travels due north at a steady speed of 3.0 m/s through calm water in which there is no current. The boat then enters an area of water in which a steady current flows at 2.0 m/s in a southwest direction, as shown in the next picture. Both the engine power and the course setting remain unchanged.
a) explain how the above paragraph gives information not only about the speed of the boat but also about its velocity.
b) draw a vector diagram showing the velocity of the boat and the velocity of the current. Use the diagram to find
i) the magnitude of the resultant velocity
ii) the angle between due north and the direction of travel of the boat.
c) calculate the distance the boat now travels in 5 mins;
d) mass of boat is 3.0 x 103kg (3000kg). Calculate the additional force that needs to be applied to give the boat an initial acceleration of 2.5 x 102m/s2 (0.025 m/s2).

a. When both speed and direction are
given, we call it velocity.
b. Draw a vector from the origin
pointihg due north to represent the velocity of the boat
Draw a vector from the origin 45o South of West(225o) to represent the velocity of the current.
1. Vr = 3m/s @ 90o + 2m/s @ 225o = Resultant velocity.
X = Hor. = 2*cos225 = 1.414 m/s.
Y = Ver. = 3 + 2*sin225 = 1.586 m/s.
(Vr)^2 = X^2 + Y^2 = 2 + 2.5 = 4.5
Vr = 2.12 m/s. = Resultant velocity.
2. cosA= X/Vr=1.586 / 2.12 = 0.66708
A = 132o. = Direction of travel of boat.
132  90 = 42o Between North and direction of travel of boat.
c. 5min. = 300 s.
d = Vr*t = 2.12m/s * 300s = 636 m.
3.