posted by Greg Knightly on .
A boat crosses a river of width 122 m in which the current has a uniform speed of 2.39 m/s. The pilot maintains a bearing (i.e., the direction in which the boat points) perpendicular to the river and a throttle setting to give a constant speed of 2.81 m/s relative to the water.
A.) What is the magnitude of the speed of the boat relative to a stationary shore observer? Answer in units of m/s.
B.)How far downstream from the initial position is the boat when it reaches the opposite shore? Answer in units of m
X = 2.81 m/s. = Speed of boat.
Y = -2.39 m/s. = Speed of current.
A. V^2 = X^2 + Y^2
V^2 = (2.81)^2 + (@.39)^2 = 13.61
V = 3.69 m/s.
B. tanA = Y/X = -2.39/2.81 = -0.85053
A = -40.4o = 40.4o South of East.
d = 122*tan(-40.4) = -103.8 m. = 103.8 m
X = 2.39 m/s = Speed of current.
Y = 2.81m/s = Speed of boat.
B. Tan A = Y/X = 2.81/2.39 = 1.17573
A = 49.6o = 40.4o E. of N.
Tan40.4 = d/122
d = 122*Tan40.4 = 103.8 m. Downstream.