A boat, whose speed in still water is 3.00 m/s , must cross a 285-m-wide river and arrive at a point 118 m upstream from start.

Question

To do so, the pilot must head the boat at a 45∘ upstream angle. What is the speed of the river's current?

Answer 1

Let's set up the river so that you are crossing from left to right (the starting shore is the y axis) and the current is moving downwards. You know that the distance traveled in the time to cross will be such that d_y/d_x = v_yt/v_xt = v_y/v_x. We can find the angle of the actual velocity of the boat with respect to earth by taking the tan^-1(d_y/d_x) = tan^-1(118 m/285 m) = 22.49°.

We want to find v_C so v = v_B + v_C where B=boat in still water, C=current and bold type mean vector variable, so this is a vector sum. We need to make the x and y components lead to the resultant vector that results in the right angle from above.

x component: v_Bcos(45°) = vcos(22.49°) we can find the magnitude, v, by substituting:

v = 3.0 m/s * cos(45°)/cos(22.49°) = 2.296 m/s (not surprising as you are heading partially upstream (against the current)) (There is no component of v_C in the x direction or v_Ccos(270°) = 0)

y component: v_Bsin(45) - v_C = vsin(22.49°), solve for v_C (formally v_C term is v_Csin(270°))

v_C = (3.0 m/s )sin(45) - (2.296 m/s)sin(22.49°) = 1.24 m/s

Please check all the arithmetic. Take care.