Question

Dave rows a boat across a river at 4.0 m/s. the river flows at 6.0 m/s and is 360 m across.
a. in what direction, relative to the shore, does dave’s boat go?
b. how long does it take dave to cross the river?
c. how far downstream is dave’s landing point?
d. how long would it take dave to cross the river if there were no current?

Answer 1

a) Let's call x the direction parallel to the river and y the direction perpendicular to the river.

Dave's velocity of 4.0 m/s corresponds to the velocity along y (across the river), while 6.0 m/s corresponds to the velocity of the boat along x. Therefore, the drection of Dave's boat is given by:

$\theta= arctan(\frac{v_y}{v_x})=arctan(\frac{4.0 m/s}{6.0 m/s})=arctan(0.67)=33.7^{\circ}$

relative to the direction of the river.

b) The distance Dave has to travel it S=360 m, along the y direction. Since the velocity along y is constant (4.0 m/s), this is a uniform motion, so the time taken to cross the river is given by

$t=\frac{S_y}{v_y}=\frac{360 m}{4.0 m/s}=90 s$

c) The boat takes 90 s in total to cross the river. The displacement along the y-direction, during this time, is 360 m. The displacement along the x-direction is

$S_x = v_x t =(6.0 m/s)(90 s)=540 m$

so, Dave's landing point is 540 m downstream.

d) If there were no current, Dave would still take 90 seconds to cross the river, because its velocity on the y-axis (4.0 m/s) does not change, so the problem would be solved exactly as done at point b).