# To cross a river that is 1 km wide, with a current of 6km/h, anovice boatskipper holds the bow of the boat perpindicularto thefar river bank,intending to cross to a point directly across the river from thelaunch point. At what point will the boat actually contactthe far back? What direction should the boat have been headedto actually reach a point directly across from the launchdock? The boat is capable of 10km/h.

To cross a river that is 1 km wide, with a current of 6km/h, anovice boatskipper holds the bow of the boat perpindicularto thefar river bank,intending to cross to a point directly across the river from thelaunch point. At what point will the boat actually contactthe far back?
What direction should the boat have been headedto actually reach a point directly across from the launchdock? The boat is capable of 10km/h.
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Gillian Howell
Since distance = rate * time, and the distance the boat has tocross is 1km and the rate is 10km/hr, we can solve for the time ittakes as follows:
1 km = 10km/hr * t so t = 0.1 hrs.
Then, since the current is pushing the boat downstream at arate of 6km/hr, and the boat takes 0.1 hrs to cross, the boat willbe pushed downstream 6km/hr * 0.1 hrs = 0.6 km.
Thus the boat will actually contact the bank at 0.6 kmdownstream.
For the next part, let's assume that the boat takes off at anangle of 'x' from upstream. (In other words if the boat wereheading completely upstream x would equal 0). It might helpto draw yourself a little diagram at this point.
Now, the key is that the component of the boat's velocityheading upstream should counteract the 6km/hr current pushing itdownstream. Since the boat moves at 10km/hr, the component ofthe motion heading upstream is actually 10km/hr * cos(x). Wewant this to equal 6 km/hr. Hence we set the two equal andsolve for x:
10 km/hr * cos(x) = 6 km/hr
cos(x) = 0.6
x = 53.13o from upstream.
Thus if the boat takes off at the angleof 53.13o from upstream, it will reach thepoint directly across from the launch dock.
Cheers!
53.13o