Answer:
If the canoe heads upstream the speed is zero. And directly across the river is 8.48 [km/h] towards southeast
Explanation:
When the canoe moves upstream, it is moving in the opposite direction of the normal river current. Since the velocities are vector (magnitude and direction) we can sum each vector:
Vr = velocity of the river = 6[km/h}
Vc = velocity of the canoe = -6 [km/h]
We take the direction of the river as positive, therefore other velocity in the opposite direction will be negative.
Vt = Vr + Vc = 6 - 6 = 0 [km/h]
For the second question, we need to make a sketch of the canoe and we are watching this movement at a high elevation. So let's say that the canoe is located in point 0 where it is located one of the river's borders.
So we are having one movement to the right (x-direction). And the movement of the river to the south ( - y-direction).
Since the velocities are vector we can sum each vector, so using the Pythagoras theorem we have:
![Vt = \sqrt{(6)^{2} +(-6)^{2} } \\Vt=8.48[km/h]](https://tex.z-dn.net/?f=Vt%20%3D%20%5Csqrt%7B%286%29%5E%7B2%7D%20%2B%28-6%29%5E%7B2%7D%20%7D%20%5C%5CVt%3D8.48%5Bkm%2Fh%5D)