Answer:
Vw = 0.98m/s due 67.16° north of east
Step-by-step explanation:
The distance the canoeist wants to travel is:
And the angle of the destination point is
![\alpha =atan((38)/(100) )=20.8\°](https://img.qammunity.org/2020/formulas/physics/college/7s4ohlkiu1ln2jg93konnx0wwrwvozepfc.png)
With this trajectory and the time of 42s, we get the velocity of the canoe respect to ground:
![Vc = (106.98<20.8\°)/(42s)=(2.547<20.8\°)m/s](https://img.qammunity.org/2020/formulas/physics/college/oiw5slc2320rrpdyrbpf9syp4n52laumcj.png)
Now we can calculate the velocity of the water:
![V_(c/w) = V_c - V_w](https://img.qammunity.org/2020/formulas/physics/college/4lzcuak079bell9bjhuzeq5ols5v5w2qro.png)
![V_w = V_c - V_(c/w)=(2.547<20.8\°)-(2<0\°)](https://img.qammunity.org/2020/formulas/physics/college/qg92w0yrgon2kwtdzs44zw4ic41m963urc.png)
![V_w = (0.98<67.16\°)m/s](https://img.qammunity.org/2020/formulas/physics/college/v7sg3f2zk2lazan8hqlp1n9v89qp7ew6r5.png)