1. **State the problem:** You want to swim so that your overall direction is North relative to a stationary observer, but the river current flows East at 0.40 m/s, and you can swim at 1.5 m/s relative to the water. 2. **Define variables:** - River current velocity vector: $\vec{v}_c = 0.40$ m/s East (bearing 090°). - Your swimming velocity relative to water: $v_s = 1.5$ m/s at an unknown bearing $\theta$ degrees East of North. - Resultant velocity relative to ground must be due North (bearing 000°). 3. **Set up velocity components:** - Let North be the $y$-axis and East be the $x$-axis. - Your swimming velocity components: $$v_{sx} = v_s \sin(\theta)$$ $$v_{sy} = v_s \cos(\theta)$$ - River current velocity components: $$v_{cx} = 0.40$$ $$v_{cy} = 0$$ 4. **Resultant velocity components:** - Total velocity in $x$ direction: $$v_x = v_{sx} + v_{cx} = 1.5 \sin(\theta) + 0.40$$ - Total velocity in $y$ direction: $$v_y = v_{sy} + v_{cy} = 1.5 \cos(\theta) + 0$$ 5. **Condition for traveling North:** - The resultant velocity must have zero East component: $$v_x = 0 \implies 1.5 \sin(\theta) + 0.40 = 0$$ 6. **Solve for $\theta$:** $$1.5 \sin(\theta) = -0.40$$ $$\sin(\theta) = -\frac{0.40}{1.5} = -0.2667$$ 7. **Calculate $\theta$:** $$\theta = \arcsin(-0.2667) \approx -15.47^\circ$$ 8. **Interpret angle:** - Negative angle means $15.47^\circ$ West of North. - Bearing is measured clockwise from North, so: $$\text{Bearing} = 360^\circ - 15.47^\circ = 344.53^\circ$$ 9. **Round to 3 significant figures:** $$\boxed{345^\circ}$$ **Final answer:** You must swim on a bearing of approximately $345^\circ$ to travel due North relative to a stationary observer.