# A ferry is crossing a river. The ferry is headed due north with a speed of 2.8 m/s relative to...

## Question:

A ferry is crossing a river. The ferry is headed due north with a speed of 2.8 m/s relative to the water. The river's velocity is 4.9 m/s to the east. What is the magnitude of the boat's velocity relative to Earth?

## Vector Sum:

If a ferry crosses a river, the ferry is displaced according to the ferry's effort and the water currents. The velocity relative to the earth is the vector sum of the velocity due to the ferry's effort and the velocity of the water currents.

Taking the velocity of the river currents east i.e. {eq}v_r = 4.9\ \text{m/s} {/eq} and the velocity of the ferry's effort north {eq}v_f = 2.8\ \text{m/s} {/eq}, the resultant velocity v is the vector sum which is the Pythagorean hypotenuse of the velocities since north and east are perpendicular, i.e.

{eq}\begin{align*} v &= \sqrt{v_r^2+v_f^2}\\ &= \sqrt{(4.9\ \text{m/s})^2+(2.8\ \text{m/s})^2}\\ &\approx\color{blue}{\boxed{5.6\ \text{m/s}}} \end{align*} {/eq}

The angle associated with the resultant velocity can be computed using suitable a trigonometric ratio i.e.

{eq}\begin{align*} \tan\theta &= \frac{4.9\ \text{m/s}}{2.8\ \text{m/s}}\\ \theta &= \arctan{\frac{4.9\ \text{m/s}}{2.8\ \text{m/s}}}\\ &\color{blue}{\approx \boxed{60.3\ ^\circ}} \end{align*} {/eq} 