Determine the speed of the boat with respect to the shore in Example 3–10.

The relative velocity of an object is its velocity with respect to a frame of reference or an observer.

For example, if the velocity of an object A with respect to a frame of reference B is ${\overrightarrow{v}}_{\mathrm{AB}}$and the relative velocity of reference frame B with respect to another frame of reference C is ${\overrightarrow{v}}_{\mathrm{BC}}$, then the relative velocity of object A with respect to reference frame C can be written as:

${\overrightarrow{v}}_{\mathrm{AC}}={\overrightarrow{v}}_{\mathrm{AB}}+{\overrightarrow{v}}_{\mathrm{BC}}$

The magnitude of the velocity of the boat in still water is $v_{\mathrm{BW}}=1.85{\mathrm{ms}}^{-1}$.

The magnitude of the velocity of the water current with respect to the shore of the river is $v_{\mathrm{WS}}=1.50{\mathrm{ms}}^{-1}$.

Let the speed of the boat with respect to the shore be ${\overrightarrow{v}}_{\mathrm{BS}}$.

The velocity of the boat relative to the shore is equal to the vector sum of the velocity of the boat with respect to the water current and velocity of the water current with respect to the shore, i.e.,

${\overrightarrow{v}}_{\mathrm{BS}}={\overrightarrow{v}}_{\mathrm{BW}}+{\overrightarrow{v}}_{\mathrm{WS}}$

Since the water current is directed westward, it will drag the boat westward. Therefore, if the boat wants to travel north directly, then it must have velocity in the north-east direction with respect to the water such that the velocity of the boat with respect to shore is directed across the river. This is shown in the figure below:

Here,$\theta$ is the angle which the velocity vector of the boat with respect to the water makes with the velocity vector of the boat with respect to the shore.

From the figure, the speed of the boat with respect to the shore can be found using Pythagoras theorem as:

$\begin{array}{c}{v}_{\mathrm{BS}}=\sqrt{v_{\mathrm{BW}}^2-v_{\mathrm{WS}}^2}\\ =\sqrt{{\left(1.85{\mathrm{ms}}^{-1}\right)}^2+{\left(1.20{\mathrm{ms}}^{-1}\right)}^2}\\ =\sqrt{\left(3.42-1.44\right){\mathrm{m}}^2{\mathrm{s}}^{-2}}\\ =\sqrt{1.98{\mathrm{m}}^2{\mathrm{s}}^{-2}}\\ =1.41{\mathrm{ms}}^{-1}\end{array}$

Thus, the speed of the boat with respect to the shore is $1.41{\mathrm{ms}}^{-1}$.