A boat, whose speed in still water is 2.50m/s, must cross a 285 m wide river and arrive at a point 118 m upstream from where it starts (Fig. 3–45). The pilot must head the boat at a 45.0° upstream angle to do so. What is the speed of the river’s current?

FIGURE 3-45 Problem 48

An object’s speed may be defined as the rate at which the object changes its position from one location to another. Its value can be positive or zero, but it cannot be negative.

Here, ${\overrightarrow{v}}_{bw}$is the velocity of the boat relative to the water, ${\overrightarrow{v}}_{ws}$is the velocity of the water relative to the shore, and ${\overrightarrow{v}}_{bs}$is the velocity of the boat relative to the shore.

Given data:

The velocity of the boat relative to the water is${\overrightarrow{v}}_{bw}=2.50\mathrm{m}/\mathrm{s}$.

The angle is $\theta =45°$.

The horizontal distance is $d_x=118\mathrm{m}$.

The vertical distance is $d_y=285\mathrm{m}$.

The horizontal component of the boat’s velocity relative to the water can be calculated as

$\begin{array}{l}{\left({\overrightarrow{v}}_{bw}\right)}_x={\overrightarrow{v}}_{bw}\sin \theta \\ {\left({\overrightarrow{v}}_{bw}\right)}_x=\left(2.50\mathrm{m}/\mathrm{s}\right)\sin \left(45°\right)\\ {\left({\overrightarrow{v}}_{bw}\right)}_x=1.77\mathrm{m}/\mathrm{s}\end{array}$

The vertical component of the boat’s velocity relative to the water can be calculated as

$\begin{array}{l}{\left({\overrightarrow{v}}_{bw}\right)}_y={\overrightarrow{v}}_{bw}\cos \theta \\ {\left({\overrightarrow{v}}_{bw}\right)}_y=\left(2.50\mathrm{m}/\mathrm{s}\right)\cos \left(45°\right)\\ {\left({\overrightarrow{v}}_{bw}\right)}_y=1.77\mathrm{m}/\mathrm{s}\end{array}$

The velocity of the boat relative to the shore can be calculated as

$\begin{array}{c}{\left({\overrightarrow{v}}_{bs}\right)}_x={\left({\overrightarrow{v}}_{bw}\right)}_x-\left({\overrightarrow{v}}_{ws}\right)\\ {\left({\overrightarrow{v}}_{bs}\right)}_x=\left(1.77\mathrm{m}/\mathrm{s}\right)-\left({\overrightarrow{v}}_{ws}\right)\end{array}$

The expression for the horizontal distance is given by

role="math" localid="1644905632831" $\begin{array}{c}{d}_x={\left({\overrightarrow{v}}_{bs}\right)}_{x}t\\ \left(118\mathrm{m}\right)=\left[\left(1.77\mathrm{m}/\mathrm{s}\right)-\left({\overrightarrow{v}}_{ws}\right)\right]t\dots \left(i\right)\end{array}$

The time taken by the boat can be calculated as

$\begin{array}{c}{d}_y={\left({\overrightarrow{v}}_{bw}\right)}_{y}t\\ \left(285\mathrm{m}\right)=\left(1.77\mathrm{m}/\mathrm{s}\right)t\\ t=161/\mathrm{s}\end{array}$

Substituting the value of time in equation (i) to calculate the velocity of the water relative to the shore,

$\begin{array}{c}\left(118\mathrm{m}\right)=\left[\left(1.77\mathrm{m}/\mathrm{s}\right)-\left({\overrightarrow{v}}_{ws}\right)\right]t\\ \left(118\mathrm{m}\right)=\left[\left(1.77\mathrm{m}/\mathrm{s}\right)-\left({\overrightarrow{v}}_{ws}\right)\right]\left(161/\mathrm{s}\right)\\ \left(1.77\mathrm{m}/\mathrm{s}\right)-\left({\overrightarrow{v}}_{ws}\right)=\left(0.73\mathrm{m}/\mathrm{s}\right)\\ \left({\overrightarrow{v}}_{ws}\right)=1.04\mathrm{m}/\mathrm{s}\end{array}$

Thus, the velocity of the water relative to the shore is$1.04\mathrm{m}/\mathrm{s}$.