A passenger on a boat moving at 1.70 m/s on a still lake walks up a flight of stairs at a speed of 0.60 m/s (Fig. 3–43). The stairs are angled at 45°, pointing in the direction of motion as shown. What is the velocity of the passenger relative to the water?

FIGURE 3-43 Problem 42

First, draw the second vector by joining the tail of the second vector to the tip of the first vector. The result of these two vectors is the vector joining its tail to the tail of the first vector and its tip to the tip of the second vector.

The figure given below shows the vector addition of the velocity $\left({\overrightarrow{v}}_{bw}\right)$of the boat with respect to the water and the velocity $\left({\overrightarrow{v}}_{pb}\right)$of the passenger with respect to the boat.

Given data:

The velocity of the passenger with respect to the boat is $v_{pb}=0.60\mathrm{m}/\mathrm{s}$.

The angle of inclination of the stairs is $\theta =45°$.

The velocity vector of the boat with respect to the water is localid="1644815143534" ${\overrightarrow{v}}_{bw}=\left(1.70\mathrm{m}/\mathrm{s}\right)\hat{i}$.

The horizontal component of the velocity of the passenger with respect to the boat can be calculated as

$\begin{array}{l}{\left(v_{pb}\right)}_x=v_{pb}\cos \theta \\ {\left(v_{pb}\right)}_x=\left(0.60\mathrm{m}/\mathrm{s}\right)\cos \left(45°\right)\\ {\left(v_{pb}\right)}_x=0.424\mathrm{m}/\mathrm{s}\end{array}$

The vertical component of the velocity of the passenger with respect to the boat can be calculated as

$\begin{array}{l}{\left(v_{pb}\right)}_y=v_{pb}\sin \theta \\ {\left(v_{pw}\right)}_y=\left(0.60\mathrm{m}/\mathrm{s}\right)\sin \left(45°\right)\\ {\left(v_{pb}\right)}_y=0.424\mathrm{m}/\mathrm{s}\end{array}$

The velocity vector for the velocity of the passenger with respect to the boat can be calculated as

$\begin{array}{l}{\overrightarrow{v}}_{pb}={\left(v_{pb}\right)}_x\hat{i}+{\left(v_{pb}\right)}_y\hat{j}\\ {\overrightarrow{v}}_{pb}=\left(0.424\mathrm{m}/\mathrm{s}\right)\hat{i}+\left(0.424\mathrm{m}/\mathrm{s}\right)\hat{j}\end{array}$

The velocity vector for the velocity of the passenger with respect to the water can be calculated as

$\begin{array}{l}{\overrightarrow{v}}_{pw}={\overrightarrow{v}}_{bw}+{\overrightarrow{v}}_{pb}\\ {\overrightarrow{v}}_{pw}=\left(1.70\mathrm{m}/\mathrm{s}\right)\hat{i}+\left(0.424\mathrm{m}/\mathrm{s}\right)\hat{i}+\left(0.424\mathrm{m}/\mathrm{s}\right)\hat{j}\\ {\overrightarrow{v}}_{pw}=\left(2.12\mathrm{m}/\mathrm{s}\right)\hat{i}+\left(0.424\mathrm{m}/\mathrm{s}\right)\end{array}$

The magnitude of the velocity vector for the velocity of the passenger with respect to the water can be calculated as

$\begin{array}{l}\left|{\overrightarrow{v}}_{pw}\right|=\sqrt{A_x^2+A_y^2}\\ \left|{\overrightarrow{v}}_{pw}\right|=\sqrt{{\left(2.12\mathrm{m}/\mathrm{s}\right)}^2+{\left(0.424\mathrm{m}/\mathrm{s}\right)}^2}\\ \left|{\overrightarrow{v}}_{pw}\right|=2.17\mathrm{m}/\mathrm{s}\end{array}$

The direction of the velocity vector for the velocity of the passenger with respect to the water can be calculated as

$\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{A_y}{A_x}\right)\\ \theta ={\tan }^{-1}\left(\frac{0.424\mathrm{m}/\mathrm{s}}{2.12\mathrm{m}/\mathrm{s}}\right)\\ \theta =11.3°\end{array}$

Thus, the relative velocity of the passenger with respect to the water is$2.17\mathrm{m}/\mathrm{s}$, and the direction is $11.3°$above the horizontal in the counterclockwise direction.