91Ó°ÊÓ

Relative velocity may be defined as the vector difference between the velocities of two particles. Its value can be positive, negative, or zero.

Here, ${\stackrel{→}{v}}_{ws}$is the velocity of the water relative to the shore, ${\stackrel{→}{v}}_{sw}$is the velocity of the swimmer relative to the water, and ${\stackrel{→}{v}}_{ss}$is the velocity of the swimmer relative to the shore.

(a)

The x component of the velocity of the swimmer relative to the water is

${\left(v_{sw}\right)}_x=0$.

The y component of the velocity of the swimmer relative to the water is

${\left(v_{sw}\right)}_y=0.60\mathrm{m}/\mathrm{s}$.

The x component of the velocity of the water relative to the shore is

${\left(v_{ws}\right)}_x=0.50\mathrm{m}/\mathrm{s}$.

The y component of the velocity of the water relative to the shore is

${\left(v_{ws}\right)}_y=0$.

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

$\begin{array}{l}{\stackrel{→}{v}}_{ss}={\stackrel{→}{v}}_{sw}+{\stackrel{→}{v}}_{ws}\\ {\stackrel{→}{v}}_{ss}=\left(0,0.60\mathrm{m}/\mathrm{s}\right)+\left(0.50\mathrm{m}/\mathrm{s},0\right)\\ {\stackrel{→}{v}}_{ss}=\left(0.50\mathrm{m}/\mathrm{s},0.60\mathrm{m}/\mathrm{s}\right)\end{array}$

The direction of the velocity of the swimmer relative to the shore can be calculated as

$\begin{array}{l}\mathrm{θ}={\sin }^{-1}\left(\frac{v_x}{v_y}\right)\\ \mathrm{θ}={\sin }^{-1}\left(\frac{0.50\mathrm{m}/\mathrm{s}}{0.60\mathrm{m}/\mathrm{s}}\right)\\ \mathrm{θ}≈56°\end{array}$

Thus, the upstream angle must be $56°$.

The speed of the swimmer relative to the shore can be calculated as

$\begin{array}{l}{v}_{ss}={\left(v_{sw}\right)}_y\cos \mathrm{θ}\\ v_{ss}=\left(0.60\mathrm{m}/\mathrm{s}\right)\cos \left(56°\right)\\ v_{ss}=0.335\mathrm{m}/\mathrm{s}\end{array}$

(b)

The required time can be calculated as

$\begin{array}{l}t=\frac{w}{v_{ss}}\\ t=\frac{\left(45\mathrm{m}\right)}{\left(0.335\mathrm{m}/\mathrm{s}\right)}\\ t=134.3/\mathrm{s}\end{array}$

Thus, the required time is $134.3/\mathrm{s}$.