91Ó°ÊÓ

The width of the boat is , the river is flowing to the east at a speed of , Mary can row with a speed of .

Since, these velocities are not in the same direction, we just can't add the magnitudes of them. Instead use the tip-to-tail rule for vector addition.

Below Figures show the vector diagram of the given situation.

The velocity of the boat with respect to the earth is,

${\stackrel{→}{v}}_{\text{BE}}={\stackrel{→}{v}}_{\text{BR}}+{\stackrel{→}{v}}_{\text{RE}}=(1.0\text{m}/\text{s})i+(2.0\text{m}/\text{s})j$

The magnitude of velocity is given as,

$\left|v_{\text{BE}}\right|=\sqrt{{(1.0\text{m}/\text{s})}^2+{(2.0\text{m}/\text{s})}^2}=\sqrt{5\text{m}/\text{s}}$

The angle made by resultant is given as follows:

$\mathrm{θ}={\tan }^{−1}\left(\frac{{\stackrel{→}{v}}_{\text{BR}}}{{\stackrel{→}{v}}_{\text{RE}}}\right)={\tan }^{−1}\left(\frac{1.0\text{m}/\text{s}}{2.0\text{m}/\text{s}}\right)={26.56}^{°}$

The component of velocity of boat with respect to the earth is calculated as,

Along north direction,

${\left(v_x\right)}_{\text{BE}}=(\sqrt{5}\text{m}/\text{s})\cos {26.6}^{°}=2.0\text{m}/\text{s}$

Along east direction,

${\left(v_y\right)}_{\text{BE}}=(\sqrt{5}\text{m}/\text{s})\sin {26.6}^{°}=1.0\text{m}/\text{s}$

The time taken to travel the width of the river is given as,

$x={\left(v_x\right)}_{\text{BE}}t$

$t=\frac{x}{{\left(v_x\right)}_{\text{BE}}}=\frac{100\text{m}}{2.0\text{m}/\text{s}}=50\text{s}$

The distance traveled by boat along the river is given as,

$y={\left(v_y\right)}_{\text{BE}}t=(1.0\text{m}/\text{s})(50\text{s})=50\text{m}$

Hence, when she reaches the opposite shore, the distance she would be away from her intended landing spot is .$50\text{m}$

Use Pythagorean Theorem on the vector-triangle drawn above.

$\begin{array}{l}\left|{\stackrel{→}{v}}_{\text{BE}}\right|=\sqrt{{\left|{\stackrel{→}{v}}_{\text{BR}}\right|}^2+{\left|{\stackrel{→}{v}}_{\text{RE}}\right|}^2}\\ =\sqrt{{(1.0\text{m}/\text{s})}^2+{(2.0\text{m}/\text{s})}^2}\\ =\sqrt{5}\text{m}/\text{s}\end{array}$

Therefore, her speed with respect to the shore is $\sqrt{5}\text{m}/\text{s}$, in a direction ${26.6}^0$clock-wise from the north.