Chapter 9.4, Problem 83AYU

To determine

Tofind:the angle so that the boat dock increased at constant speed of $20km/hr$ and to find theactual time so that it will take to cross the river.

Answer to Problem 83AYU

  $\begin{array}{l}\theta ={8.6}^{\circ }\\ t=1.52\min \end{array}$ .

Explanation of Solution

Given:

Constant current of riveris $3km/hr$

Width of the river is $\frac{1}{2}km$ .

Speed of current is increased at $20\text{km}/hr$

Calculation:

According to the question,

  $\begin{array}{l}\Vert v_w\Vert =20\\ v_w=a_i+b_j\\ \therefore v_w=\sqrt{a^2+b^2}=20\\ \Rightarrow a^2+b^2=400\end{array}$ .

  $\begin{array}{l}{v}_g=v_w+v_c\\ kj=ai+bj+3i=\left(a+3\right)i+bj\\ a+3=0\Rightarrow a=-3\end{array}$

Therefore,

  $\begin{array}{l}{\left(-3\right)}^2+b^2=400\\ b^2=400-9=391\\ k=b=\sqrt{391}=19.8\end{array}$

  $\begin{array}{l}\cos \theta =\frac{v_w\cdot j}{\Vert v_w\Vert \cdot \Vert j\Vert }\\ =\frac{-3\left(0\right)+\sqrt{391}}{20}=\frac{\sqrt{391}}{20}=0.9887\\ \theta ={8.6}^{\circ }\end{array}$

Therefore, time is $t=\frac{0.5}{19.8}=0.025hrs=1.52\min$

Hence, the angle at which the boat dock increased at constant speed and the time taken to cross the river is $\theta ={8.6}^{\circ };t=1.52\min$ .