91Ó°ÊÓ

The velocity contains two components, one is a horizontal component and another one is vertical.

The newton’s laws of motion show the relationship between the moving particle or object’s displacement, initial and final velocity, acceleration, and time.

According to the newton’s laws of motion,

$\mathit{s}\mathbf{=}\mathit{u}\mathit{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{a}{\mathit{t}}^{\mathbf{2}}$

Where, s ,u , t , and a are displacement, initial velocity, time, and acceleration respectively.

From the formula of relative velocity of any object,

If any object B is moving related to W and W is moving related to E, then the velocity of B related to E is,

$V_{\raisebox{1ex}{$B$}\!\left/ \!\raisebox{-1ex}{$E$}\right.}=V_{\raisebox{1ex}{$B$}\!\left/ \!\raisebox{-1ex}{$W$}\right.}+V_{\raisebox{1ex}{$W$}\!\left/ \!\raisebox{-1ex}{$E$}\right.}$

Speed of the boat relative to the river due north = 6.00m/s

Time taken by the boat when travel due north = 20.1 s

Speed of the boat relative to the river due south = 9.00 m/s

Time taken by the boat when travel due south = 11.2 s

From the Pythagoras theorem with the help of figure a and b,

$$\begin{gathered} For{V}_{B/W}=6.00m/s,t=20.1s \\ {\left(V_{\raisebox{1ex}{$B$}\!\left/ \!\raisebox{-1ex}{$E$}\right.}\right)}^2={\left(V_{\raisebox{1ex}{$B$}\!\left/ \!\raisebox{-1ex}{$W$}\right.}\right)}^2+{\left(V_{\raisebox{1ex}{$W$}\!\left/ \!\raisebox{-1ex}{$E$}\right.}\right)}^2 \\ {\left(\frac{d}{20.1s}\right)}^2={\left(6.00m/s\right)}^2+{\left(V_{\raisebox{1ex}{$W$}\!\left/ \!\raisebox{-1ex}{$E$}\right.}\right)}^2 \\ And \\ For{V}_{B/W}=9.00m/s,t=11.2s \\ {\left(V_{\raisebox{1ex}{$B$}\!\left/ \!\raisebox{-1ex}{$E$}\right.}\right)}^2={\left(V_{\raisebox{1ex}{$B$}\!\left/ \!\raisebox{-1ex}{$W$}\right.}\right)}^2+{\left(V_{\raisebox{1ex}{$W$}\!\left/ \!\raisebox{-1ex}{$E$}\right.}\right)}^2 \\ {\left(\frac{d}{11.2s}\right)}^2={\left(9.00m/s\right)}^2+{\left(V_{\raisebox{1ex}{$W$}\!\left/ \!\raisebox{-1ex}{$E$}\right.}\right)}^2 \end{gathered}$$

$$\begin{gathered} For{V}_{B/W}=6.00m/s,t=20.1s \\ {\left(V_{B/E}\right)}^2={\left(V_{B/W}\right)}^2-{\left(V_{W/E}\right)}^2 \\ {\left(\frac{d}{20.1s}\right)}^2={\left(6.00m/s\right)}^2-{\left(V_{W/E}\right)}^2 \\ And \\ For{V}_{B/W}=9.00m/s,t=11.s \\ {\left(V_{B/E}\right)}^2={\left(V_{B/W}\right)}^2-{\left(V_{W/E}\right)}^2 \\ {\left(\frac{d}{20.1s}\right)}^2={\left(9.00m/s\right)}^2-{\left(V_{W/E}\right)}^2 \\ For{V}_{B/W}=6.00m/s,t=20.1s \\ {\left(V_{B/E}\right)}^2={\left(V_{B/W}\right)}^2-{\left(V_{W/E}\right)}^2 \end{gathered}$$

By solving the above two equations,

d= 90.48 m , it is the width of the river

And

V_W/E= 3.90 m/s , it is the speed of the current.

The shortest time will be there when the boat will be headed perpendicular to the current, which is in north direction. The time it will take to cross will be,

So the shortest time to cross the river and landing distance from the bank will be,

$$\begin{gathered} v_{B/W}=6.00,m/s\text{d = 90.48m} \\ t=\frac{d}{v_{B/E}} \\ t=\frac{\text{90.48m}}{6.00m/s} \\ t=15.1\text{s} \end{gathered}$$

The shortest time to cross the river and landing distance from the bank is .

$$\begin{gathered} v_{B/W}=3.967\text{,}t=15.1\text{s} \\ t=\frac{x}{v_{B/W}} \\ x=3.967m/s×15.1\text{s} \\ x=59.9\text{m} \end{gathered}$$