Q78P - University Physics with Modern Physics

Question

Question: You are a member of a geological team in Central Africa. Your team comes upon a wide river that is flowing east. You must determine the width of the river and the current speed (the speed of the water relative to the earth). You have a small boat with an outboard motor. By measuring the time it takes to cross a pond where the water isn’t flowing, you have calibrated the throttle settings to the speed of the boat in still water. You set the throttle so that the speed of the boat relative to the river is a constant  . Traveling due north across the river, you reach the opposite bank in  . For the return trip, you change the throttle setting so that the speed of the boat relative to the water is 9.00m/s  . You travel due south from one bank to the other and cross the river in  . (a) How wide is the river, and what is the current speed? (b) With the throttle set so that the speed of the boat relative to the water is 6.00 m/s , what is the shortest time in which you could cross the river, and where on the far bank would you land?

Step-by-Step Solution

Answer

Answer

 

(a) The width of the river is d = 90.48 m  and the speed of the current is  $V_{\raisebox{1ex}{$W$}\!\left/ \!\raisebox{-1ex}{$E$}\right.}=390 m/s$

 

(b) The shortest time to cross the river and landing distance from the bank is x =59.9 m  and the landing time will be  t =15.1.

 

Step 1: Definition of Newton equations of motion

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$}\right.}$

 

Step 2: Given data

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

Step 3: (a) Determining the width and current speed of the river

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

 

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

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

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.

Step 4: (b) Determining the shortest time to cross the river and landing distance from the bank

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,

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

 

                                                                                          

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

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