Question

A woman can row a boat at 6.40 km/hin still water. (a) If she is crossing a river where the current is 3.20 km/h , in what direction must her boat be headed if she wants to reach a point directly opposite her starting point? (b) If the river is 6.40 km wide, how long will she take to cross the river? (c) Suppose that instead of crossing the river she rows 3.20 km/h downthe river and then back to her starting point. How long will she take? (d) How long will she take to row3.20 km upthe river and then back to her starting point? (e) In what direction should she head the boat if she wants to cross in the shortest possible time, and what is that time?

Short answer

a) Head of the boat to reach exactly the opposite point should be $26.56°$with x axis.

b) Time taken by boat to cross the river is 1.12 h .

c) Time taken by boat to row downward 3.20 km in the river and back to the starting point is 1.33 h .

d) Time taken by boat to row up 3.20 km in the river and back to the starting point is 1.33 h .

e) Shortest time to cross the riveris 1.00 h .

Step-by-step solution

The given data

• Speed of the boat in still water,$v_{xb}=6.40\mathrm{km}/\mathrm{h}$.
• Speed of water current,$v_{yr}=3.20\mathrm{km}/\mathrm{h}$.
• Width of the river, $d=6.40\mathrm{km}$.

Understanding the concept of relative velocity

When there is a relative motion between the two frames of reference Aand Bat a constant velocity, the velocity of a particle P is different for each observer in frames A and B. The two measured velocities are related by

${\overrightarrow{\mathbf{v}}}_{\mathbf{PA}}\mathbf{=}{\overrightarrow{\mathbf{v}}}_{\mathbf{PB}}\mathbf{+}{\overrightarrow{\mathbf{v}}}_{\mathbf{BA}}$

Here${\overrightarrow{\mathbf{v}}}_{\mathbf{B}\mathbf{A}}$ is the velocity of B with respect to A.

We can find the time by using the velocity equation. We can consider the velocities of boat and river as resultant velocity components and then we can find resultant velocity. We use trigonometric functions to find angles.

Formula:

The velocity of a body in motion,$v=\frac{d}{t}$ (i)

a) Calculation for the heading of the boat to reach an exactly opposite point

Assuming that the velocityof the boat is along x and the velocityof the river is along y, we can find the directionof the boat using tan(y/x). Hence, the direction is given as:

$$\begin{gathered} \tan \theta =-\frac{3.20}{6.40} \\ \theta ={\tan }^{-1}\left(-\frac{3.20}{6.40}\right) \\ =-26.56° \end{gathered}$$

So, the boat should be $26.56°$with an x axis to reach the opposite point.

b) Calculation for time taken by boat to cross the river

From equation (i), we get the time taken to cross the river by the boat. Substituting the values of the given quantities in equation.

$$\begin{gathered} t=\frac{\mathrm{d}}{{\mathrm{V}}_{\mathrm{ix}}} \\ =\frac{6.4\mathrm{km}}{\left(6.4\mathrm{km}/\mathrm{h}\right)\left(\cos \left(26.56°\right)\right)} \\ =1.12\mathrm{h} \end{gathered}$$

Hence, the time taken by the boat to cross the river is $1.12\mathrm{h}$.

c) Calculation time taken by boat to row downward and back to original point

Resultant velocity of the boat for downstream,

$$\begin{gathered} v_d=v_{xb}+v_{yr} \\ =6.4\mathrm{km}/\mathrm{h}+3.2\mathrm{km}/\mathrm{h} \\ =9.6\mathrm{km}/\mathrm{h} \end{gathered}$$

Resultant velocity for upstream,

$$\begin{gathered} v_u=v_{xb}+v_{yr} \\ =6.4\mathrm{km}/\mathrm{h}-3.2\mathrm{km}/\mathrm{h} \\ =3.2\mathrm{km}/\mathrm{h} \end{gathered}$$

Total time taken by boat to row down and row back to starting point can be found by using equation (i). Substitute the values of the quantities in equation (i), we get,

$$\begin{gathered} t=\frac{d}{v_d}+\frac{d}{v_u} \\ =\left(\frac{3.2\mathrm{km}}{9.6\mathrm{km}/\mathrm{h}}\right)+\left(\frac{3.2\mathrm{km}}{3.2\mathrm{km}/\mathrm{h}}\right) \\ =1.33\mathrm{h} \end{gathered}$$

Hence, the total time taken for going both upstream and downstream is $1.33\mathrm{h}$.

d) Calculation for time taken by boat to row up 3.20km in river and back to starting point

Using same equation above, we get the total time taken for going up and returning back. Therefore, substitute the value of the quantities in the equation (i).

$$\begin{gathered} t=\left(\frac{3.2\mathrm{km}}{3.2\mathrm{km}/\mathrm{h}}\right)+\left(\frac{3.2\mathrm{km}}{9.6\mathrm{km}/\mathrm{h}}\right) \\ =1.33\mathrm{h} \end{gathered}$$

Hence, the total time taken for going up and returning back is $1.33\mathrm{h}$.

e) Calculation for shortest time taken by boat to reach opposite point

For the shortesttime the distanceshould be shortest, so the angle should be 0.

Hence, using equation (i) and$\theta =0°$we get

$$\begin{gathered} t=\frac{6.4\mathrm{km}}{6.4\mathrm{km}/\mathrm{h}·\cos \left(0°\right)} \\ =1.0\mathrm{h} \end{gathered}$$

Hence, the shortest time taken is $1.0\mathrm{h}$.