Question

A river flows with a steady speed v. A student swims upstream a distanced and then back to the starting point. The student can swim at speedc in still water.

(a) In terms of d, vand c, what time interval is required for the round trip?

(b) What time interval would be required if the water were still?

(c) Which time interval is larger? Explain whether it is always larger.

Short answer

(a) The expression for the time taken by the student to complete the round trip is$\frac{2d}{c\left(1-\frac{v^2}{c^2}\right)}$.

(b) The expression for the time taken by the student to complete the round trip in still water is$\frac{2d}{c}$ .

(c) The time taken to complete the trip is greater when there is a current because of the resistance it offers.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The speed of river flow is v.
• The speed of student in still water is c.
• The student swims upstream a distance of d .

The upstream and downstream time

The upstream time is given by

${\mathit{t}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{d}}{\mathbf{c}\mathbf{-}\mathbf{v}}$

Here,${\mathit{t}}_{\mathbf{1}}$is the upstream time.

Write the expression to calculate the downstream time

${\mathit{t}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{d}}{\mathbf{c}\mathbf{+}\mathbf{v}}$

Here,${\mathit{t}}_{\mathbf{2}}$is the upstream time.

Determination of the time interval

(a)

The speed of river flow is v , the speed of student in still water is c and the student swims upstream a distance of d .

Write the expression to calculate the upstream time

$t_1=\frac{d}{c-v}$

Here, $t_1$is the upstream time.

Write the expression to calculate the downstream time

$t_2=\frac{d}{c+v}$

Here, $t_2$is the upstream time.

Write the formula to calculate the total time

$t=t_1+t_2$

Here, t is the total time of the trip

Substitute the values in above expression, and we get,

$\begin{array}{rcl}t& =& \left(\frac{d}{c-v}\right)+\left(\frac{d}{c+v}\right)\\ & =& d\left(\frac{1}{c-v}+\frac{1}{c+v}\right)\\ & =& d\left(\frac{c+v+c-v}{c^2-v^2}\right)\\ & =& d\left(\frac{2c}{c^2-v^2}\right)\end{array}$

Further, solve by taking c² common.

$\begin{array}{rcl}t& =& \frac{d}{c^2}\left(\frac{2c}{1-\frac{v^2}{c^2}}\right)\\ & =& \frac{2d}{c\left(1-\frac{v^2}{c^2}\right)}\\ & & \end{array}$

Therefore, the expression for the time taken by the student to complete the round trip is $\frac{2d}{c\left(1-\frac{v^2}{c^2}\right)}$.

The expression for the time taken by the student to complete the round trip in still water

(b)

If the water is considered to be still, its relative velocity is zero.

Substitute 0 for v in equation (I).

$\begin{array}{rcl}t\text{'}& =& \frac{2d}{c\left(1-\frac{0}{c^2}\right)}\\ & =& \frac{2d}{c}\\ & & \end{array}$

Here, $t\text{'}$is the time for the round trip in still water.

Conclusion:

Therefore, the expression for the time taken by the student to complete the round trip in still water is $\frac{2d}{c}$.

Comparison of both time intervals

(c)

The time interval for round trip is more in part (a), since the denominator is small in the expression. The time interval is always larger than the part (b).

$\begin{array}{rcl}\left(\frac{2d}{c\left(1-\frac{v^2}{c^2}\right)}\right)& >& \left(\frac{2d}{c}\right)\\ t& >& t^{\text{'}}\end{array}$

The resistance is caused by the water flow.

Therefore, the time taken to complete the trip is greater when there is a current because of the resistance it offers.