Question

A river has a steady speed of $\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{ }\mathbf{\text{m/s}}$. A student swims upstream a distance of 1.00 kmand swims back to the starting point.

(a) If the student can swim at a speed of 1.20 m/sin still water, how long does the trip take?

(b) How much time is required in still water for the same length swim?

(c) Intuitively, why does the swim take longer when there is a current?

Short answer

(a) The time taken by the student to complete the trip at a speed of$1.20\mathrm{m}/\mathrm{s}$ is$2.02\times {10}^3 \text{s}$ .

(b) The time taken by the student to complete the trip in still water is$27.78 \text{min}$ .

(c) The student takes longer time to swim when there is a current because his relative velocity decreases.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The steady speed of river is 0.500 m/s.
• The student swims upstream a distance of 1.00 km .
• The student swim at a speed of 1.20 m/s .

The formula for upstream time

Write the calculate the upstream time

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

Here,t₁ is the upstream time, dis the distance,${\mathbf{v}}_{\mathbf{s}}$is the speed of the student and${\mathbf{v}}_{\mathbf{r}}$is the speed of the river.

Determination of the how long does the trip take

(a)

Substitute values in the above expression and we get,

$\begin{array}{rcl}{t}_1& =& \frac{(1.00 \text{km})\left(\frac{{10}^3 \text{m}}{1 \text{km}}\right)}{(1.20 \text{m/s})-(0.500 \text{m/s})}\\ & =& 1428.57 \text{s}\end{array}$

Write the expression to calculate the downstream time

$t_2=\frac{d}{v_{\text{s}}+v_{\text{r}}}$

Substitute values in the above expression and we get,

$\begin{array}{rcl}{t}_2& =& \frac{(1.00 \text{km})\left(\frac{{10}^3 \text{m}}{1 \text{km}}\right)}{(1.20 \text{m/s})+(0.500 \text{m/s})}\\ & =& 588.23 \text{s}\\ & & \end{array}$

Write the formula to calculate the total time

$t=t_1+t_2$

Substitute values in the above expression and we get,

$\begin{array}{rcl}t& =& (1428.57 \text{s})+(588.23 \text{s})\\ & =& 2016.8 \text{s}\\ & =& 2.02\times {10}^3 \text{s}\\ & & \end{array}$

Therefore, the time taken by the student to complete the trip at a speed of 1.20 m/s is $2.02\times {10}^3 \text{s}$.

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

(b)

Write the formula to calculate the total time in still water

$t^{\text{'}}=\frac{2d}{v_{\text{s}}}$

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

Substitute values in the above expression and we get,

$\begin{array}{rcl}t\text{'}& =& \frac{2(1.00 \text{km})\left(\frac{{10}^3 \text{m}}{1 \text{km}}\right)}{1.20 \text{m/s}}\\ & =& 1666.67 \text{s}\times \frac{1 \text{min}}{60 \text{s}}\\ & =& 27.78 \text{min}\\ & & \end{array}$

Therefore, the time taken by the student to complete the trip in still water is$\begin{array}{rcl}& & 27.78 \text{min}\end{array}$ .

Reason why does the swim take longer when there is a current

(c)

The velocity of an object relative to a reference frame is known as relative velocity. The resistance is caused by the water flow.

When the student swims upstream it takes more time to cover the same distance than when he goes downstream.

When the student swim upstream, his relative velocity decreases so, he will take longer time to swim. Hence, he will take longer time to swim when there is a current.

Conclusion:

Therefore, student takes longer time to swim when there is a current because his relative velocity decreases.