Chapter 3, Problem 29P

A river has a steady speed of 0.500 m/s. A student swims upstream a distance of 1.00 km and swims back to the starting point, (a) If the student can swim at a speed of 1.20 m/s in 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?

To determine

The time taken for the trip.

Answer to Problem 29P

The time taken for the trip is $2.02\times {10}^3\text{s}$.

Explanation of Solution

The relation between the velocity of the student relative to the water, the velocity of the student relative to Earth and the water relative to Earth is,

$v_{\text{SW}}=v_{\text{SE}}-v_{\text{WE}}$

Here,

$v_{\text{SE}}$ is the velocity of the student relative to the Earth

$v_{\text{WE}}$ is the velocity of the water relative to the Earth

$v_{\text{SW}}$ is the velocity of the student relative to the water

If downstream is the positive direction,

$v_{\text{WE}}=0.500\text{m/s}$

$v_{\text{SW}}=-1.20\text{m/s}$ when the student is going upstream.

$v_{\text{SW}}=1.20\text{m/s}$ when the student is going downstream.

The velocity of the student relative to Earth going upstream is,

${\left(v_{\text{SE}}\right)}_{\text{upstream}}=v_{\text{WE}}+{\left(v_{\text{SW}}\right)}_{\text{upstream}}$

Substitute $0.500\text{m/s}$ for $v_{\text{WE}}$ and $-1.20\text{m/s}$ for ${\left(v_{\text{SW}}\right)}_{\text{upstream}}$.

$\begin{array}{c}{\left(v_{\text{SE}}\right)}_{\text{upstream}}=0.500\text{m/s}+\left(-1.20\text{m/s}\right)\\ =-0.700\text{m/s}\end{array}$

The distance for one leg of the trip is, $1.00\text{km}$.

The time taken for the upstream journey is,

$t_{\text{upstream}}=\frac{d}{{\left(v_{\text{SE}}\right)}_{\text{upstream}}}$

Here,

$d$ is the distance travelled

Substitute $1.00\text{km}$ for $d$ and $0.700\text{m/s}$ for ${\left(v_{\text{SE}}\right)}_{\text{upstream}}$.

$\begin{array}{c}{t}_{\text{upstream}}=\frac{\left(1.00\text{km}\right)\left(\frac{{10}^3\text{m}}{1\text{km}}\right)}{0.700\text{m/s}}\\ =1.43\times {10}^3\text{s}\end{array}$

The velocity of the student relative to Earth going downstream is,

${\left(v_{\text{SE}}\right)}_{\text{downstream}}=v_{\text{WE}}+{\left(v_{\text{SW}}\right)}_{\text{downstream}}$

Substitute $0.500\text{m/s}$ for $v_{\text{WE}}$ and $1.20\text{m/s}$ for ${\left(v_{\text{SW}}\right)}_{\text{downstream}}$.

$\begin{array}{c}{\left(v_{\text{SE}}\right)}_{\text{upstream}}=0.500\text{m/s}+1.20\text{m/s}\\ =1.70\text{m/s}\end{array}$

The distance for one leg of the trip is, $1.00\text{km}$.

The time taken for the upstream journey is,

$t_{\text{downstream}}=\frac{d}{{\left(v_{\text{SE}}\right)}_{\text{downstream}}}$

Substitute $1.00\text{km}$ for $d$ and $1.70\text{m/s}$ for ${\left(v_{\text{SE}}\right)}_{\text{downstream}}$.

$\begin{array}{c}{t}_{\text{downstream}}=\frac{\left(1.00\text{km}\right)\left(\frac{{10}^3\text{m}}{1\text{km}}\right)}{1.70\text{m/s}}\\ =5.88\times {10}^2\text{s}\end{array}$

The total time for the trip is,

$t_{\text{total}}=t_{\text{upstream}}+t_{\text{downstream}}$

Substitute $1.43\times {10}^3\text{s}$ for $t_{\text{upstream}}$ and $5.88\times {10}^2\text{s}$ for $t_{\text{downstream}}$.

$\begin{array}{c}{t}_{\text{total}}=1.43\times {10}^3\text{s}+5.88\times {10}^2\text{s}\\ =2.02\times {10}^3\text{s}\end{array}$

Conclusion:

Thus, the time taken for the trip is $2.02\times {10}^3\text{s}$.

To determine

The time required for the trip in still water.

Answer to Problem 29P

The time taken for the trip in still water is $1.67\times {10}^3\text{s}$.

Explanation of Solution

If the water is sill, the speed will be same for each leg.

$v_{\text{SE}}={\left(v_{\text{SE}}\right)}_{\text{upstream}}={\left(v_{\text{SE}}\right)}_{\text{downstream}}=1.20\text{m/s}$

The time for each leg is,

$t_{\text{leg}}=\frac{d}{v_{\text{SE}}}$

Substitute $1.00\text{km}$ for $d$ and $1.20\text{m/s}$ for $v_{\text{SE}}$.

$\begin{array}{c}{t}_{\text{leg}}=\frac{\left(1.00\text{km}\right)\left(\frac{{10}^3\text{m}}{1\text{km}}\right)}{1.20\text{m/s}}\\ =8.33\times {10}^2\text{s}\end{array}$

Conclusion:

The total time taken is twice the time taken for one leg.

$t_{\text{total}}=2t_{\text{leg}}$

Substitute $8.33\times {10}^2\text{s}$ for $t_{\text{leg}}$.

$\begin{array}{c}{t}_{\text{total}}=2\left(8.33\times {10}^2\text{s}\right)\\ =1.67\times {10}^3\text{s}\end{array}$

The time taken for the trip in still water is $1.67\times {10}^3\text{s}$.

To determine

Why does the swim takes longer where there is a current.

Answer to Problem 29P

The time takes to go against the current is longer.

Explanation of Solution

The time taken going downstream is a lot less compared to the time taken to go upstream the same distance. The current will slow the swimmer going up and it will aid the swimmer going along with the current.

Conclusion:

The time savings going downstream with the current is always less than the extra time required to go the same distance against the current.