Chapter 4, Problem 27CCSR

Solution

To find: The speeds of the swimmer and the current in which he/she swims.

The speed of swimmer and current is 50 meter/minute 16.67 meter/minute respectively.

Given data:

The swimmer swims 1000 meters downstream in 15 minutes and against the current in 30 minutes.

Method/Formula used:

Let the speed of swimmer is S and that of current C . Then the relative or effective downstream speed of swimmer is $S+C$ and against the current $S-C$ . If swimmer swims one way distance D then,

Time taken swimming downstream $t_{down}=\frac{D}{S+C}$ ... (1)

Time taken swimming against current $t_{against}=\frac{D}{S-C}$ … (2)

Calculation:

The swimmer swims 1000 meters downstream in 15 minutes, so substitute 1000 meter for $D$ and 15 min. for $t_{down}$ in result (1),

  $\begin{array}{c}15=\frac{1000}{S+C}\\ S+C=\frac{1000}{15}\\ S+C=\frac{200}{3}\end{array}$

Therefore,

  $S+C=\frac{200}{3}$ ... (3)

The swimmer swims 1000 meters against the current in 30 minutes, so substitute 1000 meter for $D$ and 30 min. for $t_{against}$ in result (1),

  $\begin{array}{c}30=\frac{1000}{S-C}\\ S-C=\frac{1000}{30}\\ S-C=\frac{100}{3}\end{array}$

Therefore,

  $S-C=\frac{100}{3}$ ... (4)

Adding equation (3) and (4):

  $\begin{array}{c}S+C+S-C=\frac{200}{3}+\frac{100}{3}\\ S+C+S-C=\frac{200+100}{3}\\ 2S=\frac{300}{3}\\ S=\frac{300}{3\cdot 2}\end{array}$

Solve further,

  $\begin{array}{c}S=\frac{300}{6}\\ =50\end{array}$

Subtract equation (4) from equation (3),

  $\begin{array}{c}\left(S+C\right)-\left(S-C\right)=\frac{200}{3}-\frac{100}{3}\\ S+C-S+C=\frac{200-100}{3}\\ 2C=\frac{100}{3}\\ S=\frac{100}{3\cdot 2}\end{array}$

Solve further,

  $\begin{array}{c}S=\frac{100}{6}\\ =16.67\end{array}$

Thus, the speed of the swimmer is 50 meter/minute and that of current 16.67 meter/minute.