Chapter 8.4, Problem 4E

To determine

To infer : the speed of current is greater or less than $2\text{ mi/h}$ .

Answer to Problem 4E

The speed of the current is $\text{1 mi/h}$ , which is less than $\text{2 mi/h}$ .

Explanation of Solution

Given information :

Consider the Problem $20$ on page $134$ .

The canoeist travel two miles upstream from the starting point before passing a log floating in the river’s current. After paddling upstream for one more hour, he paddles back to the starting point just as the log arrived.

Formula used :

  $\text{speed}=\frac{\text{distance}}{\text{time}}$

Calculation :

As per the problem,

Assume:

The speed of the current is $s\text{ mi/h}$ .

The speed at which the canoeist peddles is $r\text{ mi/h}$

The distance travelled by the canoeist for one hour after seeing the log is $x$ miles.

The time from the moment the canoeist sees the log till he reaches the starting point is $t\text{ h}$

The log floats at $s\text{ mi/h}$ and covers two miles in $t\text{ h}$ .

So,

  $\begin{array}{l}s=\frac{2}{t}\\ t=\frac{2}{s}\end{array}$

The canoeist first travels x miles against the current for one hour after seeing the log then travels x miles back with the current and then two miles along the current for reaching the starting point in a total time of $t\text{ h}$

Assuming speed of peddling is greater than the speed of current,

  $t=\frac{x}{r-s}+\frac{x}{r+s}+\frac{2}{r+s}$

The time in which the canoeist peddles travels x milesagainst the currentis $\text{1 h}$ .

So,

  $\begin{array}{l}\frac{x}{r-s}=\text{1 }\\ x=r-s\end{array}$

Substitute $t=\frac{2}{s}$ and $x=r-s$ in the above equation.

  $\begin{array}{l}\frac{2}{s}=1+\frac{r-s}{r+s}+\frac{2}{r+s}\text{ [}t=\frac{2}{s},x=r-s]\\ \frac{2}{s}=\frac{r+s+r-s+2}{r+s}\text{ [Take LCM]}\\ \frac{2}{s}=\frac{2r+2}{r+s}\text{ [Simplify]}\\ \frac{2}{s}=\frac{2(r+1)}{r+s}\text{ [Factor]}\\ \frac{1}{s}=\frac{r+1}{r+s}\text{ [Divide both sides by 2]}\\ r+s=rs+s\text{ [Cross-multiply]}\\ r=rs\\ 1=s\end{array}$

Therefore, speed of the current is $\text{1 mi/h}$ .

Hence, the speed of the current is less than $\text{2 mi/h}$ .