Chapter 3.1, Problem 68PPS

(a)

To determine

To Find: the rowing speed of A and speed of current.

Answer to Problem 68PPS

Speed of A is $3.558\text{mi/hr}$ and the speed of current is $1.058\text{mi/hr}$ .

Explanation of Solution

Given:

A cover $1\text{mile}$ upstream in $24\text{ min}\text{.}$ and same distance downstream in $\text{13min}\text{.}$

Total number people playing the matches are $38$ .

Assuming rowing speed and speed of current is constant.

Calculation:

Let $x$ be the speed of Alison and $y$ be the speed of current.

Equation of speed for upstream motion.

  $x-y$

Equation of speed for upstream motion.

  $x+y$

Equation for upstream motion.

  $\begin{array}{l}\text{Speed}\times \text{Time}=\text{Distance}\\ \left(x-y\right)\times \frac{24}{60}=1\therefore \left(1\text{hr=60min}\text{.}\right)\\ x-y=\frac{60}{24}\\ x=y+\frac{60}{24}\mathrm{......}\left(1\right)\end{array}$

Equation for downstream motion.

  $\begin{array}{l}\text{Speed}\times \text{Time}=\text{Distance}\\ \left(x+y\right)\times \frac{13}{60}=1\therefore \left(1\text{hr=60min}\text{.}\right)\\ x+y=\frac{60}{13}\end{array}$

Substituting $x=y+\frac{60}{24}$ from equation (1).

  $\begin{array}{l}x+y=\frac{60}{13}\\ y+\frac{60}{24}+y=\frac{60}{13}\\ 2y=\frac{60}{13}-\frac{60}{24}\\ y=1.058\text{mi/hr}\end{array}$

Substituting $y=1.058\text{mi/hr}$ in equation (1).

  $\begin{array}{l}x=y+\frac{60}{24}\\ x=1.058+\frac{60}{24}\\ x=3.558\text{mi/hr}\end{array}$

Therefore, the speed of A is $3.558\text{mi/hr}$ andthe speed of current is $1.058\text{mi/hr}$ .

To determine

(b)To find: A can meet her friend in $1\text{hr}$ at a distance of $3\text{ miles}$ upstream.

Answer to Problem 68PPS

No, she will not be on time.

Explanation of Solution

Given; Distance to be covered by A is $3\text{ miles}$ upstream in $1\text{hr}$ .

Speed of A is $3.558\text{mi/hr}$ and the speed of current is $1.058\text{mi/hr}$.

Calculation;

Using relation,

Equation of speed for upstream motion.

  $x-y$

And,

  $\begin{array}{l}\text{Speed}\times \text{Time}=\text{Distance}\\ \left(x-y\right)\times 1=\\ 3.558-1.058=2.5\end{array}$

Therefore, A would not be able to reach on time because in $1\text{hr}$ ,he only completes $2.5\text{miles}$ .