1. ## A mathematical adventure down the river...!

Q Tom was floating down the river on a raft, when 1 km lower down, Michael took to the water in a rowing boat. Michael rowed downstream at his fastest pace. Then he turned around and rowed back, arriving at his starting point just as Tom drifted by. If Michael's rowing speed in still water is ten times the speed of the current in the river, what distance had Michael covered before he turned his boat around?

Okay... First I drew a diagram, and defined some variables...

$\displaystyle V_{M} =$Michael's rowing velocity in still water

$\displaystyle V_{C} =$Velocity of the current in the river

$\displaystyle T_{1} =$Time taken for Michael to cover x km downstream

$\displaystyle T_{2} =$Time taken for Michael to cover x km upstream

$\displaystyle V_{T} =$Time taken for Tom to cover 1 km

We know $\displaystyle V_{M} = 10V_{C}$. We also know that $\displaystyle S = \frac{D}{T}$

Also, the velocity of the current will affect Michael's rowing velocity. So, Michael's speed downstream will be $\displaystyle V_{M} + V_{C} = 11V_{C}$, and $\displaystyle V_{M} - V_{C} = 9V_{C}$ when he turns around and moves upstream.

So, I got three equations, as follows:

$\displaystyle 11V_{C} = \frac{x}{T_{1}}...[1]$

$\displaystyle 9V_{C} = \frac{x}{T_{2}}...[2]$

For the third equation, I used the fact that $\displaystyle T_{1} + T_{2}$ equals the time taken for Tom to cover 1 km. So:

$\displaystyle T_{1} + T_{2} = \frac{1}{V_{T}}...[3]$

$\displaystyle \Rightarrow\frac{x}{11V_{C}} + \frac{x}{9V_{C}} = \frac{1}{V_{T}}$

$\displaystyle \Rightarrow\frac{9V_{C}x + 11V_{C}x}{99V_{C}^{2}} = \frac{1}{V_{T}}$

There are too many variables... How do I proceed? Or am I on the wrong track... and there's an easier way to get this done?

Help!

Thanks.

ILoveMaths07.

2. let $\displaystyle v$ = current speed
$\displaystyle 10v$ = rowing speed
$\displaystyle t_1$ = time downstream for the rower
$\displaystyle t_2$ = time back upstream for the rower

for the "floater" ...

$\displaystyle v(t_1+t_2) = 1 \, km$

for the rower ...

$\displaystyle 11v(t_1) - 9v(t_2) = 0 \, km$

from the rower's equation ...

$\displaystyle \frac{11}{9}t_1 = t_2$

sub into the floater's equation ...

$\displaystyle v\left(t_1 + \frac{11}{9}t_1\right) = 1$

$\displaystyle vt_1 = \frac{9}{20}$

rower's downstream distance = $\displaystyle 11vt_1 = \frac{99}{20} \, km$