Thread: Help please - distance

"Calum was floating down a river on a raft, when, half a mile downstream, his brother duncan set off in a canoe. Duncan paddled downstream as quickly as he could, then turned round and paddles back again, still at his best pace. He arrived back at his starting point just as Calum floated by. Assuming Duncans best pace in still water is 10 times that of the river current, how far did he paddle ?"

distance = rate*time = speed*time = velocity*time.

Let x = distance down river that Duncan paddled to, in miles.
And c = rate of current or stream, in mi/hr.

So, Duncan's best rate in still water is 10c mi/hr.

Calum:
When Calum and Duncan met at Duncan's starting/finishing point, Calum travelled half a mile, or (1/2)mi.
The time he spent doing that is
distance = rate*time
1/2 = c *T
T = (1/2)/c = 1/(2c) hrs. -----------**

Duncan:
Total time of Duncan paddling downstream to x and then paddling upstream to his starting point is the same as Calum's T.

Downstream, Duncan's net rate is 10c +c = 11c
So,
distance = rate*time,
x = (11c)(t1)
t1 = x/(11c) hr. --------time spent paddling x miles downstream.

Upstream, Duncan's net rate is 10c -c = 9c
So,
distance = rate*time,
x = (9c)(t2)
t2 = x/(9c) hr. --------time spent paddling x miles upstream.

------------------------------
Now,
t1 +t2 = T
x/(11c) +x/(9c) = 1/(2c)
Clear the fractions, multiply both sides by (11*9*2*c),
x(9*2) +x(11*2) = 1(11*9)
18x +22x = 99
40x = 99
x = 99/40 = 2.475 mi. ----------------answer.

[In feet, that is (2.475mi)(5280ft/1mi) = 13,068 ft.]