jamal and alex spent 3 hours canoeing on a stream. They went 5 miles upstream and 5 miles back downstream. In still water the average speed of the canoe was 6 miles per hour.

Question

Jamal and Alex spent 3 hours canoeing on a stream. They went 5 miles upstream and 5 miles back downstream. In still water the average speed of the canoe was 6 miles per hour. The time T spent canoeing can be modeled by the function

T(s)= 5/6-s + 5/6+s = 60/(6-s)(6+s)

where s is the average speed of the current

what is the average speed of the current during their trip?

Answer 1

Hi Victoria V.,

T is equal to the total time spent canoeing, 3 hours, therefore T = 3, (T(s) = 3, same thing).

We can write the equation given with T(s) = 3:

3 = 60/[(6 - s)*(6 + s)];

3 = 60/(36 - s²), foil;

3*(36 - s²) = 60, multiply both sides by (36 - s²);

108 - 3s² = 60, multiply 3 through;

48 - 3s² = 0, subtract 60 from both sides;

3s² - 48 = 0, multiply both sides by -1;

(3s - 12)*(s + 4) = 0, factor;

3s - 12 = 0 and s + 4 = 0, set both factors to zero;

3s = 12 and s = -4, add 12 and subtract 4 respectively;

s = 4, divide by 3.

So s = ± 4, and for average speed of the current we choose +4 mi/hr.

I hope this helps, Joe.