Question

A stream flows at a rate of 4mph. A boat travels 70 miles downstream and returns in a total time of 6 hours. What is the speed of the boat in still water

Answer 1

\[v_w = 4\]\[d = vt\]Boat travels downstream in time \[T_d = 70/(v_s+v_w)\]Boat travels upstream in time \[T_u = 70/(v_s - v_w)\]\[T_d+T_u = 6\]Combine the equations and solve for \(v_s\) which is the speed of the boat in still water.

Answer 2

I am coming up with two answers somehow the numbers seem very large

Answer 3

show me your work. you will get two answers, but one of them is clearly not applicable.

Answer 4

Is this for 6.05 in Algebra 1?

Answer 5

70(v-4)+70(v+6)= 6 (v^2-16) which become

\[140\pm \sqrt 17296 /12\]

Answer 6

Well that should be
\[\frac{-140\pm\sqrt{140^2-4(96)(-6)}}{2(-6)} = \frac{-140\pm148}{-12}\]

Answer 7

what does that evaluate to?

Answer 8

so the speed of the boat in still water would be 24 mph because .666mph wouldnt be applicable

Answer 9

you dropped a minus sign under the radical which is why you had 17296 instead of 21904

Answer 10

One of the answers is negative: \[\frac{-140 + 148}{-12} = \frac{8}{-12} = -\frac{2}{3}\]

Answer 11

That's the solution to the quadratic that doesn't apply...

Answer 12

thank you so much!

Answer 13

Let's check if 24 works for the speed of the boat:

70/(24+4) = 70/28 = 5/2
70/(24-4) = 70/20 = 7/2
7/2 + 5/2 = 12/2 = 6

Answer 14

especially with "story problems" it is vital to check the answers in the original equations and statements to make sure they do solve the problem correctly. -2/3 is a valid mathematical solution to the equation we set up, but it isn't a valid solution to this problem.

Answer 15

(it's also a good opportunity to verify that you have in fact understood the problem correctly and solved the right problem!)