SOLUTION: I really need help. A tugboat goes 140miles upstream in 10 hours. The return trip downstream takes 5 hours. Find the speed. the speed pf the tugboat is ___mph, and the speed of

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: I really need help. A tugboat goes 140miles upstream in 10 hours. The return trip downstream takes 5 hours. Find the speed. the speed pf the tugboat is ___mph, and the speed of      Log On


   

Question 655526: I really need help.
A tugboat goes 140miles upstream in 10 hours. The return trip downstream takes 5 hours. Find the speed.
the speed pf the tugboat is ___mph, and the speed of the current is ___mph.
this is substitution to solve linear equations.
Found 2 solutions by jim_thompson5910, MathTherapy:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Let c = speed of the current and s = speed of the boat

Going upstream, the boat is fighting against the current. So its speed goes from 's' mph to s-c mph

Therefore, the equation going 140 miles upstream in 10 hrs is

d = rt

140 = (s-c)*10

140/10 = s-c

14 = s-c

Solve for s to get

14 + c = s

s = 14 + c

Returning downstream, you pick up speed. So you go from 's' mph to s+c mph

So going 140 miles downstream in 5 hours means

d = rt

140 = (s+c)*5

140/5 = s+c

28 = s+c

s+c = 28

Now plug in s = 14 + c and solve for c

s+c = 28

14+c+c = 28

14+2c = 28

2c = 28-14

2c = 14

c = 14/2

c = 7

So the speed of the current is 7 mph

s = 14+c

s = 14+7

s = 21

and the speed of the boat is 21 mph

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

I really need help.
A tugboat goes 140miles upstream in 10 hours. The return trip downstream takes 5 hours. Find the speed.
the speed pf the tugboat is ___mph, and the speed of the current is ___mph.
this is substitution to solve linear equations.

Let the speed of the boat, in still water be S, and the the speed of the current, C

In going 140 miles in 10 hours, the boat's speed, against the current = 140%2F10, or 14 mph. This means that S - C = 14
C = S - 14 ----- eq (i)

In going 140 miles in 5 hours, the boat's speed, with the current = 140%2F5, or 28 mph. This means that S + C = 28 ----- eq (ii)

S + S - 14 = 28 ----- Substituting S - 14 for C in eq (ii)

2S = 28 + 14

2S = 42

S, or speed of boat in still water = 42%2F2, or highlight_green%2821%29 mph

C = 21 - 14 ------ Substituting 21 for S in eq (i)

C, or speed of current = highlight_green%287%29 mph

You can do the check!!!

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