Question

​Bob's boat travels at 11 mph in still water. Find the speed of the current if he can go 6 miles upstream in the same time that it takes to go 9 miles downstream. Write an equation that could be used to model this problem. Let c represent the speed​?

word problem

Answer 1

The speed of water current is `2.2` miles/hour.
Model equation is `s_x= (s*(d_d-d_u))/(d_d+d_u)`

Explanation:

Let the speed of water current is `s_x=x` miles/hour . and

the speed of the boat in still water is `s=11` mph

Then the speed of the boat upstream is `s_u=(11-x)` mph

and the speed of the boat downstream is `s_d=(11+x)` mph.

Time taken by the boat `d_u=6` miles upstream is `t_u=6/(11-x)`

hour and time taken by the boat `d_d=9` miles downstream is

`t_d=9/(11+x)` hour. In given codition time is same.

`:. 6/(11-x) = 9/(11+x) or 66+6x= 99-9x` or

`6x+9x= 99-66 or 15x =33 or x= 33/15=2.2` mph

The speed of water current is `2.2` miles/hour

Model equation is `s_x= (s*(d_d-d_u))/(d_d+d_u)`[Ans]

Answer 2

`6/(11 -c) = 9/(11 +c)`

and

`c = 2.2` miles per hour

Explanation:

The speed of the boat in still water = `11` miles per hour

And The speed of the current is `c` miles per houer. [As stated above]

If we are going upstream, then it means that we are going against the current.

So, the effective speed of the boat will be `(11 - c)` miles per hour.

So, the time taken by the boat to go 6 miles = `6/(11 - c)` miles per hour.

And If we are going downstream, then it means that we are going along the current.

So, the effective speed of the boat will be `(11 + c)` miles per hour.

So, the time taken by the boat to go 9 miles = `9/(11 + c)` miles per hour.

Now,

According to the question,

`6/(11 -c) = 9/(11 +c)` [As the time is said to be same]

`rArr 6(11 + c) = 9(11 - c)` [Multiplying Both sides by #(11 + c) (11- c)#]

`rArr 66 + 6c = 99 - 9c`

`rArr 15c = 33`

`rArr c = 33/15 = 11/5 = 2.2` miles per hour

Answer 3

Model equation explanation:

Time is same for downstream and upstrem, so equating the

`t_u and t_d` we get `d_u/(s-s_x)= d_d/(s+s_x)`

`:. d_u*(s+s_x)= d_d*(s-s_x)` or

`d_u*s+d_u*s_x= d_d*s-d_d*s_x` or

`d_d*s_x+d_u*s_x= d_d*s-d_u*s` or

`s_x(d_d+d_u)= s(d_d-d_u)`

`:.s_x= (s(d_d-d_u))/((d_d+d_u))`

`s=11 , d_d=9 , d_u=6 , :. s_x=(11(9-6))/(9+6)=33/15=2.2`mph