Question

HALLP QUICKKKK

A tourist sailed against the current on a river for 6 km, and then he sailed in a lake for 15 km. In the lake he sailed for 1 hour longer than he sailed in the river. Knowing that the current of the river is 2 km/hour, find the speed of the boat while it is traveling in the lake.

Answer 1

To solve this we are going to use the formula for speed:
`S= (d)/(t)`
where

`S` is the speed

`d` is the distance

`t` is the time

Let
`S_(l)` be the speed of the boat in the lake,
`S_(a)` the speed of the boat in the river,
`t_(l)` the time of the boat in the lake, and
`t_(a)` the time of the boat in the river.

We know for our problem that the current of the river is 2 km/hour, so the speed of the boat in the river will be the speed of the boat in the lake minus 2km/hour:

`S_(a)=S_(l)-2`
We also know that in the lake the boat sailed for 1 hour longer than it sailed in the river, so:

`t_(l)=t_(a)+1`

Now, we can set up our equations.
Speed of the boat traveling in the river:

`S_(a)= (6)/(t_(a) )`
But we know that
`S_(a)=S_(l)-2`, so:

`S_(l)-2= (6)/(t_(a) )` equation (1)

Speed of the boat traveling in the lake:

`S_(l)= (15)/(t_(l) )`
But we know that
`t_(l)=t_(a)+1`, so:

`S_(l)= (15)/(t_(a)+1)` equation (2)

Solving for
`t_(a)` in equation (1):

`S_(l)-2= (6)/(t_(a) )`

`t_(a)= (6)/(S_(l)-2)` equation (3)

Solving for
`t_(a)` in equation (2):

`S_(l)= (15)/(t_(a)+1)`

`t_(a)+1= (15)/(S_(l))`

`t_(a)=(15)/(S_(l))-1`

`t_(a)= (15-S_(l))/(S_(l))` equation (4)

Replacing equation (4) in equation (3):

`t_(a)= (6)/(S_(l)-2)`

`(15-S_(l))/(S_(l))=(6)/(S_(l)-2)`

Solving for
`S_(l)`:

`(15-S_(l))/(S_(l))=(6)/(S_(l)-2)`

`(15-S_(l))(S_(l)-2)=6S_(l)`

`15S_(l)-30-S_(l)^2+2S_(l)=6S_(l)`

`S_(l)^2-11S_(l)+30=0`

`(S_(l)-6)(S_(l)-5)=0`

`S_(l)=6` or
`S_(l)=5`

We can conclude that the speed of the boat traveling in the lake was either 6 km/hour or 5 km/hour.