Question

You decide to enter in a rowing competition. To train, you go to the boat house and begin rolling down stream. You row for the same amount of time upstream and downstream, travel down stream for 2400 m, but only one sixth that distance upstream. The current travels at a rate of 10 m/s. What is the time you spend rowing down stream? How fast would you be rolling in Stillwater?

Answer 1

Time spent rowing down stream
`=100\ seconds`

Speed of boat in still water
`=14\ ms^(-1)`

Explanation:

Let speed of boat in still water be
`= x\ ms^(-1)`

Speed of current
`= 10\ ms^(-1)`

Speed of boat down stream =
`\textrm{Speed of boat in still water +Speed of current}= (x+10)\ ms^(-1)`

Distance rowed down stream = 2400 m

Time spent rowing down stream =
`(Distance)/(Speed)=(2400)/(x+10)\ s` =
`(Distance)/(Speed)=(2400)/(x+10)\ s`

Speed of boat up stream =
`\textrm{Speed of boat in still water -Speed of current}= (x-10)\ ms^(-1)`

Distance rowed up stream =
`(1)/(6) \textrm{ of distance rowed downstream}=(1)/(6)* 2400 = 400\ m`

Time spent rowing up stream =
`(Distance)/(Speed)=(400)/(x-10)\ s`

We know that,

`\textrm{Time spent rowing down stream =Time spent rowing up stream}`

So,

`(2400)/(x+10)=(400)/(x-10)`

Cross multiplying

`2400(x-10)=400(x+10)`

Dividing both sides by
`400`

`(2400(x-10))/(400)=(400(x+10))/(400)`

`6(x-10)=x+10`

`6x-60=x+10`

Adding 60 to both sides.

`6x-60+60=x+10+60`

`6x=x+70`

Subtracting both sides by
`x`

`6x-x=x+70-x`

`5x=70`

Dividing both sides by 5.

`(5x)/(5)=(70)/(5)`

∴
`x=14`

Speed of boat in still water
`=14\ ms^-1`

Time spent rowing down stream =
`(2400)/(14+10)=(2400)/(24)=100\ s`