Question

A boat can travel from its dock to a town downriver with a 7 mph current in 10 hours. The return trip against the same takes 15 hours. What is the speed of the boat in still water?

Answer 1

I got `35"mi"/h`

Explanation:

Have a look:

where I used:

`"speed"="distance"/"time"`

Answer 2

`v_b = 35 "mi"/h`

Explanation:

The river helps on the downriver part of the trip and hinders on the upriver part of the trip. So the boat's progress downriver to the town is at a rate of `v_b + v_r`.

And the boat's progress upriver back to the dock is at a rate of `v_b - v_r`.

Multiplying speed by time (for example, downriver: (`v_b + v_r)*t_"dr"` gives the distances

`"distance downriver" = (v_b + 7 "mi"/h)*10 hrs`

`"distance upriver" = (v_b - 7 "mi"/h)*15 hrs`

and the distance is the same in both directions. So

`(v_b + 7 "mi"/h)*10 hrs = (v_b - 7 "mi"/h)*15 hrs`

`v_b*10 hrs + 7 "mi"/cancel(h)*10 cancel(hrs) = v_b*15 hrs - 7 "mi"/cancel(h)*15 cancel(h)`

`v_b*10 hrs - v_b*15 hrs = - 7*15 "mi" - 7*10 "mi"`

`- v_b*5 hrs = - 175 "mi"`

`v_b = (175 "mi")/(5 hrs) = 35 "mi"/h`

To verify, plug that in for `v_b` in the first 2 equations and check to see if `"distance downriver " = " distance upriver"`.

I hope this helps,
Steve