Question

A boat traveled 210 miles downstream and back. The trip downstream took 10 hours. The trip back took 70 hours. What is the speed of the the boat in still water? What is the speed of the current?

Answer 1

`v = 6["miles/hour"]`
`v_s = 9/2["miles/hour"]`

Explanation:

A boat traveled 210 miles downstream and back, covering the same distance twice. So

`2d=210`

Now, calling `Delta t_d` and `Delta t_u` the required times to cover downstream and upstream trips,

`d = (v + v_s)Delta t_d` and
`d = (v-v_s)Delta t_u`

where `v, v_s` are respectively the relative boat speed regarding the stream, and `v_s` the absolute stream speed.

Solving for `v,v_s` we obtain

`v = (d(Delta t_u + Delta t_d))/(2(Delta t_d Delta t_u))` and
`v_s = (d(Delta t_u - Delta t_d))/(2(Delta t_d Delta t_u))`

Substituting values we get at

`v = 6["miles/hour"]`
`v_s = 9/2["miles/hour"]`

Answer 2

Speed of the the boat in still water`=6mph`
Speed of the current`=4.5mph`

Explanation:

Total distance traveled by the boat is given `=210mil es`
One way travel is `210/2=105 mil es`
Let `v_band v_s` be speed of boat in still water and speed of stream respectively.

We know that `"Distance" ="Speed" xx "time"`
For downstream travel time `=10hours`
Downstream Speed is `=v_b+v_s`
`:.` `105=(v_b+v_s)xx10`
`=>(v_b+v_s)=10.5` .....(1)
The upstream speed being `(v_b-v_s)`
For upstream travel time `=70hours`
`:.` `105=(v_b-v_s)xx70`
`=>(v_b-v_s)=1.5` ......(2)
Adding (1) and (2) we obtain
`2v_b=12`
`=>v_b=6mph`
Subtracting (2) from (1) we get
`2v_s=9`
`=>v_s=4.5mph`