Question

Ben's boat will take 1½ hours longer to go 12 miles up a river than to return. What is the rate of his boat if the rate of the current is 2 miles an hour?

Answer 1

Rate of the boat `v=6 " "`miles/hour

Explanation:

From the given data:
Let `d=12` miles
Let `t=`time going down with the river current
Let `v =`the rate of Ben's boat
Let `v_c=2` miles/hour-rate of the current

Let `v-v_c=`the rate going up the river
Let `v+v_c=`the rate going down the river

`(v+v_c)(t)=12" "` the equation for distance going down
`(v-v_c)(t+3/2)=12" "`the equation for distance going up

Let us do the simultaneous solution of these equations

From `(v+v_c)(t)=12" "` the equation for distance going down

`(v+v_c)(t)=12" "`

`t=color(red)(12/(v+v_c)" ")`substitute this in the equation for distance going up

`(v-v_c)(t+3/2)=12" "`equation for distance going up
`(v-v_c)(color(red)(12/(v+v_c))+3/2)=12" "`
Now change `v_c=2`

`(v-2)(color(red)(12/(v+2))+3/2)=12" "`

We can now solve for v:
Multiply both sides by (v+2)

`(v-2)(12/(v+2)+3/2)(v+2)=12(v+2)`

`(12(v-2)(v+2))/(v+2)+3/2*(v-2)(v+2)=12v+24`

Simplify

`(12(v-2)cancel(v+2))/cancel(v+2)+3/2*(v-2)(v+2)=12v+24`

`12(v-2)+3/2*(v-2)(v+2)=12v+24`
Expand
`12v-24+3/2(v^2-4)=12v+24`

Simplify

`3/2(v^2-4)=48`

`v^2-4=32`

`v^2=32+4`

`v^2=36`

`v=6" "`miles/hour