Question

A boat takes 8 hours to go 6 miles upstream and come back to the starting point. If the speed of the boat is 4 miles per hour in still water, what is the rate of the current? Hint: total time taken = time taken to go upstream + time taken to go downstream `"Time taken to go upstream" = "distance"/"speed of boat in still water + rate of current"` ` ` `"Time taken to go downstream" = "distance"/"speed of boat in still water - rate of current"`

a. 10miles per hour
b.8.63 miles per hour
c.3.16 miles per hour
d.2.24 miles per hour

Answer 1

Let the speed of the water current be v.

total travel time = time to go upstream + time to return downstream to the starting point

Hence, 8 = time in hours = 6 / (4-v) + 6 / (4 +v) = 6*8/(16-v^2)

Hence, 16 - v^2 = 6, or v^2 = 10 or v = √10 miles / hour

Answer 2

The correct option is: c. 3.16 miles per hour.

Explanation:

Suppose, the rate of the current
`= x` mph.

Speed of the boat is
`4` mph in still water.

So, speed of the boat in upstream
`= (4-x)` mph and speed of the boat in downstream
`=(4+x)` mph

We know,
`Time= (Distance)/(Speed)`

For upstream, distance
`= 6` miles and speed
`=(4-x)` mph

So, the time taken in upstream
`=(6)/(4-x)` hours

For downstream, distance
`= 6` miles and speed
`=(4+x)` mph

So, the time taken in downstream
`= (6)/(4+x)` hours

Now the total time taken for upstream and downstream is
`8` hours, so the equation will be.....

`(6)/(4-x)+(6)/(4+x)= 8\\ \\ (24+6x+24-6x)/((4-x)(4+x))=8\\ \\ (48)/(16-x^2)=8\\ \\ 8(16-x^2)=48\\ \\ 16-x^2=(48)/(8)\\ \\ 16-x^2=6\\ \\ -x^2=6-16\\ \\ -x^2=-10\\ \\ x^2=10\\ \\ x=√(10)\approx 3.16`

So, the rate of the current is 3.16 miles per hour.