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

Answer:

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

Step-by-step 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,  $Time= \frac{Distance}{Speed}$

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

So, the time taken in upstream $=\frac{6}{4-x}$ hours

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

So, the time taken in downstream $= \frac{6}{4+x}$ hours

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

$\frac{6}{4-x}+\frac{6}{4+x}= 8\\ \\ \frac{24+6x+24-6x}{(4-x)(4+x)}=8\\ \\ \frac{48}{16-x^2}=8\\ \\ 8(16-x^2)=48\\ \\ 16-x^2=\frac{48}{8}\\ \\ 16-x^2=6\\ \\ -x^2=6-16\\ \\ -x^2=-10\\ \\ x^2=10\\ \\ x=\sqrt{10}\approx 3.16$

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