# Going against the current, a boat takes 6 hours to make a 120-mile trip. When the boat travels with the current on the return trip, it takes 5 hours. What is the rate of the boat in still water?

Jun 13, 2017

Assuming the current is constant, the average of the rates of travel with and against the current should cancel each out to get the rate of travel in still water.

#### Explanation:

Against the current, the boat travels 120 miles in 6 hours. to get the rate of travel, in mph, you need to divide the distance traveled by the time it took.

120 miles/6 hours = 20 mph.

You need to do the same for the boat going with the current.

120 miles/ 5 hours = 24 mph.

Now, you need to average the two rates. The "against current" and "with current" should cancel each other out (again, assuming the current is constant) to get the "still water" rate.

$\frac{20 + 24}{2} =$ $\frac{44}{2} =$22 mph.

The rate that the boat travels in still water is 22 mph.

Jun 13, 2017

$22$ $\text{mi/h}$

#### Explanation:

I'll assume the boat and current move in one dimension, and that the speed of the current is consistent in both routes.

This question is actually simpler than it looks; we can say that the speed (rate) of the boat in still water is the average of the speed with the current and against the current

The speed ${v}_{1 x}$ against the current is

${v}_{1 x} = \left(120 \text{mi")/(6"h}\right) = 20$ $\text{mi/h}$

The speed ${v}_{2 x}$ with the current is

${v}_{2 x} = \left(120 \text{mi")/(5"h}\right) = 24$ $\text{mi/h}$

The velocity in still water is thus

v_"still" = (v_(1x) + v_(2x))/2 = (20"mi/h" + 24"mi/h")/2 = color(red)(22 color(red)("mi/h"