Question

A motorboat used as a river taxi can travel at 8 mph in calm water. the amount of time it takes to travel 9 miles against the current is the same amount of time it takes to travel 15 miles with the current. what is the rate of the current?

Answer 1

Hello

1) Let's call
`v_0 = 8~m/h` the velocity of the motorboat in calm water. When it travels against the current, the relative velocity will be
`v_0-v_c`, where
`v_c` is the current velocity. Similarly, when the motorboat travels with the current, its relative velocity will be
`v_0+v_c`.

2) Given
`v_r= (S)/(t)`, where
`v_r` is the relative velocity, when the motorboat travels against the current the time to cover a distance of
`S_1=9~m` is

`t_1= (S_1)/(v_0-v_c)`
Similarly, when the motorboat travels with the current, the time to cover a distance of
`S_2=15~m` is

`t_2=(S_2)/(v_0+v_c)`

3) The problem says that these two times are equal. So we can write:

`(S_1)/(v_0-v_c)=(S_2)/(v_0+v_c)`
And we can solve this to find
`v_c`, the current rate:

`v_c=v_0 (S_2-S_1)/(S_2+S_1) = 8~mph (15~m-9~m)/(15~m+9~m)=2 mph`