A man takes his motor boat down the river to his pub. Going with the current, he can cover the 2 miles in 4 minutes. Returning against the current, which is steady, it takes him 8 minutes.
How long would it take him at still water, when there is no current?
2006-08-02 20:35:39 · 18 answers · asked by brainyandy 6 in Science & Mathematics ➔ Mathematics
If there was no current, it would take 5 min 20 sec to go the 2 miles.
Let the speed of the current = C
Let V = the current independent speed of the motor boat.
So, the current aided speed = V+C; and the current reduced speed = V-C. The distance to the pub is 2 miles. 4mins = 1/15 hr and 8 mins = 2/15 hr. Speed = distance / time. So,
going to the pub his speed = V+C = 2 / (1/15) = 30 mph
The return trip speed = V-C = 1/ (1/15) = 15 mph
So V+C +V-C = 2V = 30+15 = 45 mph, and so V = 22.5 mph
So with no current he can make the 2 mile trip to the pub in (2mi /22.5 mph) x 60 min = 5 1/3 min = 5 min 20 sec. A round trip would take 10 min 40 sec. (And C = 7.5 mph)
2006-08-02 21:30:31 · answer #1 · answered by Jimbo 5 · 1⤊ 0⤋
6
2006-08-02 23:32:55 · answer #2 · answered by blind_chameleon 5 · 0⤊ 0⤋
I am going to assume for this question that he is using the same amount of force to try and get home, and the current is of the same force etc.
If the current is the same both ways, just one time against him and one with him then his time when there is no current will be half way between the with and against times.
The half-way point between 4 and 8 minutes is 6 minutes.
Therefore it would take 6 minutes with no current.
2006-08-03 04:52:07 · answer #3 · answered by Anonymous · 0⤊ 0⤋
With the current he is travelling at 15 mph. Against the current he is travelling at 7.5 mph.
Without know the drag co-efficient of the boat, it is difficult to calculate.
As an estimate, I'm guessing at 3 minutes.
2006-08-02 20:42:09 · answer #4 · answered by Boris 5 · 0⤊ 0⤋
8 minutes
2006-08-02 21:39:23 · answer #5 · answered by cute-t 2 · 0⤊ 0⤋
6 minutes
2006-08-02 20:49:50 · answer #6 · answered by Anonymous · 0⤊ 0⤋
It will take 5 minutes 20 seconds.
2006-08-02 20:51:14 · answer #7 · answered by sharanan 2 · 0⤊ 0⤋
(4 min + 8 min) / 2 = 6 min, assuming the motor speed and current speeds the same on both trips.
2006-08-02 21:03:59 · answer #8 · answered by hellbent 4 · 0⤊ 0⤋
let the speed od boat =x
river current speed =y
then 2/4=x+y
and 2/8=x-y
solving these two, we get,
the motor boat speed is 0.375 miles/min.
current speed is 0.125 miles/min.
If motor boat goes with same speed in still water also then it will take 2/0.375 mins i.e. 6 mins.
2006-08-02 21:31:51 · answer #9 · answered by neeraj j 1 · 0⤊ 0⤋
Jimbo is right 5 mins 20 secs
2006-08-03 08:35:24 · answer #10 · answered by Jam 1 · 0⤊ 0⤋