a boat travel at a speed of 20 km/h in still water the current in a river flows at 5 km/h
so that downstream the boat can travel at 25 km/h and upstream it travels at only 15 km/h the boat has only enough fuel for 3 hours.after what time does the boat turn around so that it has enough fuel to return to base
Let us say boat travel xkm distance from base in t1 hours down stream when it turn around and comes upstream.
`t1 = x/25`
For the return travel let us say it takes t2 hours.
`t2 = x/15`
How ever at the critical time limit;
`t1+t2 = 3`
`x/25+x/15 = 3`
`(x(3+5))/75 = 3`
`x = (75xx3)/8`
`x = 28.125`
`t1 = x/25 = 28.125/25 = 1.125`
So the boat should turn around in 1.125 hours to return to base without running out of fuel.
The boat travels at a speed of 20 km/h in still water and water in the river flows at 5 km/h. When the boat travels downstream its speed is 20+5 = 25 km/h and when it travels upstream the speed of the boat is 20 - 5 = 15 km/h.
The fuel in the boat is enough for it to travel for 3 hours. It has to travel along the river, turn around and return to base.
Assume the boat travels downstream first and travels distance D.
The time required for this is D/25.
It then has to travel back upstream in time 3 - D/25.
3 - D/25 is equal to D/15
3 - D/25 = D/15
=> 3 = D(1/15 + 1/25)
D = 3/(1/15 + 1/25) = 225/8
If the boat travels upstream first and has to turn back after distance D, the time spent traveling in this direction is D/15. It has to now complete D traveling at 25 km/h in 3 - D/15 hours.
D/25 = 3 - D/15
D(1/25 + 1/15) = 3
D = 3/(1/25 + 1/15)
D = 225/8
The distance the boat can travel is 28.125 km irrespective of which direction it travels in first.