two fishing boats have the same average speed in still water. they leave a dock at the same time. one boat heads upstream and the other heads downstream. at a certain point, boat a is 56 km down stream ans boat b is 24 km upstream. the speed of the current is 8kn/h what is the average speed of the boats in still water? textbook answer 20km/h
You say " at a certain point"? I think it should say "after 2 hours".........etc.
Boat "a" average speed per hour=56/2=28 - 8(current's speed)=20 Km/h.
Boat "b" average speed per hour=24/2=12 + 8 ,,,,,,,,,,,,,,,,,,,,,,,,,=20 Km/h
+33653 Let v be the boats speed in still water.
Downstream speed = v + 8 km/hr
Time to get to 56 km: t = 56/(v + 8)
Upstream speed = v - 8 km/hr
Time to get to 24 km: t = 24/(v - 8)
The times are the same so: 56/(v + 8) = 24/(v - 8)
56(v - 8) = 24(v + 8)
32v = 80*8
v = 80*8/32
v = 80/4
v = 20 km/ hr
Boat "a" distance travelled without the water current=56 - 8=48.
Boat "b" ditance travelled without the water current =24 + 8=32
48 - 32 =16 Km that the two boats are apart due to water current
16/8 =2 hours since the two boats left.
56/2 =28 Km/h - 8 Km/h =20 Km/h speed of boat "a"
24/2=12Km/h + 8 Km/h =20 Km/h speed of boat "b"
+129840 Let their rate in still water = x
So....the downstream boat travels at [ x + 8 ] km/h
And the upstream boat travels at [ x - 8 ] km/h
And Distance / Rate = Time .......so....since they travel the same time, we have
56/ [ x + 8] = 24/ [x - 8] cross-multiply
56 [ x - 8] = 24 [ x + 8] simplify
56x - 448 = 24x + 192 subtract 24x from both sides, add 448 to both sides
32x = 640 divide both sides by 32
x = 20 km/h = their rate in still water
