108/2= 54 km an hour downstream.
54-36=18 km in still water i believe.
hope this ; )
The speed of the motorboat in still water is 45mph
The current rate is 9mph.
Let u be the motorboat speed in still water and v be the current rate.
The effective speed going upstream is
Distance÷Time = 108÷3 = 36mph
It is the DIFFERENCE of the motorboat speed in still water and the rate of the current. It gives you your first equation
u - v = 36. (1)
The effective speed going downstream is
Distance÷Time = 108÷2 = 54mph
It is the SUM of the motorboat speed in still water and the rate of the current. It gives you your second equation
u + v = 54. (2)
Thus you have this system of two equations in 2 unknowns
u - v = 36, (1) and
u + v = 54. (2)
Add the two equations. You will get
2u = 36 + 54 = 90 > u = 90÷2 = 45mph
So, you just found the speed of the motorboat in still water. It is 45mph
Then from the equation (2) you get v = 54 = 45 = 9mph is the current rate.
Answer. The speed of the motorboat in still water is 45mph
The current rate is 9mph
let's say the boat has a still water rate of "b", and the current has a a rate of "c", ok.... when the boat is going upstream, is not really going "b" fast, is going slower at "b - c", due to the current going in the opposite direction.
when the boat is coming downstream, is not going "b" fast either, is going faster, is going "b + c", due to the current adding speed to it.
we know the trip up was 108 kms, thus the return trip is also 108 kms.
and surely you know how much that is.
what's the boat's speed? well, 36 + c = b.
the answer is c
hope this !