A swimmer swam 1000 meters downstream in 15 minutes and swam back in 30 minutes against the current. what was the rate of the swimmer in still water? how fast was the current?
1000 = (v+c)15 = 15 v + 15 c
1000 = (v-c)30 = 30 v - 30 c
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2000 = 30 v + 30 c
1000 = 30 v - 30 c
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1000 = 60 c etc
To find the rate of the swimmer in still water and the speed of the current, we can use the concept of relative speed.
Let's assume the rate of the swimmer in still water is 'S' meters per minute, and the speed of the current is 'C' meters per minute.
When swimming downstream, the swimmer gets a boost from the current, so the effective speed is the sum of the swimmer's speed in still water and the speed of the current: S + C.
Similarly, when swimming upstream against the current, the swimmer's effective speed is the difference between the swimmer's speed in still water and the speed of the current: S - C.
We are given that the swimmer swam 1000 meters downstream in 15 minutes. Using the formula distance = speed × time, we can write the equation:
1000 = (S + C) × 15
We are also given that the swimmer swam back in 30 minutes. Using the same formula, we can write another equation:
1000 = (S - C) × 30
Now, we have two equations with two unknowns (S and C). We can solve this system of equations to find the values of S and C.
Let's solve the equations:
From the first equation, we have:
1000 = 15S + 15C --(1)
From the second equation, we have:
1000 = 30S - 30C --(2)
Simplifying equation (1), we get:
S + C = 1000/15
S + C = 66.67 --(3)
Simplifying equation (2), we get:
S - C = 1000/30
S - C = 33.33 --(4)
Now, we can solve equations (3) and (4) simultaneously.
Adding equations (3) and (4), we eliminate the 'C' term:
(S + C) + (S - C) = 66.67 + 33.33
2S = 100
S = 100/2
S = 50
Substituting the value of S in equation (3), we can find 'C':
50 + C = 66.67
C = 66.67 - 50
C = 16.67
Therefore, the rate of the swimmer in still water is 50 meters per minute, and the speed of the current is 16.67 meters per minute.