In a canoe race, a team paddles downstream 480 m. in 60 s. The same team travels up stream in 80 s. Find the team's rate in still water and the rate of the current.
Please show how the equation is set up to solve
Let the distance paddled (one way) be L = 480 m
Let V be the canoe speed in still water and v be the stream speed.
L/(V+v) = 60 s
L/(V-v) = 80 s
(V+v)/(V-v) = 80/60 = 4/3
V + v = (4/3)V - (4/3)v
V/3 = (7/3)v
v = V/7
480/(8V/7) = 60
V = 7 m/s
v = 1 m/s
Let's assume the speed of the canoe in still water is represented by "x" meters per second, and the speed of the current (or river flow) is represented by "y" meters per second.
When the team paddles downstream, their effective speed is the sum of the speed in still water and the current, so it is (x + y).
According to the problem, the team paddles downstream (with the current) for 480 meters in 60 seconds, which gives us the equation:
480 / 60 = x + y
Simplifying, we have:
8 = x + y .....(1)
When the team paddles upstream, their effective speed is the difference between the speed in still water and the current, so it is (x - y).
According to the problem, the team paddles upstream (against the current) for 480 meters in 80 seconds, which gives us the equation:
480 / 80 = x - y
Simplifying, we have:
6 = x - y .....(2)
Now, we have a system of two equations with two variables (x and y). We can solve these equations simultaneously to find the values of x and y, which represent the team's rate in still water and the rate of the current.
Let's solve the system of equations:
Equation (1): 8 = x + y
Equation (2): 6 = x - y
Add equation (1) and equation (2) together:
8 + 6 = (x + y) + (x - y)
14 = 2x
Divide both sides by 2:
14 / 2 = 2x / 2
7 = x
Now substitute the value of x = 7 into either Equation (1) or Equation (2) to solve for y:
Equation (1): 8 = x + y
8 = 7 + y
Subtract 7 from both sides:
8 - 7 = 7 - 7 + y
1 = y
Therefore, the team's rate in still water is 7 meters per second, and the rate of the current is 1 meter per second.
To solve this problem, let's assume that the rate of the canoe in still water is represented by c (in meters per second) and the rate of the current is represented by r (also in meters per second).
When the team is paddling downstream, their effective speed is increased by the speed of the current. Therefore, their rate downstream is given by (c + r). We are told that they cover a distance of 480 meters in 60 seconds, so we can set up the equation:
480 m = (c + r) * 60 s
Similarly, when the team paddles upstream, their effective speed is decreased by the speed of the current. Therefore, their rate upstream is given by (c - r). We are told that they cover the same distance (480 meters) in 80 seconds, so we can set up another equation:
480 m = (c - r) * 80 s
Now, we have a system of two equations that we can solve simultaneously to find the values of c and r.
Let's simplify each equation by dividing both sides by the time:
1) 480 m / 60 s = c + r
2) 480 m / 80 s = c - r
Simplifying further, we get:
1) 8 m/s = c + r
2) 6 m/s = c - r
Now, we can solve this system of equations using a method like substitution or elimination.
Let's add equation 1) and equation 2) to eliminate r:
(8 m/s) + (6 m/s) = (c + r) + (c - r)
14 m/s = 2c
Divide both sides by 2:
7 m/s = c
Now we have the value for c, which represents the team's rate in still water, but we also need to find the value for r, which represents the rate of the current.
Let's substitute the value of c back into either equation 1) or equation 2):
8 m/s = (7 m/s) + r
Subtract 7 m/s from both sides:
8 m/s - 7 m/s = r
1 m/s = r
Therefore, the team's rate in still water is 7 m/s, and the rate of the current is 1 m/s.