A boat travels 45 miles upstream in 5h. The boat travels the same distance downstream in 3hr. What is the rate of the boat in still water? What is the rate of the current? (thank you)

still-water speed: s

river speed: r

45 = 5(s-r) = 3(s+r)

5s - 5r = 45
3s + 3r = 45
multiply to match coefficients of r:
15s-15r=135
15s+15r=225

30s = 360
s=12
r=3

check:
45mi/9mph = 5 hr
45mi/15mph = 3 hr

To find the rate of the boat in still water, we can use the concept of relative motion. Let's assume the rate of the boat in still water is 'B' mph and the rate of the current is 'C' mph.

When the boat travels upstream, against the current, its effective speed decreases. So, the boat's actual speed is given by (B - C) mph.

Similarly, when the boat travels downstream, with the current, its effective speed increases. So, the boat's actual speed is given by (B + C) mph.

We are given that the boat travels 45 miles upstream in 5 hours, so we can write:

45 = (B - C) * 5 ---(1)

And the boat travels the same distance downstream in 3 hours, so we can write:

45 = (B + C) * 3 ---(2)

We can now solve these equations to find the values of B and C.

First, let's simplify equation (1):

45 = 5B - 5C
9 = B - C ---(3)

Now, let's simplify equation (2):

45 = 3B + 3C
15 = B + C ---(4)

Now, we can solve equations (3) and (4) simultaneously. Adding equation (3) and equation (4), we get:

9 + 15 = B - C + B + C
24 = 2B
B = 12

Substituting the value of B into equation (4), we can find the value of C:

15 = 12 + C
C = 3

Therefore, the rate of the boat in still water is 12 mph, and the rate of the current is 3 mph.

To find the rate of the boat in still water, we need to consider the rate of the current as well. Let's assume the rate of the boat in still water is 'b' miles per hour and the rate of the current is 'c' miles per hour.

When the boat is traveling upstream, it is going against the current. The effective rate is reduced by the rate of the current. So, the equation for this situation would be:

b - c = (distance traveled upstream) / (time taken upstream)

Given that the distance traveled upstream is 45 miles and the time taken is 5 hours, we can substitute these values into the equation:

b - c = 45/5

Simplifying this equation gives us:

b - c = 9

Similarly, when the boat is traveling downstream, it is going with the current. The effective rate is increased by the rate of the current. So, the equation for this situation would be:

b + c = (distance traveled downstream) / (time taken downstream)

Given that the distance traveled downstream is also 45 miles and the time taken is 3 hours, we can substitute these values into the equation:

b + c = 45/3

Simplifying this equation gives us:

b + c = 15

Now, we have a system of two equations with two variables:

b - c = 9
b + c = 15

To solve this system of equations, we can add the two equations together. This eliminates the 'c' variable:

(b - c) + (b + c) = 9 + 15

Simplifying this equation gives us:

2b = 24

Dividing both sides by 2, we get:

b = 12

So, the rate of the boat in still water is 12 miles per hour.

To find the rate of the current, we can substitute this value back into one of the equations. Let's use the equation b + c = 15:

12 + c = 15

Subtracting 12 from both sides, we get:

c = 3

Therefore, the rate of the current is 3 miles per hour.

So, the rate of the boat in still water is 12 miles per hour, and the rate of the current is 3 miles per hour.