A motorboat takes 5 hours to travel 100mi going upstream. The return trip takes 2 hours going downstream. What is the rate of the boat in still water and what is the rate of the current?
To find the rate of the boat in still water and the rate of the current, we can use the concept of relative speed. Let's assume the rate of the motorboat in still water is "b" and the rate of the current is "c."
Going upstream:
The speed of the boat relative to the water is reduced by the speed of the current, so the effective speed is (b - c). We know that the boat takes 5 hours to cover 100 miles upstream, so we can set up the equation:
Time = Distance / Speed
5 = 100 / (b - c)
Going downstream:
The speed of the boat relative to the water is increased by the speed of the current, so the effective speed is (b + c). We know that the boat takes 2 hours to cover 100 miles downstream, so we can set up the equation:
Time = Distance / Speed
2 = 100 / (b + c)
Now, we have a system of two equations with two unknowns. We can solve these equations simultaneously to find the values of "b" and "c."
First, let's rearrange the first equation:
5(b - c) = 100
b - c = 20 (Equation 1)
Next, let's rearrange the second equation:
2(b + c) = 100
b + c = 50 (Equation 2)
Now, we have a system of two equations:
b - c = 20 (Equation 1)
b + c = 50 (Equation 2)
We can solve this system of equations by adding Equation 1 and Equation 2:
( b - c) + (b + c) = 20 + 50
2b = 70
b = 35
Substituting the value of "b" in Equation 2, we can find "c":
35 + c = 50
c = 50 - 35
c = 15
Therefore, the rate of the boat in still water is 35 mph and the rate of the current is 15 mph.
distance = time * speed
Let
b = boat's speed
c = current's speed
5(b-c) = 100
2(b+c) = 100
b-c = 20
b+c = 50
I guess you can take it from there, yes?