A tugboat is going 100 miles upstream in 10 hours and returning downstream in 5 hours. What is the boats speed and the speed of the current?
Vmin = 100mi / 10hrs =10 mi/hr.
Vmax = 100mi / 5hrs = 20 mi/hr.
Vb = Vavg = (Vmin+Vmax)/2.
Vb = ((10+20) / 2 = 15 mi/hr.= Velocity
of boat.
Vc = (Vmax-Vmin)/2.
Vc = (20-10) / 2 = 5 mi/hr. = Velocity of current.
To find the boat's speed and the speed of the current, we can set up a system of equations. Let's denote the boat's speed as 'b' and the speed of the current as 'c'.
When the boat is traveling upstream, it is moving against the current, so its effective speed is reduced. The equation for the upstream journey can be written as:
100 miles = (b - c) * 10 hours
Similarly, when the boat is traveling downstream, it is moving with the current, so its effective speed is increased. The equation for the downstream journey can be written as:
100 miles = (b + c) * 5 hours
Now we have a system of two equations:
1. (b - c) * 10 = 100
2. (b + c) * 5 = 100
To solve this system, we can use the method of substitution or elimination.
Method of Substitution:
1. Rearrange the first equation to solve for b-c: 10(b - c) = 100
Simplify: 10b - 10c = 100
Rearrange: b - c = 10
2. Substitute b - c = 10 into the second equation: (b + c) * 5 = 100
Simplify: 5b + 5c = 100
Rearrange: b + c = 20
3. Solve the system of equations:
Multiply equation 1 by 5: 5b - 5c = 50
Multiply equation 2 by 10: 10b + 10c = 200
Add the equations together: 15b = 250
Divide by 15: b = 16.67
4. Substitute the value of b back into equation 1: 16.67 - c = 10
Simplify: -c = -6.67
c = 6.67
So, the boat's speed is approximately 16.67 miles per hour, and the speed of the current is approximately 6.67 miles per hour.