A tugboat goes 180 miles upstream in 15 hours. The return trip downstream takes 10 hours. find the speed of the tugboat without a current and the speed of the current
distance=(velocityboat+vstream)time
180=(vb+Vs)*10
180=(Vb-Vs)*15
divide both sides of a) by 10, and b) by 15
18=vb+vs
12=Vb-Vs
add the equations
30=2vb solve for vboat
then put that into either equation and solve for vstream
To find the speed of the tugboat without a current and the speed of the current, we can use a system of equations.
Let's denote the speed of the tugboat without a current as "x" and the speed of the current as "y".
When the tugboat is going upstream (against the current), its effective speed is reduced by the speed of the current. Therefore, the equation for the upstream trip can be written as:
180 = (x - y) * 15
Similarly, when the tugboat is going downstream (with the current), its effective speed is increased by the speed of the current. So, the equation for the downstream trip can be written as:
180 = (x + y) * 10
Now, we can solve this system of equations to find the values of x and y.
First, let's simplify the equations:
15x - 15y = 180 --> Equation 1
10x + 10y = 180 --> Equation 2
Now, we can solve the system of equations using any method we prefer. One straightforward way is to eliminate y by multiplying both sides of Equation 1 by 2 and adding it to Equation 2:
30x - 30y + 10x + 10y = 360 + 180
40x = 540
x = 13.5
Substituting the value of x back into Equation 1:
15(13.5) - 15y = 180
202.5 - 15y = 180
-15y = 180 - 202.5
-15y = -22.5
y = 1.5
Therefore, the speed of the tugboat without a current (x) is 13.5 mph, and the speed of the current (y) is 1.5 mph.