tugboat goes 160 miles upstream in 10 hours. The return trip downstream takes 5 hours. Find the speed of the tugboat without a current and the speed of the current.
speed of tugboat is
speed of the current is
Well, if the tugboat goes 160 miles upstream in 10 hours, that means it is traveling at a speed of 16 miles per hour.
Now, for the return trip downstream, it takes the tugboat only 5 hours to cover the same 160 miles.
So, if we subtract the downstream speed from the upstream speed, we can find the speed of the current.
16 miles per hour (upstream speed) minus x miles per hour (current speed) equals 160 miles divided by 5 hours (downstream speed):
16 - x = 160/5
Simplifying, we have:
16 - x = 32
Now, we can solve for x by subtracting 16 from both sides:
-x = 32 - 16
-x = 16
Dividing both sides by -1, we find:
x = -16
Wait a minute...negative speed for the current? That doesn't make sense! I think we need to take a closer look at the problem, because I'm pretty sure water doesn't flow backwards.
To find the speed of the tugboat without a current, let's assume it is represented by 'x' (in miles per hour).
Now, let's calculate the speed of the current. Let's assume it is represented by 'y' (in miles per hour).
When the tugboat is going upstream, it faces the current, so its effective speed is reduced by the speed of the current. Therefore, the speed of the tugboat going upstream is (x - y) miles per hour.
Similarly, when the tugboat is going downstream, it moves with the current, so its effective speed is increased by the speed of the current. Therefore, the speed of the tugboat going downstream is (x + y) miles per hour.
We are given that the tugboat goes 160 miles upstream in 10 hours, so we can write the equation:
160 = (x - y) * 10
We are also given that the return trip downstream takes 5 hours, so we can write the equation:
160 = (x + y) * 5
Now we have a system of two equations:
10(x - y) = 160
5(x + y) = 160
Simplifying the first equation:
10x - 10y = 160
Divide by 10:
x - y = 16 [Equation 1]
Simplifying the second equation:
5x + 5y = 160
Divide by 5:
x + y = 32 [Equation 2]
Now we can solve this system of equations. Adding Equation 1 and Equation 2 together:
(x - y) + (x + y) = 16 + 32
2x = 48
Divide by 2:
x = 24
Substitute the value of x in any of the two equations to find y:
24 + y = 32
y = 32 - 24
y = 8
Therefore,
The speed of the tugboat without a current is 24 miles per hour.
The speed of the current is 8 miles per hour.
To find the speed of the tugboat without the current and the speed of the current, we need to use the formula relating distance, speed, and time:
Distance = Speed × Time
Let's denote the speed of the tugboat as "T" and the speed of the current as "C".
In the upstream trip, the effective speed of the tugboat is reduced by the current, so we have the equation:
160 = (T - C) × 10
In the downstream trip, the current adds to the effective speed of the tugboat, so we have:
160 = (T + C) × 5
We now have a system of two equations with two unknowns. We can solve this system of equations to find the values of T and C.
First, let's simplify the equations:
10T - 10C = 160 (Equation 1)
5T + 5C = 160 (Equation 2)
Now we can solve this system of equations.
Multiply Equation 1 by 2 to eliminate the variable "C":
20T - 20C = 320 (Equation 3)
Add Equation 3 to Equation 2:
20T - 20C + 5T + 5C = 320 + 160
25T - 15C = 480 (Equation 4)
Next, let's multiply Equation 2 by 2 to eliminate the variable "T":
10T + 10C = 320 (Equation 5)
Subtract Equation 5 from Equation 4:
25T - 15C - (10T + 10C) = 480 - 320
Simplifying:
15T - 25C = 160
Divide both sides of the equation by 5:
3T - 5C = 32 (Equation 6)
Now we have a new equation relating T and C.
To isolate T in Equation 6, we can multiply it by 3:
9T - 15C = 96 (Equation 7)
Now, multiply Equation 1 by 5 to eliminate T:
50T - 50C = 800 (Equation 8)
Add Equation 8 to Equation 7:
50T - 50C + 9T - 15C = 800 + 96
Simplifying:
59T - 65C = 896
Divide both sides of the equation by 59:
T - (65/59)C = 896/59
To obtain a simpler equation, let's multiply both sides by 59:
59T - 65C = 896
Now we have a simplified equation:
59T - 65C = 896 (Equation 9)
Now, we can solve this equation for T:
T = (896 + 65C) / 59 (Equation 10)
We can substitute this value of T into Equation 7 to solve for C:
9T - 15C = 96
Substitute T from Equation 10:
9[(896 + 65C) / 59] - 15C = 96
Simplify:
5664 + 585C - 885C = 5664
Combine like terms:
-300C = 0
Divide both sides by -300:
C = 0
Therefore, the speed of the current is 0 mph.
Substituting this value of C into Equation 10, we can solve for T:
T = (896 + 65(0)) / 59
T = 896 / 59
T ≈ 15.1864
Therefore, the speed of the tugboat without the current (T) is approximately 15.1864 mph, and the speed of the current (C) is 0 mph.
I don't really understand what I have to do.
speed of tugboat in still water -- x
speed of current ----- y
the key is that the distance is the same
so 10(x-y) = 160 or
x - y = 16
and 5(x+y) = 160
x+y = 32
add them:::
2x = 48
x = 24 , then y = 8