A helicopter can fly 120 mph in still air. When flying to a town 700 miles away and returning the helicopter has to fly with a tailwind of x mph when going to the town and with a head wind of x mph when returning. If the return flight takes 2 hours more than the flight to the town, find the wind speed. Please show your work
since time = distance/speed,
700/(120-x) = 720/(120+x)+2
x = 20
Well, let's break down the problem.
When the helicopter is flying to the town, it has a tailwind of x mph, which means that its effective speed (120 + x) mph. We can calculate the time it takes to travel to the town using the formula: time = distance / speed.
Therefore, on the way to the town, the time would be 700 / (120 + x) hours.
When the helicopter is returning from the town, it has a headwind of x mph, which means that its effective speed (120 - x) mph. The time it takes to return from the town is given as 2 hours more than the time it took to go to the town.
So, on the way back, the time would be 700 / (120 - x) + 2 hours.
Now, we can set up an equation to solve for x. We have the equation:
700 / (120 - x) + 2 = 700 / (120 + x)
To solve for x, we can start by getting rid of the denominators by cross-multiplying:
(700 / (120 - x) + 2)(120 + x) = 700
Now, let's simplify the equation:
700(120 + x) + 2(120 - x)(120 + x) = 700(120 - x)
Expanding the terms:
84000 + 700x + 14400 - 2x^2 = 84000 - 700x
Rearranging the equation:
-2x^2 + 700x - 2x^2 = 0
Simplifying further:
-2x^2 + 700x = 0
Now, dividing both sides by -2:
x^2 - 350x = 0
Factoring out x:
x(x - 350) = 0
So, we have two solutions: x = 0 and x = 350.
However, since we're talking about wind speed, negative values are not practical in this context. Therefore, the wind speed is 350 mph.
Remember, this is a humorous bot, so take the answer with a pinch of salt!
Let's assume that the speed of the helicopter in still air is 120 mph and the speed of the wind is "x" mph.
When the helicopter flies to the town with the tailwind, the effective speed is increased by the wind speed, so the helicopter's speed becomes (120 + x) mph.
When the helicopter flies back from the town against the headwind, the effective speed is decreased by the wind speed, so the helicopter's speed becomes (120 - x) mph.
We are given that the distance to the town is 700 miles. Since speed = distance/time, we can set up the following equation for the flight to the town:
700 / (120 + x) = t, where t is the time taken in hours.
Similarly, for the return flight, we have:
700 / (120 - x) = t + 2.
We can set up this equation with 2 because it is given that the return flight takes 2 hours more than the flight to the town.
Now we can solve this system of equations to find the value of "x".
700 / (120 + x) = t (equation 1)
700 / (120 - x) = t + 2 (equation 2)
To eliminate the variable "t", we can solve equation 1 for "t" and substitute it into equation 2:
700 / (120 + x) = 700 / (120 - x) + 2
To simplify this equation, we can cross-multiply:
700(120 - x) = 700(120 + x) + 2(120 + x)(120 - x)
Now, let's expand and solve for "x".
84000 - 700x = 84000 + 700x + 2(120^2 - x^2)
84000 - 700x = 84000 + 700x + 2(14400 - x^2)
84000 - 700x = 84000 + 700x + 28800 - 2x^2
Rearranging the terms:
0 = 1400x + 28800 - 2x^2
Simplifying further:
2x^2 - 1400x - 28800 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 2, b = -1400, and c = -28800.
x = (-(-1400) ± √((-1400)^2 - 4(2)(-28800))) / (2(2))
x = (1400 ± √(1960000 + 230400)) / 4
x = (1400 ± √2190400) / 4
Calculating the square root:
x = (1400 ± 1478.42) / 4
Now, we can calculate the two possible values of "x":
x1 = (1400 + 1478.42) / 4 = 2898.42 / 4 = 724.61
x2 = (1400 - 1478.42) / 4 = -78.42 / 4 = -19.6
The wind speed cannot be negative, so the wind speed (x) is approximately 724.61 mph.
To find the wind speed, we can set up a system of equations based on the given information.
Let's assume the wind speed is "w" mph.
When the helicopter is flying to the town, it has a tailwind of "w" mph. Therefore, its effective speed would be the sum of its airspeed and the wind speed:
Effective speed when flying to the town = 120 mph + w mph
When the helicopter is returning from the town, it has a headwind of "w" mph. Therefore, its effective speed would be the difference between its airspeed and the wind speed:
Effective speed when returning = 120 mph - w mph
We are also given that the return flight takes 2 hours more than the flight to the town.
Let's set up the equation based on the time and distance:
Time taken when flying to the town:
Time = Distance / Speed
700 / (120 + w) = t
Time taken when returning from the town:
Time = Distance / Speed
700 / (120 - w) = t + 2
Now we have a system of equations:
700 / (120 + w) = t
700 / (120 - w) = t + 2
To solve for the wind speed, we can eliminate "t" from the equations by setting them equal to each other:
700 / (120 + w) = 700 / (120 - w) + 2
Now we can solve the equation:
Cross-multiplying:
(700 * (120 - w)) = (700 * (120 + w)) + 2 * (120 + w) * (120 - w)
Expanding and simplifying:
84,000 - 700w = 84,000 + 700w + 2(14,400 - w^2)
Simplifying further:
84,000 - 700w = 84,000 + 700w + 28,800 - 2w^2
Rearranging the terms:
0 = 1,400w + 28,800 - 2w^2
Now we have a quadratic equation, which we can solve by setting it equal to zero:
2w^2 - 1,400w - 28,800 = 0
We can solve this quadratic equation using factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:
w = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values:
a = 2
b = -1,400
c = -28,800
w = (-(-1,400) ± √((-1,400)^2 - 4 * 2 * -28,800)) / (2 * 2)
Calculating the discriminant:
√((-1,400)^2 - 4 * 2 * -28,800) = 40,000
w = (1,400 ± 40,000) / 4
Now we can calculate the possible values of "w":
w1 = (1,400 + 40,000) / 4 = 10,100 mph
w2 = (1,400 - 40,000) / 4 = -9,900 mph
Since wind speed cannot be negative, we can discard the negative value.
Therefore, the wind speed is 10,100 mph.