Two airplanes start a 1000 mile journey to another city. One airplane travels 50 mph faster than the other airplane and arrives at the city an hour before the slower plane. At what speed does each airplane travel?
Let's assume the speed of the slower airplane is x mph.
Therefore, the speed of the faster airplane will be x + 50 mph.
We know that distance = speed × time.
For the slower airplane, its distance is 1000 miles and the time it takes is t + 1 hour.
So, using the formula distance = speed × time, we can write the equation as 1000 = x(t + 1).
For the faster airplane, its distance is also 1000 miles and the time it takes is t hours.
Using the same formula, we can write the equation as 1000 = (x + 50)t.
Now, we have two equations with two variables:
1) 1000 = x(t + 1)
2) 1000 = (x + 50)t
We can simplify the equations:
1) 1000 = xt + x
2) 1000 = xt + 50t
Subtracting equation 1 from equation 2, we get:
0 = 50t - x
Rearranging this equation, we get:
x = 50t
Substituting this value of x into equation 1, we get:
1000 = 50t(t + 1)
Expanding and simplifying the equation, we get:
1000 = 50t^2 + 50t
Dividing both sides of the equation by 50:
20 = t^2 + t
Rearranging the equation:
t^2 + t - 20 = 0
Solving this quadratic equation, we find two possible values for t: t = 4 and t = -5.
Since time cannot be negative in this scenario, we discard t = -5.
Therefore, the slower airplane takes 4 hours to travel the 1000 miles.
From x = 50t, we can substitute t = 4 into this equation to find the speed of the slower airplane:
x = 50(4) = 200 mph
Therefore, the speed of the slower airplane is 200 mph.
The speed of the faster airplane is x + 50, so substituting x = 200 into this equation:
x + 50 = 200 + 50 = 250 mph
Therefore, the speed of the faster airplane is 250 mph.