An airline runs a commuter flight between Portland and Seattle, which are 145 miles apart. If the average air speed could be increased by 40 miles per hour, the travel time would be decreased by 12 minutes. What air speed is required to obtain this decrease in travel time?
-I know I have to set up equations and use the quadratic equation, however I don't know how to do this. I believe the answer should be 190. Thanks for any help in advance!
how did s^2 + 40s - 29000 = 0
become s = 151.46 mi/hr?
Just think about what you have and what you need.
distance (miles) is speed(mi/hr) * time (hr)
If t is the current travel time, and s is the current speed, then
st = 145
t = 145/s
At the new speed (s+40), the new time would be t-12/60 hours. The distance is the same
(s+40)(t - 1/5) = 145
(s+40)(145/s - 1/5) = 145
Putting it all over 5s, we get for the numerator:
(s+40)(725-s) = 725s
725s + 29000 - s^2 - 40s - 725s = 0
(We can disregard the denominator of 5s, since it is not zero here, and the fraction is zero when the numerator is zero)
s^2 + 40s - 29000 = 0
s = 151.46 mi/hr
The new required speed is s+40 = 191.46 mi/hr
Well, well, well, looks like someone needs some help with math! Don't worry, I'm here to clown around and help you out. Let's make this journey together!
First, let's set up the equation. We have two scenarios here - the original speed and the increased speed. Let's call the original speed "x" (because I feel like being fancy) and the increased speed "x + 40".
Now, let's talk about time. The original travel time is 145 miles divided by x mph. The decreased travel time is 12 minutes less than the original travel time, which means it's 145 miles divided by (x + 40) mph minus 12 minutes.
Now, here's where the math fun comes in. We can set up an equation using these two scenarios:
145 / x = 145 / (x + 40) - 12/60
Let's simplify this equation by multiplying everything by x times (x + 40) to get rid of those pesky denominators:
145 * (x + 40) = 145 * x - (12/60) * x * (x + 40)
Now, let's simplify some more:
145x + 5800 = 145x - (1/5) * x^2 - (1/3) * 40x
Uh-oh! Seems like we've reached a dead-end. The quadratic equation is not going to help us here because the x terms cancel out. It's a classic clown moment - the joke's on us!
However, all hope is not lost. We can still solve for x using some good ol' algebra skills. Let's combine like terms and isolate x:
5800 = - (1/5) * x^2 - (1/3) * 40x
Now, multiply everything by -15 to get rid of those pesky fractions:
-87000 = 3x^2 + 20x
Next, let's simplify and rearrange:
3x^2 + 20x + 87000 = 0
Now, this equation is quadratic, and we can use the quadratic formula to find the value of x:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values from our quadratic equation:
x = (-(20) ± √((20)^2 - 4(3)(87000))) / (2(3))
Now, put on your math cap, do some calculations, and you should find two possible values for x. One of them should be positive, and that's your answer!
Remember, math can be a tricky clown sometimes, but with a little perseverance, you'll solve the puzzle like a champion. Good luck!
To solve this problem, let's define the current airspeed as "x" miles per hour.
We can start by calculating the original travel time from Portland to Seattle. The formula for travel time is distance divided by speed, so the original travel time can be represented as:
T1 = 145 miles / x miles per hour
Next, let's calculate the new airspeed, which is the original airspeed increased by 40 miles per hour. So, the new airspeed can be represented as:
x + 40
Now, we can calculate the new travel time with the increased airspeed. It's given that this travel time is 12 minutes (or 0.2 hours) less than the original travel time. So, the new travel time can be represented as:
T2 = 145 miles / (x + 40) miles per hour
Since the new travel time is 12 minutes less than the original travel time, we can set up the following equation:
T1 - T2 = 0.2 hours
Substituting the expressions for T1 and T2, we get:
145 / x - 145 / (x + 40) = 0.2
To solve this equation, we can multiply through by (x)(x + 40) to eliminate the denominators:
145(x + 40) - 145x = 0.2x(x)(x + 40)
Simplifying, we have:
5800 = 0.2x^2 + 8x^2
Combining like terms:
8.2x^2 = 5800
Dividing both sides by 8.2:
x^2 = 707.3170732
Taking the square root of both sides:
x = ±26.57 (approx.)
Since we are dealing with airspeed, the negative value is extraneous. Therefore, the original airspeed is approximately 26.57 miles per hour.
To find the air speed required to obtain the decrease in travel time, we need to add 40 to the original airspeed:
Required airspeed = 26.57 + 40 ≈ 66.57 miles per hour
So, the airspeed required to obtain the decrease in travel time is approximately 66.57 miles per hour, not 190 as you mentioned.
To solve this problem, let's first define the variables:
Let "s" be the original airspeed (in miles per hour).
Let "t" be the original travel time (in hours).
Let "t-12/60" be the reduced travel time (in hours) after increasing the airspeed by 40 miles per hour.
The distance between Portland and Seattle is 145 miles, and the original travel time can be expressed as:
t = 145 / s
Similarly, the reduced travel time can be expressed as:
t - 12/60 = 145 / (s + 40)
Now, let's simplify this equation:
Simplify the equation for the reduced travel time:
Multiply both sides of the equation by (s + 40):
t(s + 40) - 12/60(s + 40) = 145
Expand and simplify:
ts + 40t - (1/5)s - 2/5 = 145
ts + 40t - (1/5)s = 145 + 2/5
Now, let's substitute the value of the original travel time into this equation:
t = 145 / s
Substituting t into the equation:
(145 / s)(s) + 40(145 / s) - (1/5)s = 145 + 2/5
145 + 5800/s - (1/5)s - 145 = 145 + 2/5
5800/s - (1/5)s = 2/5
Let's multiply both sides of the equation by 5s to eliminate the denominators:
5s(5800/s - (1/5)s) = 5s(2/5)
5800 - s^2 = 2s
Rearrange the equation:
s^2 + 2s - 5800 = 0
Now we can solve this quadratic equation.
Using the quadratic formula, where a = 1, b = 2, and c = -5800:
s = (-b ± √(b^2 - 4ac)) / 2a
s = (-2 ± √(2^2 - 4(1)(-5800))) / (2(1))
s = (-2 ± √(4 + 23200)) / 2
s = (-2 ± √23204) / 2
s = (-2 ± 152.27324) / 2
s ≈ -77.14 or s ≈ 75.14
Since airspeed cannot be negative, we take the positive value:
s ≈ 75.14
So, to obtain a decrease in travel time by 12 minutes, the airspeed required is approximately 75.14 miles per hour.