a plane goes to an airport that is 4200km away

on the way back (4200km) it goes 100km/h faster than before

if the total time taken (there and back) is 13 hours, how fast was the airplane going originally (x)

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- Oct 5th 2008, 05:16 PMattack4uusing quadratic equation
a plane goes to an airport that is 4200km away

on the way back (4200km) it goes 100km/h faster than before

if the total time taken (there and back) is 13 hours, how fast was the airplane going originally (x) - Oct 5th 2008, 05:21 PMicemanfan
Use the formula

$\displaystyle r_{avg} = \frac{2r_1r_2}{r_1 + r_2}$

to obtain

$\displaystyle \frac{2(4200)}{13} = \frac{2x(x + 100)}{x + (x + 100)}$ - Oct 5th 2008, 05:27 PMattack4u
and where do you go from there??

+ we never learned that equation - Oct 5th 2008, 05:36 PMicemanfan
If you would like, I can derive the equation. You would have to derive it (in some sense) in order to solve the problem.

Continuing on:

$\displaystyle \frac{8400}{13} = \frac{2x^2 + 200x}{2x + 100}$

$\displaystyle \frac{8400(2x + 100)}{13} = 2x^2 + 200x$

$\displaystyle 8400(2x + 100) = 13(2x^2 + 200x)$

$\displaystyle 16800x + 8400 = 26x^2 + 2600x$

$\displaystyle 26x^2 - 14200x - 8400 = 0$

$\displaystyle x = \frac{14200 \pm \sqrt{14200^2 + 4(26)(8400)}}{52}$

Recognizing that the negative answer for x makes no sense, we have

$\displaystyle x = \frac{14200 + \sqrt{14200^2 + 4(26)(8400)}}{52}$

which is about 546.74 km/h. - Oct 6th 2008, 01:06 AMticbol
distance = rate *time,

so, time = distance /rate -----**

time in going 4200km away = 4200/x hrs.

time in going back = 4200 /(x +100) hrs.

4200/x +4200/(x+100) = 13

Clear the fractions, multiply both sides by x(x+100),

4200(x+100) +4200(x) = 13(x)(x+100)

4200x +420,000 +4200x = 13x^2 +1300x

13x^2 -7100x -420,000 = 0

Using the Quadratic formula,

x = {7100 +,-sqrt[7100^2 -4(13)(-420,000))]} /2(13)

x = 600 km/hr, or -53.85 km/hr

Therefore, the plane was going originally at 600 km/hr. -----answer.