+4 "An airplane makes a 990 km flight with a tailwind and returns, flying into the same wind. The total flying time is 3 hours 20 minutes, and the airplane’s sped in still air is 600 km/h. What is the speed of the wind?"
I have no idea on how to even approach this question! I know I have to use quadratics but this is nothing like any of the example questions on my online course! If anyone can help it would be much appreciated!
+26388 An airplane makes a 990 km flight with a tailwind and returns,
flying into the same wind. The total flying time is 3 hours 20 minutes,
and the airplane’s speed in still air is 600 km/h. What is the speed of the wind?
velocity airplane: \(v_a\)
velocity wind: \(v_w\)
time for the journey there: \(t_1\)
time for way back: \(t_2\)
distance: d = 990 km
total time: \(t = t_1 + t_2\)
( t = 3 hours 20 minutes )
\(\begin{array}{|rcll|} \hline d &=& (v_a+v_w)\cdot t_1 \quad \text{ for the journey there} \\ d &=& (v_a-v_w)\cdot t_2 \quad \text{ for way back} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline t_1 &=& \dfrac{d}{v_a+v_w} \\ t_2 &=& \dfrac{d}{v_a-v_w} \\\\ t &=& t_1 + t_2 \\\\ t &=& \dfrac{d}{v_a+v_w} + \dfrac{d}{v_a-v_w} \\\\ t &=& d\cdot \left( \dfrac{1}{v_a+v_w} + \dfrac{1}{v_a-v_w} \right) \\\\ \dfrac{t}{d} &=& \dfrac{1}{v_a+v_w} + \dfrac{1}{v_a-v_w} \\\\ \dfrac{t}{d} &=& \dfrac{v_a-v_w+v_a+v_w}{(v_a+v_w)\cdot (v_a-v_w)} \\\\ \dfrac{t}{d} &=& \dfrac{2v_a}{(v_a+v_w)\cdot (v_a-v_w)} \\\\ \dfrac{t}{d} &=& \dfrac{2v_a}{v_a^2-v_w^2} \\\\ \dfrac{d}{t} &=& \dfrac{v_a^2-v_w^2}{2v_a} \quad &| \quad \cdot 2v_a\\ 2v_a\cdot \dfrac{d}{t} &=& v_a^2-v_w^2 \quad &| \quad +v_w^2\\ v_w^2 + 2v_a\cdot \dfrac{d}{t} &=& v_a^2 \quad &| \quad -2v_a\cdot \dfrac{d}{t}\\ v_w^2 &=& v_a^2 -2v_a\cdot \dfrac{d}{t}\\ v_w^2 &=& v_a\cdot \left( v_a - \dfrac{2d}{t} \right) \quad &|d=990 \quad v_a = 600 \quad t=3\frac{1}{3}=\frac{10}{3}\ \text{hours}\\ v_w^2 &=& 600\cdot \left( 600 - \dfrac{2\cdot 990}{\frac{10}{3}} \right) \\ v_w^2 &=& 600\cdot \left( 600 - \dfrac{2\cdot 990\cdot 3}{10} \right) \\ v_w^2 &=& 600\cdot \left( 600 - 6\cdot 99 \right) \\ v_w^2 &=& 600\cdot \left( 600 - 594 \right) \\ v_w^2 &=& 600\cdot 6 \\ v_w^2 &=& 3600 \quad &| \quad \sqrt{} \\ v_w & = & \sqrt{3600} \\ \mathbf{v_w} & \mathbf{=} & \mathbf{60} \\ \hline \end{array}\)
The speed of the wind is \( \mathbf{60 \ \frac{km}{h}}\) .
+26388 An airplane makes a 990 km flight with a tailwind and returns,
flying into the same wind. The total flying time is 3 hours 20 minutes,
and the airplane’s speed in still air is 600 km/h. What is the speed of the wind?
velocity airplane: \(v_a\)
velocity wind: \(v_w\)
time for the journey there: \(t_1\)
time for way back: \(t_2\)
distance: d = 990 km
total time: \(t = t_1 + t_2\)
( t = 3 hours 20 minutes )
\(\begin{array}{|rcll|} \hline d &=& (v_a+v_w)\cdot t_1 \quad \text{ for the journey there} \\ d &=& (v_a-v_w)\cdot t_2 \quad \text{ for way back} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline t_1 &=& \dfrac{d}{v_a+v_w} \\ t_2 &=& \dfrac{d}{v_a-v_w} \\\\ t &=& t_1 + t_2 \\\\ t &=& \dfrac{d}{v_a+v_w} + \dfrac{d}{v_a-v_w} \\\\ t &=& d\cdot \left( \dfrac{1}{v_a+v_w} + \dfrac{1}{v_a-v_w} \right) \\\\ \dfrac{t}{d} &=& \dfrac{1}{v_a+v_w} + \dfrac{1}{v_a-v_w} \\\\ \dfrac{t}{d} &=& \dfrac{v_a-v_w+v_a+v_w}{(v_a+v_w)\cdot (v_a-v_w)} \\\\ \dfrac{t}{d} &=& \dfrac{2v_a}{(v_a+v_w)\cdot (v_a-v_w)} \\\\ \dfrac{t}{d} &=& \dfrac{2v_a}{v_a^2-v_w^2} \\\\ \dfrac{d}{t} &=& \dfrac{v_a^2-v_w^2}{2v_a} \quad &| \quad \cdot 2v_a\\ 2v_a\cdot \dfrac{d}{t} &=& v_a^2-v_w^2 \quad &| \quad +v_w^2\\ v_w^2 + 2v_a\cdot \dfrac{d}{t} &=& v_a^2 \quad &| \quad -2v_a\cdot \dfrac{d}{t}\\ v_w^2 &=& v_a^2 -2v_a\cdot \dfrac{d}{t}\\ v_w^2 &=& v_a\cdot \left( v_a - \dfrac{2d}{t} \right) \quad &|d=990 \quad v_a = 600 \quad t=3\frac{1}{3}=\frac{10}{3}\ \text{hours}\\ v_w^2 &=& 600\cdot \left( 600 - \dfrac{2\cdot 990}{\frac{10}{3}} \right) \\ v_w^2 &=& 600\cdot \left( 600 - \dfrac{2\cdot 990\cdot 3}{10} \right) \\ v_w^2 &=& 600\cdot \left( 600 - 6\cdot 99 \right) \\ v_w^2 &=& 600\cdot \left( 600 - 594 \right) \\ v_w^2 &=& 600\cdot 6 \\ v_w^2 &=& 3600 \quad &| \quad \sqrt{} \\ v_w & = & \sqrt{3600} \\ \mathbf{v_w} & \mathbf{=} & \mathbf{60} \\ \hline \end{array}\)
The speed of the wind is \( \mathbf{60 \ \frac{km}{h}}\) .
+129840 Let the speed of the wind in km/ hr = S
The total flight time [ in hours] with the wind =T1 = D/R = 990 / (600 + S)
The total flight time [ in hours ] against the wind = T2 = 990 / (600 - S)
And the total flight time for both trips = Tt = 3 hrs, 20 min = 3 + 1/3 hrs = 10/3 hrs ......so
T1 + T2 = Tt
990 / (600 + S) + 990 / (600 - S) = 10/3 simplify ..... multiply through by 3
2970 / (600 + S) + 2970 /( 600- S) = 10
[2970 (600 - S) + 2970(600 + S) ] / [ (600 + S) (600 - S) ] = 10 cross-multiply
]2970 (600 - S) + 2970(600 + S) ] = 10 [ (600 + S) (600 - S) ]
2970 * 600 * 2 = 10 (360,000 - S^2) divide both sides by 10
297*1200 = 360,000 - S^2 rearrange
S^2 = 360,000 - 297*1200
S^2 = 3600 take the positive square root of both sides
S = 60km / hr
