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A plane leaves San Francisco for New York, 4600 km away. With a strong tailwind, its speed is 1100 km/h. At the same time, a second jet leaves New York for San Francisco (taking the same route). Flying into the wind, it makes only 700km/h. When and where do the two planes pass?

Answer :

boffeemadrid

Answer:

9200 seconds

2811111.11 m, 1788888.89 m

Explanation:

The combined distance the planes will travel is 4600 km

Time taken when they will meet will be the same for both the planes

Distance is given by

[tex]Distance=Speed\times Time[/tex]

[tex]4600=1100t+700t\\\Rightarrow t=\dfrac{4600}{1800}\times 60\times 60\\\Rightarrow t=9200\ seconds\\\Rightarrow t=2\ hours\ 33\ minutes\ 19\ seconds[/tex]

The planes will meet after 9200 seconds

Distance the first plane travels is

[tex]\dfrac{1100}{3.6}\times 9200=2811111.11\ m[/tex]

Distance the second plane travels is

[tex]4600000-2811111.11=1788888.89\ m[/tex]

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