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icang [17]
3 years ago
15

A time/motion word problem

Mathematics
2 answers:
andriy [413]3 years ago
3 0
The two planes are flying on the two legs of a right triangle.
The straight distance between them is the hypotenuse of the triangle.

Since the speeds are in mph, let's work the time in hours.
Call the time 'H' that we're looking for.
It's the number of hours after they both take off that they're 650 miles apart.

After 'H' hours, the first plane has gone 500H miles north.
After 'H' hours, the second plane has gone 1200H miles east.
After 'H' hours, they are 650 miles apart.

Do you remember this for a right triangle ? ==>    A² + B² = C²

(500H)² + (1200H)² = (650)²

250,000H² + 1,440,000H² = 422,500

1,690,000 H² = 422,500

H² = (422,500) / (1,690,000) = 0.25

H = √0.25 = 1/2 hour = 30 minutes
bezimeni [28]3 years ago
3 0
x^2=500^2+1200^2\\\\x^2=250000+1440000\ \ \ \Rightarrow\ \ \ \ x^2=1690000\ \ \ \Rightarrow\ \ \ \ x=1300\ [mph]\\\\the\ distance=650\ miles\\\\the\ speed= \frac{the\ distance}{the\ time} \ \ \ \Rightarrow\ \ \ 1300\ [mph]= \frac{650\ [miles]}{the\ time}\\\\the\ time= \frac{650}{1300} \ [hr]=0.5\ [hr]=30\ [min]\\\\Ans.\ after\ 30\ minutes.

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