A plane takes off from Winnipeg and files east as 620km/hr. at the same time a second plane off from the surface of Lake Winnipeg 310 km due north of Winnipeg and files due north at 560 km/hr for how man hours are the planes less than 1000 km apart

1
pls answer the question immediately
and the answer is 0lessthan or equal to t lessthan 0.92 h
0<=t<0.92h

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2014-08-19T13:32:31+05:30

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See figure.
Distance of plane 1 (East) at time t hours  is  (620 * t)  km

Distance of Plane2 at time t=0 is 310 km    as it is at Lake Winnipeg

Distance of Plane 2  at time t hours is =  310 + 560 * t  km

The triangle formed by Plane2, plane1 and winnipeg is a right angle triangle.

Let the Distance between two planes be  D

So  D²  =  ( 620 t )² + (310 + 560 t)²

As time t increases from 0, then D increases.  Let us see when D
becomes 1000 km.  Substitute D = 1000 km  in the above equation.

1000² = 620² t² + 310² + 560² t² + 2 * 310 * 560 t

1000000  = 384400 t² + 96100 + 313600 t² + 347200 t
   Let us strike of two zeroes of each number. cancel 100
10000 = (3844 + 3136) t² + 3472 t + 961

So  6980 t² + 3472  t - 9039 = 0

   solving quadratic equation, roots are
Δ = 3472² + 4*6980*9039 = 16261.1
t = ( -3472 +- 16261.1 )/ 2*6980  =  0.916 hours  OR  - 1.413 hour
We neglect the negative answer.
So t = 0.916 hours

2 5 2
Good explanation
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