# A helicopter travels east at 45 mi/hr, then turns north at 40 mi/hr. if the total trip takes 5.0h and the helicopter ends at a point 150 mi north of east of the starting point, how long was each part of the tripand the answer is 2.0h,3.0h;2.4h,2.6h

1
by sweetysiri92

2014-08-10T19:09:29+05:30

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Let helicopter start at A, travel east at 45 mi/hr for time t1 hrs till B. It then turns north and travels at 40 mi/hr for t2 hrs till C.
Now AC  =  150 mi    and ABC is a right angle triangle.
t1 + t2 = 5 hrs
Displacement is AC = 150 miles
AB² + BC² = AC²

(45 t1)² + (40 t2)² = 150²      simplify by canceling 5²
81 t1² + 64 t2² = 900
81 t1² + 64 (5 - t1)² = 900
145 t1² - 640 t1 + 700 = 0        29 t1² - 128 t1 +140 = 0
Δ = 128² - 4 * 29 * 140 = 144

t1 = (128 +- 15) / 2*29        = 1.95 hr      or    2.46 hr
t2 =  3.05 hrs    or  2.54 hrs

t1, t2 =  (1.95 hrs, 3.05  hrs)            or    (  2.46 hrs , 2.54 hrs)

Either of the two combinations you get to C.  Both sets are correct.