In a flight of 600 , km,an aircraft was slowe | vedclass

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(A) Let the original duration of the flight be $t$ hours.Let the original speed be $v = \frac{600}{t} \, km/hr$.According to the problem,the new speed

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(A) Let the original duration of the flight be $t$ hours.Let the original speed be $v = \frac{600}{t} \, km/hr$.According to the problem,the new speed is $v' = v - 200 = \frac{600}{t} - 200$.The new time is $t' = t + 0.5$ hours.Since distance is constant $(600 \, km)$,we have $v' 	imes t' = 600$.$(\frac{600}{t} - 200)(t + 0.5) = 600$.$600 + \frac{300}{t} - 200t - 100 = 600$.$\frac{300}{t} - 200t - 100 = 0$.Divide by $100$: $\frac{3}{t} - 2t - 1 = 0$.$3 - 2t^2 - t = 0 \implies 2t^2 + t - 3 = 0$.Factoring the quadratic equation: $2t^2 + 3t - 2t - 3 = 0$.$t(2t + 3) - 1(2t + 3) = 0$.$(t - 1)(2t + 3) = 0$.Since time cannot be negative,$t = 1$ hour.

In a flight of $600 \, km$,an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by $200 \, km/hr$ and the time of flight increased by $30 \, minutes$. The duration of the flight is (in $hour$):

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