Question

A flight, traveling to a destination $11,200\text{ kms}$ away, was supposed to take off at 6:30 AM. Due to bad weather, the departure of the flight got delayed by three hours. The pilot increased the average speed of the airplane by $100\text{ km/hr}$ from the initially planned average speed, to reduce the overall delay to one hour.

Had the pilot increased the average speed by $350\text{ km/hr}$ from the initially planned average speed, when would have the flight reached its destination?

Accepted Answer

D. Let initial speed be $F	ext{ km/hr}$ and initial planned time be $t	ext{ hours}$. $F \cdot t = 11200 \implies t = \frac{11200}{F}$. Delayed departure by 3 hours, overall delay reduced to 1 hour $\implies$ travel time reduced by $3 - 1 = 2	ext{ hours}$. So new travel time is $t - 2$. $(F + 100)(t - 2) = 11200 \implies (F + 100)\left(\frac{11200}{F} - 2ight) = 11200$ $11200 - 2F + \frac{1120000}{F} - 200 = 11200$ $-2F^2 - 200F + 1120000 = 0 \implies F^2 + 100F - 560000 = 0$. Solving: $(F - 700)(F + 800) = 0 \implies F = 700	ext{ km/hr}$. If speed is increased by $350	ext{ km/hr}$, new speed $= 700 + 350 = 1050	ext{ km/hr}$. New travel time $= \frac{11200}{1050} = \frac{32}{3}	ext{ hours} = 10	ext{ hours } 40	ext{ minutes}$. Actual departure time $= 6:30	ext{ AM} + 3	ext{ hours} = 9:30	ext{ AM}$. Arrival time $= 9:30	ext{ AM} + 10	ext{ hours } 40	ext{ minutes} = 8:10	ext{

Flight Speed and Delay Calculation

Explanation

Practise this under exam conditions