Question

A jet accelerates at a=3.92 m/s^2 from rest until it reaches its takeoff velocity of 360 km/hr. It has to travel for 5 seconds at its takeoff velocity before taking off. What's the minimum distance that the jet will travel on the runway?

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Answer #1

Accleration of jet is a = 3.92m/s^2

It starts from rest. u = 0 m/s

It reaches final velocity of 360km/hr or 100 m/s

it travels with that speed for 5 sec .

The distance travelled by the jet while accletating is

where s is the distance

The distance travelled by the jet while moving at 100 m/s for 5 secs is

Total distance travelled is 1275.51 + 500 = 1775.51m or 1.77 Km

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