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?
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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