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larisa86 [58]
2 years ago
9

A jet accelerates at a=3.92 m/s2 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?
Physics
1 answer:
Art [367]2 years ago
7 0

Answer:

\Delta s = 1775.510\,m

Explanation:

The minimum distance for takeoff is:

\Delta s = \Delta s_{1} + \Delta s_{2}

\Delta s = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot a} + v_{f}\cdot \Delta t

\Delta s = \frac{(100\,\frac{m}{s} )^{2}-(0\,\frac{m}{s} )^{2}}{2\cdot (3.92\,\frac{m}{s^{2}})}+(100\,\frac{m}{s} )\cdot (5\,s)

\Delta s = 1775.510\,m

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