Question

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?

Answer 1

Answer:

$\Delta s = 1775.510\,m$

Explanation:

The minimum distance for takeoff is:

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

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

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

$\Delta s = 1775.510\,m$