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

`\Delta s = 1775.510\,m`

Step-by-step explanation:

The minimum distance for takeoff is:

`\Delta s = \Delta s_(1) + \Delta s_(2)`

`\Delta s = (v_(f)^(2)-v_(o)^(2))/(2\cdot a) + v_(f)\cdot \Delta t`

`\Delta s = ((100\,(m)/(s) )^(2)-(0\,(m)/(s) )^(2))/(2\cdot (3.92\,(m)/(s^(2))))+(100\,(m)/(s) )\cdot (5\,s)`

`\Delta s = 1775.510\,m`