Answer:
The acceleration that the jet liner that must have is 2.241 meters per square second.
Explanation:
Let suppose that the jet liner accelerates uniformly. From statement we know the initial ([tex]v_{o}[/tex]) and final speeds ([tex]v_{f}[/tex]), measured in meters per second, of the aircraft and likewise the runway length ([tex]d[/tex]), measured in meters. The following kinematic equation is used to calculate the minimum acceleration needed ([tex]a[/tex]), measured in meters per square second:
[tex]a = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot d}[/tex]
If we know that [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]v_{f} = 82\,\frac{m}{s}[/tex] and [tex]d = 1500\,m[/tex], then the acceleration that the jet must have is:
[tex]a = \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (1500\,m)}[/tex]
[tex]a = 2.241\,\frac{m}{s^{2}}[/tex]
The acceleration that the jet liner that must have is 2.241 meters per square second.