Respuesta :
The minimum time needed for the plane to come to rest will happen if the maximum acceleration is the same throughout the landing; this means that for this situation we can look at the motion as an uniform accelerated motion and then we will be able to use the equations describing this type of motion.
For this kind of motion we know that:
[tex]a=\frac{v_f-v_0}{t}[/tex]we want the plane to stop, this means that we want the final velocity to be zero; plugging this value in the equation above and the values given by the problem we have that:
[tex]\begin{gathered} -6=\frac{0-120}{t} \\ t=\frac{-120}{-6} \\ t=20 \end{gathered}[/tex]Therefore the plane will need a minimum of 20 seconds to stop.
To determine the minimum distance needed for the plane to stop safely we have to remember that for this kind of motion:
[tex]v^2_f-v^2_0=2ax[/tex]Plugging the values we have that:
[tex]\begin{gathered} 0^2-120^2=2(-6)x \\ -120^2=-12x \\ x_{}_{}=\frac{-120^2}{-12} \\ x_{}=1200 \end{gathered}[/tex]Hence, we need at least 1200 m for the plane to safely land and therefore the plan can't land in the school parking lot.
