A jumbo jet taxiing down the runway receives word that it must return to the gate to pick up an important passenger late to his connecting flight. The jet is traveling at \( 45.0 \mathrm{~m} / \mathrm{s} \) when the pilot receives the message. What is the acceleration of the plane if it takes the pilot 5.00 s to bring the plane to a halt?

To find the acceleration of the jumbo jet, we can use the formula: \[ a = \frac{\Delta v}{\Delta t} \] Here, the initial velocity \( \Delta v = v_f - v_i \), where \( v_f = 0 \, \mathrm{m/s} \) (final velocity when the jet stops) and \( v_i = 45.0 \, \mathrm{m/s} \) (initial velocity). The time \( \Delta t \) is \( 5.00 \, \mathrm{s} \). Plugging in the values: \[ \Delta v = 0 - 45.0 = -45.0 \, \mathrm{m/s} \] Now, substituting into the acceleration formula gives: \[ a = \frac{-45.0 \, \mathrm{m/s}}{5.00 \, \mathrm{s}} = -9.0 \, \mathrm{m/s^2} \] Thus, the acceleration of the plane is \( -9.0 \, \mathrm{m/s^2} \), indicating a deceleration. The concept of acceleration in aviation is crucial for pilots, as understanding how quickly a plane can slow down impacts safety regulations and flight protocols. Pilots train extensively to master these principles, as quick reactions can make all the difference during critical situations like emergencies or sudden changes in flight plans. Moreover, this knowledge extends to the engineering of aircraft as well. Engineers design systems not just for optimal speed but for effective braking and control as planes decelerate. Learning more about aerodynamics and braking systems can be enjoyed through various aviation literature and documentaries, making the science behind flight both fascinating and accessible!