Use the time-independent kinematic equation to find the displacement while the plane is braking. Take a = -4.47 m/s and vo = 71.5 1 = = VO 2+2aAxbraking 0 - (71.5 m/s) Axbraking = 572 m = 2a 2.00 (-4.47 m/s) m/s. The negative sign on a means that the plane is slowing down. Sum the two results to find the total displacement. Ax coasting + Axbraking = 72 m +572 m = 644 m LEARN MORE Remarks To find the displacement while braking, we could have used the two kinematics equations involving time, namely, Ax = vot + at and y = vo + at, but because we weren't interested in time, the time-independent equation was easier to use. Question By how much would the answer change if the plane coasted for 2.0 s before the pilot applied the brakes? 71.5 m PRACTICE IT Use the worked example above to help you solve this problem. A typical jetliner lands at a speed of 146 mi/h and decelerates at the rate of (11.8 mi/h)/s. If the jetliner travels at a constant speed of 146 mi/h for 1.1 s after landing before applying the brakes, what is the total displacement of the jetliner between touchdown on the runway and coming to rest? 2 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m EXERCISE HINTS: GETTING STARTED | I'M STUCK! A jet lands at 26.2 m/s, applying the brakes 1.68 s after landing. Find the acceleration needed to stop the jet within 5.29 x 102 m. -0.647 X You are neglecting some of the given information. The jet must stop within a certain distance after