Problem 1:a) In this part of the problem, we will find the two accelerations of the plane. The initialvelocity of the plane was v0 = 67.4 m/sec. From t = 0 sec to t = 18 sec, theacceleration of the plane was a₁ = -a₁. From t = 18 sec to t = 29 sec, theacceleration of the plane was a₂ = -a₂. The velocity of the plane at t = 29 sec is27.3 m/sec. From t = 0 sec to t = 29 sec, the total distance travelled by the plane was1419.8 m.i.ii.iii.iv.Sketch a velocity vs. time graph for the motion.What is the velocity of the plane at t = 18 sec? Answer: 45.7 m/secWhat is the acceleration a₁? Answer: 1.2 m/sec²What is the acceleration a₂? Answer: 1.7 m/sec²b) In this part of the problem, we will find the total distance the plane would have neededto come to rest. The initial velocity of the plane was vo= 67.4 m/sec. From t = 0 secto t = 18 sec, the acceleration of the plane was a₁ = a₁ = -1.2 m/sec². Fromt = 18 sec until the plane comes to rest, the acceleration of the plane wasa₂-a₂ = -1.7 m/sec². What is the total distance traveled by the airplane, until itcomes to rest? Answer: 1635.8 m. Note that this answer is larger than 1419.8 m.c) In this part of the problem, we will find the total distance the plane would have neededto come to rest, if the reverse thrust was activated immediately on touchdown. Theinitial velocity of the plane was vo = 67.4 m/sec. From t = 0 sec until the plane comesto rest, the acceleration of the plane was a₂ = -a₂ = -1.7 m/sec². What is the totaldistance traveled by the airplane, until it comes to rest? Answer: 1336.1 m. Note thatthis answer is smaller than 1419.8 m.d) In this part of the problem, we will find the latest possible time the reverse thrust couldhave been activated so that the airplane would have safely come to rest on the runway,without going into the safety runway. The initial velocity of the plane wasVo = 67.4 m/sec. From t = 0 sec to t = t₁, the acceleration of the plane wasa₁ = a₁ = -1.2 m/sec². From t = t₁ until the plane comes to rest, the accelerationof the plane was a₂ = -a₂ = -1.7 m/sec². The total distance traveled by the airplaneis 1394.8 m. Let's define the time that the acceleration a2 is present as t₂. Thus, thetotal time the plane needs to come to rest is t = t₁ + t₂.i.What is the velocity of the plane at t = t₁? Answer: 63.75 m/secii.iii.What is the time t₁? Answer: 3.04 sec. Notice that this time is less than 18 sec.What is the time t₂? Answer: 37.5 sec. Notice that t = t₁ + t₂ = 40.04 sec. Thisis slightly larger than the time the plane needs to come to rest in part c, wherethe reverse thrust is activated immediately on touchdown (t = D = 39.65 sec).a2 x = x₁ + vot + 1²/1at², v = v₁ + at, (v²) = (v₂)² + 2a^x,(νο)221Ax = (0+¹) At, Atrapezoid = (₁+²) h, Atriangle = bh22Introduction to the ProblemOn December 8, 2005 Southwest Airlines flight 1248, a Boeing 737-700 class jet with 103persons aboard, attempted to land on a snow-covered runway at Chicago's Midway Airport.Tragically, the attempt resulted in the death of a child on the ground, as the plane slid farenough beyond the runway to leave airport property and collide with a car (in which the childwas a passenger) on the road beyond. Both the analysis and prevention of accidents such asthis rely on the kinematics of one-dimensional motion. In this problem we will examine how theNational Transportation Safety Board (NTSB) arrived at the conclusion that "the probable causeof a fatal runway overrun...was the pilots' failure to use available reverse thrust in a timelymanner to safely slow or stop the airplane after landing." (NTSB press release SB-07-48).The NTSB report does not give the full range of data available to the official accidentinvestigators, but it does yield enough details to make a basic check of their findings. Accordingto the information released to the public, the plane touched down with 4576 feet of regularrunway remaining and an 82 foot runway safety overrun area beyond. The plane was moving ata ground speed of 131 knots when it touched down. Brakes were employed essentiallyimmediately, but the pilots failed to completely reverse the engine thrust until 18 seconds aftertouchdown (for comparison, four of the five air carrier airplanes landing in the 25 minutes priorto the accident reversed engine thrust within 4 seconds of touchdown, and all did so within 6seconds). At approximately 29 seconds after touchdown the plane left the back of the runwaysafety area moving at a ground speed of about 53 knots.

Name: Problem 1:a) In this part of the problem, we will find the two accele
Uploaded: 2024-03-05T11:56:34.000Z
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Description: Problem 1:a) In this part of the problem, we will find the two accelerations of the plane. The initialvelocity of the plane was v0 = 67.4 m/sec. From t = 0 sec to t = 18 sec, theacceleration of the plane was a₁ = -a₁. From t = 18 sec to t = 29 sec,