CATAGORY: Mechanics QUESTION: A Boeing 747 Jumbo Jet has a length of 57m. The runway on which the plane lands intersect another runway. The width of the intersection is 25.0m. The plane decellerates through the intersection at a rate of 5.70 m/s^2 and xlears it with a final speed of 45.0m/s. How much time is needed for the plane to clear the intersection.

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From the middle figure below we see that the front part of the 747 has crossed the intersection and there is 25 more meters to go before the tail crosses over the intersection. The plane is regarded as being in the intersection so long as any part of the plane extends into the space of the intersction. For the nose of the plane to get to the point it is now it traveled a distance of 57 m from when it was just touching the left side of the intersection Before ____ ____/ |\_ ============\ =====>______/ \____|> \ UU OO ---------|==========|----------------|----- 0m 25m 57m Intersection During --- ____ \ \_______________/ |\_ <==========================\ -----_____<=====>______/ \____|> \ UU OO ---------|==========|----------------|----- 0m 25m 57m Intersection After --- ____ \ \_______________/ |\_ <==========================\ -----_____<=====>______/ \____|> \ UU OO ---------|==========|----------------|----- 0m 25m 57m Intersection The tail has to now travel adistance of 25 m to clear the intersection which during this time the nose also travels forward another 25 meters.the plane has thus traveled an total of 57 m +25 m = 82m The equation which relates velocity to acceleration and displacement is (1) v^2 = vo^2 +2ad Where d is the displacement, v and vo are the final and initial velocities and a is the acceleration. Solving (1) for vo gives (2) vo^2 = v^2 - 2ad so (3) vo =sqrt(v^2 - 2ad) substitution of v = 45 m/s, a = -5.7m/s^2 and d = 82 m gives (4) vo =sqrt((45 m/s)^2 - 2(-5.7m/s^2)82 m) = 54.4 m/s Next displacement is related to time by (4) d = do + vo t + (1/2)a t^2 Here do =0m, d =82m, vo=54.4m, and a=-5.7m so (5) 82m = (54.4m/s)t - (2.85m/s^2) t^2 or (6) (2.85m/s^2) t^2 - (54.4m/s)t +82m = 0m Applying the quadratic formula to (7) 2.85 t^2 - 54.4m/s t +82 = 0 gives (8) -(-54.4) +/-sqrt((-54.4)^2 - 4(2.85)82) t = ---------------------------------------- 2(2.85) 54.4 +/-sqrt(2959.36 - 934.8) = ------------------------------- 5.7 54.4 +/-45.0 = --------------- 5.7 = (17.5, 1.6) We take the smaller value because the plane does not cross the intersection, stop and then reverse its course and cross again! Only the first crossing counts. Thus ************ * * * t = 1.6s * * * ************