Taking a -O,vo-71.5 m/s, and t 1.00 s, find the displacement while the plane Is

Question

Taking a -O,vo-71.5 m/s, and t 1.00 s, find the displacement while the plane Is coasting. Aucoasting-vot + at. (71.5 m/s)(1.00 s) + O 71.5 m Use the time-Independent kinematic equation to find the displacement while the plane is braking. a.-4. 47 m/s2 and vo-71.5 v2 2a02200(1-447m/s)-5 -2 0 -(71.5 m/s) 0rel ing- braking = m/s. The negative sign on a means that the plane is slowing down. 2.00(-4.47 m/s) = 572 m Sum the two results to find the total displacement. coasting + xbraking 71.5 m + 572 m-644 LEARN MORE REMARKS To find the displacement while braking, we could have used the two kinematics equations involving time, namely, Ax- voat and v - Vo+at, but because we weren't interested ih time, the time-independent equation was easier to use. 2 QUESTION By how much would the answer change if the plane coasted for 2.0 s before the pilot applied the brakes? PRACTICE IT Use the worked example above to help you solve this problem. A typical jetliner lands at a speed of 155 mi/h and decelerates at the rate of (10.2 mi/h)/s. If the jetliner travels at a constant speed of 155 mi/h for 1.4 s after landing before applying the brakes, what is the total displacement of the jetliner between touchdown on the runway and coming to rest? EXERCISE A jet lands at 49.9 m/s, the pilot applying the brakes 2.26 s after landing. Find the acceleration needed to stop the jet within 5.85 × 102 m after touchdown. HINTS: GETTING STARTED I IM STUCK m/s? Need Help? Reaei

Explanation / Answer

Question :

the answer would change by 71.5 m

Practice it :

vox = 155 mi/h

= 155*1609/(60*60) m/s

= 69.3 m/s

ax = -10.2 (mi/h)/s

= -10.2*1609/(60*60) m/s^2

= -4.56 m/s^2

delta_x_coasting = vox*t

= 69.3*1.4

= 97.0 m

delta_x_braking = (vfx^2 - vox^2)/(2*ax)

= (0^2 - 69.3^2)/(2*(-4.56))

= 527 m

so,

delta_x_coasting + delta_x_braking = 97.0 + 527

= 624 m <<<<<<<<<-------------------Answer

Ecercise :

distance travelled before brakes are applied, d = vox*t

= 49.9*2.26

= 113 m

distance travelled when brakes are applied, d1 = 585 - 113

= 472 m

acceleration needed, a = (vfx^2 - vox^2)/(2*d1)

= (0^2 - 49.9^2)/(2*472)

= -2.64 m/s^2 <<<<<<<<<<<<---------------------Answer

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