LEARN MORE REMARKS To find the displacement while braking, we could have used th

Question

LEARN MORE REMARKS To find the displacement while braking, we could have used the two kinematics equations involving time, namely, = vot + Jt2 and v = vo + at, but because we weren't interested in 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 159 mi/h and decelerates at the rate of (10.6 mi/h)/s. If the jetliner travels at a constant speed of 159 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? EXERCISE HINTS: GETTING STARTED I I'M STUCK! A jet lands at 64 m/s, the pilot applying the brakes 2.02 s after landing. Find the acceleration needed to stop the jet within 5.55 x 102 m after touchdown. You are neglecting some of the given information. The jet must stop within a certain distance after landing, but it does not start slowing down right away. How much distance is left on the runway once the jet starts to slow down? m/s2

Explanation / Answer

Distance, d = v*t

it applies brakes after 2.2 sec so it will have travelled , d = 64*2.02 = 129.28 m

so, total distance, "s" is

s = 555 - 129.28 = 425.72 m

v2 - u2= 2as

0 - u2= 2as

- u2 = 2as
a = - u2/2s
   = - 642 / 2*425.72
   = - 4.81 m/s^2

the value is negative because its retardation

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