To find the displacement while braking, we could have used the two kinematics eq

Question

To find the displacement while braking, we could have used the two kinematics equations involving time, namely, x = v0t + ½ at2 and v = v0 + 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? NOTE: The first time i asked this question they gave me a wrong answer thank you.

Use the worked example above to help you solve this problem. A typical jetliner lands at a speed of 152mi/h and decelerates at the rate of (11.7 mi/h)/s. If the jetliner travels at a constant speed of 152 mi/h for 1.5 s after landing before applying the brakes, what is the total displacement of the jetliner between touchdown on the runway and coming to rest?

Explanation / Answer

a)

u = 152 mi/h

for the time t = 1.5s, the distance traveled, s1 = (152*1.61/3.6)1.5 = 102 m

So, now, for the time after t = 1.5s , using the equation :

v^2 = u^2 + 2as

So, 0^2 = (152)^2 - 2*11.7*3600*s <----- NOTE : 11.7 mi/h/s = 11.7*3600 mi/h2

So, s2 = 0.274 mi = 441 m

So, total distance traveled = 441 + 102 = 543 m

b)

Now, in this part time taken after 2s:

So, s1 = (152*1.61/3.6)*2 = 136 m

and for distance traveled while decelerating ,

So, 0^2 =  (152)^2 - 2*11.7*3600*s <----- NOTE : 11.7 mi/h/s = 11.7*3600 mi/h2

So, s2 = 0.274 mi = 441 m

So, total distance = 136 + 441 = 577 m

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