Question

Use the worked example above to help you solve this problem. A typical jetliner lands at a speed of 140 mi/h and decelerates at the rate of (11.6 mi/h)/s. If the jetliner travels at a constant speed of 140 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?

Answer 1

Given that landing speed of jetliner = U1 = 140 mi/hr = (140 mi/hr)*(1609.34 m/1 mi)*(1 hr/3600 sec) = 62.586 m/s

Now he travels for 1.5 sec before applying brakes, So

d1 = displacement during constant speed = Speed*time = (62.586 m/s)*1.5 s = 93.88 m

Now he applies brakes, So deceleration = (11.6 mi/hr)/s = (11.6 mi/hr/s)*(1609.34 m/1 mi)*(1 hr/3600 s) = 11.6*1609.34/3600

acceleration = -5.186 m/s^2 (-ve sign because speed is decreasing)

V1 = final speed of jetliner = 0 m/s

Using 3rd kinematic equation:

V1^2 = U1^2 + 2*a*d2

d2 = displacement during constant deceleration = (V1^2 - U1^2)/(2*a)

d2 = (0^2 - 62.586^2)/(2*(-5.186))

d2 = 377.65 m

So total displacement of jetliner before stopping will be:

d = d1 + d2

d = 93.88 + 377.65

d = 471.53 m = total displacement

(If need answer in significant figures, then use d = 470 m)