Question

A typical jetliner lands at a speed of 146 mi/h and decelerates at the rate of (10.4 mi/h)/s. If the jetliner travels at a constant speed of 146 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

jetliner is moving with speed 146 mi/h

now we will have

$v= 146 \frac{1609 m}{3600 s}$

$v = 65.25 m/s$

now if it moves with this speed for 1.5 s

so the displacement for above time

$d_1 = 1.5(65.25) = 97.9 m$

now the deceleration of jet is given as

$a = - 10.4 mi/h/s$

$a = -(10.4)(\frac{1609 m}{3600 s})\frac{1}{s}$

$a = - 4.65 m/s^2$

now by kinematics

$v_f^2 - v_i^2 = 2ad$

$0 - 65.25^2 = 2(-4.65)(d_2)$

$d_2 = 457.8 m$

so total displacement is given as

$d = d_1 + d_2$

$d = 97.9 + 457.8 = 555.7 m$

so displacement will be 555.7 m

Answer 2

Solution

In this Question, We have given,

$speed=146\frac{m}{h}$

$speed=146\times 1609.34\frac{m}{3600s}$

$speed=65.27\frac{m}{s}$

acceleration=$10.4\frac{mi}{hs}$

acceleration=$10.4\times 0.44704\frac{m}{s^2}$

acceleration=$4.65\frac{m}{s^2}$

We will first calculate displacement of jetliner when speed is constant  

From second equation of motion we know that,

$S=ut+\frac{at^2}{2}$................(1)

here, speed is constant therefore a=0

put values of u,t and a in equation (1)

$S=65.27\frac{m}{s}\times 1.5S+\frac{0\times (1.5S)^2}{2}$

now calculate displacement S' when it is decelerating  

$v^2=u^2-2aS'$

so $0^2=(65.27\frac{m}{s})^2-2\times 4.65\frac{m}{s^2}\times S'$

$9.29\frac{m}{s^2}\times S'=(65.27\frac{m}{s})^2$

$S'=458.16m$

therefore total displacement=S+S'

$Total Displacement=979.01m+458.16m$

$Total Displacement=1437.17m$