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

In this Question, We have given,

`speed=146(m)/(h)`

`speed=146* 1609.34(m)/(3600s)`

`speed=65.27(m)/(s)`

acceleration=
`10.4(mi)/(hs)`

acceleration=
`10.4* 0.44704(m)/(s^2)`

acceleration=
`4.65(m)/(s^2)`

We will first calculate displacement of jetliner when speed is constant

From second equation of motion we know that,

`S=ut+(at^2)/(2)`................(1)

here, speed is constant therefore a=0

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

`S=65.27(m)/(s)* 1.5S+(0* (1.5S)^2)/(2)`

now calculate displacement S' when it is decelerating

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

so
`0^2=(65.27(m)/(s))^2-2* 4.65(m)/(s^2)* S'`

`9.29(m)/(s^2)* S'=(65.27(m)/(s))^2`

`S'=458.16m`

therefore total displacement=S+S'

`Total Displacement=979.01m+458.16m`

`Total Displacement=1437.17m`

Answer 2

jetliner is moving with speed 146 mi/h

now we will have

`v= 146 (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)((1609 m)/(3600 s))(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