Question

The flight path of a jet aircraft as it takes off is defined by the parmetric equations x=1.25 t2 and y=0.03 t3, where t is the time after take-off, measured in seconds, and x and y are given in meters. At t=40 s (just before it starts to level off), determine at this instant (a) the horizontal distance it is from the airport, (b) its altitude, (c) its speed and (d) the magnitude of its acceleration.

Answer 1

Answer:

(a) $X=2000m$

(b)  $Y=1920m$

(c)  $v=175.3m/s$

(d)  $a=7.6m/s^2$

Explanation:

$x=1.25 t^2$

and

$y=0.03 t^3$

Kinematics formules:

$v_{x}=dx/dt=2.5t$

$a_{x}=dv_{x}/dt=2.5$

$v_{y}=dy/dt=0.09t^2$

$a_{y}=dv_{y}/dt=0.018t$

at T=40s:

(a) $X=1.25*(40)^2=2000m$

(b)  $Y=0.03*(40)^3=1920m$

(c)  $v=\sqrt{v_{x}^2+v_{y}^2} =\sqrt{(2.5*40)^2+(0.09*40^2)^2} =175.3m/s$

(d)  $a=\sqrt{a_{x}^2+a_{y}^2} =\sqrt{(2.5)^2+(0.18*40)^2} =7.6m/s^2$