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. If the plane starts to level off at t = 40 s, determine at this instant:a. The horizontal distance it is from the airport b. Its altitudec. Its speed.

A. We are given t= 40s, x=1.25t²

So The horizontal distance it is from the airport will be = 1.25(40)²= 2000meters

B. y= 0.03t³

So y= 0.03(40)³= 1920m altitude

C. To find speed

We say

dx/dt=Vx= 2.5t= 100

dy/dy = 0.09t= 144

So speed= √100²+144²= 175.31m/s

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force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2