Question

Due to their short runway, airplanes that take off from an aircraft carrier often need to utilize a "catapult system" to get them above stall speed. While attached to the catapult, an airplane has position as a function of x(t)=bt^2-ct^3, spending a total time T being launched. How fast is the airplane going at the end of the launch? you may use b=29m/s^2, c=3m/s^2, T=3s (t and T are different) please show work

Answer 1

We know the position of the plane at any time 't' while the plane is being launched. So we can say that the derivative of the position with respect to time will give us the speed of the plane at any time 't'.

X(t)= bt²-ct³

So the derivative of X(t) wrt. To time:

d(x(t))/ dt = velocity V(t) = 2bt- 3ct²

Given b=29m/s² and c=3 m/s²

V(t)= 58t- 9t² m/s

As we have to find the velocity of plane at the time of launch which is already provided ie. T= 3s

V(3)= 58*3-9*3²

V(3)= 93 m/s______Answer

Your feedback is greatly appreciated!