Answer to Question #155879 in Physics for Adams

Question

The flight path of a jet aircraft as it takes off from the Nnamdi Azikiwe airport Abuja,is defined by the parametric equations x=1.25t^2 and y=0.03t^3,Where t, is the time after take off measured in seconds and X and Y are given in meters.At t=40 (just before it starts to level off).calculate at this instant it's speed

Answer 1

Find the derivative of horizontal and vertical displacement (this will be the horizontal and vertical velocity):

$$v_x=2\cdot1.25t^{2-1}=2.5t,\\ v_y=3\cdot 0.03t^{3-1}=0.09t^2.$$

By Pythagorean theorem find the absolute velocity:

$$v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.5t)^2+(0.09t^2)^2}=\\=175.3\text{ m/s}.$$

This is equivalent to 630 km/h.