## University Calculus: Early Transcendentals (3rd Edition)

The time it takes the aircraft to be airborne is $25\ sec$. It has traveled $694.4\ m$ during that time.
$$D=\frac{10}{9}t^2(m)$$ The velocity of the aircraft is the derivative of the distance it has traveled, $D$: $$v=\frac{dD}{dt}=\frac{20}{9}t (m/s)$$ The speed of the aircraft, thus, is $|v|=|20/9t|=20/9 t$ (m/s) (because the time $t$ is positive). Convert $200km/h$ to m/s: $$200km/h=\frac{200\times1000}{3600}m/s=\frac{500}{9}m/s$$ The time it takes the aircraft to reach $200km/h$ is $$t=\frac{\frac{500}{9}}{\frac{20}{9}}=25(sec)$$ The distance it has traveled after $25sec$ is $$D=\frac{10}{9}\times(25^2)\approx694.4(m)$$