Question

. A Cessna aircraft has a liftoff speed of 120 km/h . (a) What minimum constant acceleration does the aircraft require if it is to be airborne after a takeoff run of 240 m? (b) How long does it take the aircraft to become airborne?

Answer 1

Here's what you're given:
Vinitial = 0 m/s
Vfinal = 120 km/h
d = 240m

First convert 120 km/h to m/s to get 33.33m/s.

-- Part A --
Start by using this kinematic equation to find the acceleration:\[V_f^2-V_i^2=2ad\]Solve that equation for a:\[a=\frac{V_f^2-V_i^2}{2d}\]Plug in values:\[a=\frac{(33.33m/s)^2-(0m/s)^2}{2(240m)}=2.31 m/s^2\]
-- Part B --
Use this equation to find out how long it took to get to that speed:\[\Delta V=at\]Solve that equation for t: \[t=\frac{\Delta V}{a}\]Plug in values:\[t=\frac{33.33m/s}{2.31m/s^2}=14.43s\]