Question

A Cessna aircraft has a lift-off speed of
112 km/h.
What minimum constant acceleration does
this require if the aircraft is to be airborne
after a take-off run of 253 m?
Answer in units of m/s2

Answer 1

A Cessna aircraft has a lift-off speed of $112~\mathrm{km/h}$. What minimum constant acceleration does this require if the aircraft is to be airborne after a take-off run of $253~\mathrm{m}$? Answer in units of $\mathrm{m/s^2}$

Equation of motion with uniform acceleration:

$$l = \frac{v^2}{2a}$$

where $l$ – distance, $v$ – speed, $a$ – acceleration.

Therefore:

$$a = \frac{v^2}{2l}$$ $$a = \frac{\left(112~\frac{\mathrm{km}}{\mathrm{h}}\right)^2}{2 * 253~\mathrm{m}} = \frac{\left(112~\frac{\mathrm{m}}{3.6}~\frac{\mathrm{s}}{\mathrm{s}}\right)^2}{2 * 253~\mathrm{m}} = 1.91~\mathrm{m/s^2}$$

Answer: $1.91~\mathrm{m/s^2}$