Question

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

Answer 1

Answer on Question #69396-Physics / Other

A Cessna aircraft has a liftoff speed of $v = 120\frac{\mathrm{km}}{\mathrm{hr}} = 33.3\frac{\mathrm{m}}{\mathrm{s}}$.

a.) What minimum constant acceleration does the aircraft require if it is to be airborne after a takeoff run of $S = 240 \, \text{m}$?

b.) How long does to take the aircraft to become airborne?

Solution:

a)

$$S = \frac{v^2 - u^2}{2a}.$$

$u$-initial velocity, $v$-final velocity, $a$-acceleration, $S$-distance.

So

$$a = \frac{v^2 - u^2}{2S} = \frac{33.3^2 - 0^2}{2 \times 240} = 2.3 \frac{\mathrm{m}}{\mathrm{s}^2}.$$

b)

$$a = \frac{v - u}{t}, \qquad t = \frac{v - u}{a} = \frac{33.3 - 0}{2.3} = 14.4 \, \text{s}.$$

Answer $a = 2.3\frac{\mathrm{m}}{\mathrm{s}^2}$; $t = 14.4 \, \text{s}$.

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