Question #15004

A cessna aircraft has a lift-off speed of 120km/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?

Expert's answer

(a)

As V0 = 0, then the equation of the speed is the following:

V =

at,

and the equation of the covered distance:

S = at^2 / 2

where a -

acceleration.

V = 120 km/h = 100 / 3 m/s.

After multiplying the first

equation by t and division by the second equation we get:

V^2 / S =

2a,

a = V^2 / (2S) = 10000 / (9 * 2 * 240) = 2.31 m / s^2.

(b)

From

the first equation:

t = V / a = 100 / (3 * 2.31) = 14.4 s.

As V0 = 0, then the equation of the speed is the following:

V =

at,

and the equation of the covered distance:

S = at^2 / 2

where a -

acceleration.

V = 120 km/h = 100 / 3 m/s.

After multiplying the first

equation by t and division by the second equation we get:

V^2 / S =

2a,

a = V^2 / (2S) = 10000 / (9 * 2 * 240) = 2.31 m / s^2.

(b)

From

the first equation:

t = V / a = 100 / (3 * 2.31) = 14.4 s.

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