Question

A Cessna 150 aircraft has a lift-off speed of approximately 125 km/h. What minimum constant acceleration does this require if the aircraft is to be airborne after a take-off run of 220 m? Answer in units of m/s 2

Answer 1

Mininmum acceleration required =
`2.74 m/s^(2)`

Step-by-step explanation:

Given,

Lift off speed = 125km/hr.

We know km/hr conversion factor to m/s is
`(5)/(18)`,

Therefore,

Lift off speed required =
`125 *(5)/(18)` m/s

Lift off speed required = 34.72222 m/s

Length of Takeoff Run given = 220 m.

So,

the flight needs to achieve its takeoff speed at a constant acceleration before it reaches the end of the 220 m runway.

Consider the formula from kinematics,

`v^(2)-u^(2)=2as`,

where,

v - final velocity of particle,

u - initial velocity of particle

a - the constant acceleration at which the particle is moving with

s - distance travelled by particle

So at minimum acceleration,it reaches the takeoff speed at the end of the runway.

Therefore in the formula,

we use, v = Lift off speed = 34.72222 m/s;

u = 0 m/s (plane starts from rest);

a = minimum acceleration of the plane

s = 220 m;

Substituting these values in the formula, we get,

`(34.72222)^(2)-0 = 2* a * 220`

a =
`((34.72222)^(2) )/(2*220)`

a =
`2.74 m/s^(2)`.

Therefore,the minimum acceleration of the plane required =
`2.74 m/s^(2)`.