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A Boeing 747 takes off from a runway of length 2.5km. The aircraft moves from a point A, with initial speed 2mms It moves with constant acceleration and leaves the runway, from a point B. 28 seconds later with speed 72ms ¹ Find the acceleration of the aircraft.

Respuesta :

The acceleration of the aircraft will be 2.571 m/sec².Acceleration is denoted by a.

What is acceleration?

Acceleration is defined as the rate of change of the velocity of the body. Its unit is m/sec².It is a vector quantity. It requires both magnitudes as well as direction to define.

Given data;

Initial velocity,u= 0 m/sec

Final velocity,v=72 m/sec

Time period,t= 28 sec

Acceleeration,a=?

The acceleration of the aircraft is;

[tex]\rm a = \frac{v-u}{t} \\\\\ a= \frac{72-0}{28} \\\\\ a=2.571 \ m/sec^2[/tex]

Hence, the acceleration of the aircraft will be 2.571 m/sec.

To learn more about acceleration refer to the link;

https://brainly.com/question/969842

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