An airplane begins its take-off run at a with zero velocity and a constant acceleration a. knowing that it becomes airborne 30 s later at b with a take-off velocity of 270 km/h, determine (a) the acceleration a, (b) distance ab.

a)The acceleration of the airplane is 2.5 m/s².

b)The distance AB is 1125 m.

Explanation:

Hi there!

a)The equation of velocity of an object moving in a straight line with constant acceleration is the following:

v = v0 + a · t

Where:

v = velocity of the object at time "t".

v0 = initial velocity.

a = acceleration.

t = time

We have the following information:

The airplane starts with zero velocity (v0 = 0) and its velocity after 30 s is 270 km/h (converted into m/s: 270 km/h · 1000 m/1 km · 1 h /3600 s = 75 m/s). Then, we can solve the equation to obtain the acceleration:

75 m/s = a · 30 s

75 m/s / 30 s = a

a = 2.5 m/s²

The acceleration of the airplane is 2.5 m/s².

b)The distance AB can be calculated using the equation of position of an object moving in a straight line with constant acceleration:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time "t".

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

If we place the origin of the frame of reference at A, then, x0 = 0. Since the airplane is initially at rest, v0 = 0. So, the equation gets reduced to this:

x = 1/2 · a · t²

Let´s find the position of the airplane after 30 s:

x = 1/2 · 2.5 m/s² · (30 s)²

x = 1125 m

The position of B is 1125 m away from A (the origin), then, the distance AB is 1125 m.

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000