To solve this problem, we can use the equation of motion:

d = ut + 0.5at^2

where:
d = distance covered
u = initial velocity
a = acceleration
t = time

Given in the problem:
u = 0 (since the airplane starts from rest)
a = 5.50 m/s^2
t = 20.25 s

Substituting these values into the equation, we get:

d = 0*(20.25) + 0.5*(5.50)*(20.25)^2

Solving this, we find:

d = 0 + 0.5*5.50*(20.25)^2
d = 0 + 0.5*5.50*410.0625
d = 0 + 1125.171875
d = 1125.17 m

So, the airplane covers a distance of approximately 1125.17 meters before takeoff.

Question

To solve this problem, we can use the equation of motion:

d = ut + 0.5at^2

where:
d = distance covered
u = initial velocity
a = acceleration
t = time

Given in the problem:
u = 0 (since the airplane starts from rest)
a = 5.50 m/s^2
t = 20.25 s

Substituting these values into the equation, we get:

d = 0*(20.25) + 0.5*(5.50)*(20.25)^2

Solving this, we find:

d = 0 + 0.5*5.50*(20.25)^2
d = 0 + 0.5*5.50*410.0625
d = 0 + 1125.171875
d = 1125.17 m

So, the airplane covers a distance of approximately 1125.17 meters before takeoff.

Knowee AI · Accepted Answer