An airplane requires 420 m of runway to take off from an airstrip. If the plane starts from rest and it takes 32 s to lift off, with what velocity does it leave the ground? What was the average acceleration of the plane?
Δx = vi*Δt+(1/2)aΔt²
vi=0
so
Δx = (1/2)aΔt²
a=2Δx/Δt²
=2*420/(32²)
=(105/128) m/s²
vf=aΔt
=(105/128)*32
=26.25 m/s (not extremely fast for a plane)
To find the velocity at which the plane leaves the ground, we can use the equation v = u + at, where:
- v is the final velocity,
- u is the initial velocity (which is 0 in this case because the plane starts from rest),
- a is the acceleration, and
- t is the time taken.
Given:
Initial velocity (u) = 0 m/s
Acceleration (a) = ?
Time (t) = 32 s
To find the acceleration, we can use the equation s = ut + (1/2)at^2, where:
- s is the distance covered (in this case, the length of the runway),
- u is the initial velocity (0 m/s),
- t is the time taken, and
- a is the acceleration.
Given:
Distance (s) = 420 m
Initial velocity (u) = 0 m/s
Time (t) = 32 s
Acceleration (a) = ?
Using the equation s = ut + (1/2)at^2, we can solve for a.
420 = 0(32) + (1/2)a(32)^2
420 = 0 + 16a(32)^2
420 = 16a(1024)
420 = 16384a
Dividing both sides by 16384:
420/16384 = a
The average acceleration of the plane is approximately 0.0256 m/s^2.
Now, let's calculate the velocity at which the plane leaves the ground using v = u + at.
v = 0 + (0.0256)(32)
v = 0.8192 m/s
Therefore, the velocity at which the plane leaves the ground is approximately 0.8192 m/s.
To find the velocity with which the airplane leaves the ground, we can use the formula:
v = u + at
Where:
v = final velocity (the velocity with which the airplane leaves the ground)
u = initial velocity (in this case, the airplane is initially at rest, so u = 0)
a = acceleration
t = time taken to lift off (32 s)
Now, let's find the acceleration of the airplane to use in the above equation. We can use the formula:
s = ut + (1/2)at^2
Where:
s = distance traveled (420 m)
u = initial velocity (0 m/s)
t = time taken to lift off (32 s)
a = acceleration (unknown)
Rearranging the formula to solve for a, we have:
a = 2(s - ut) / t^2
Plugging in the values, we get:
a = 2(420 m - 0 m/s * 32 s) / (32 s)^2
a = 2(420 m) / (1024 s^2)
a ≈ 0.82 m/s^2
Now that we have the acceleration (a), we can use the first formula to find the final velocity (v):
v = u + at
v = 0 + (0.82 m/s^2) * 32 s
v ≈ 26.24 m/s
Therefore, the airplane leaves the ground with a velocity of approximately 26.24 m/s.
To find the average acceleration of the plane, we can use the formula:
average acceleration (a_avg) = Δv / t
Where:
Δv = change in velocity (final velocity - initial velocity)
t = time taken to lift off (32 s)
Since the initial velocity (u) is 0 m/s, the change in velocity (Δv) is equal to the final velocity (v). Therefore:
a_avg = v / t
a_avg ≈ (26.24 m/s) / (32 s)
a_avg ≈ 0.82 m/s^2
Hence, the average acceleration of the airplane is approximately 0.82 m/s^2.