An airport has runways only 132 m long. A small plane must reach a ground speed of 40 m/s before it can become airborne.
What average acceleration must the plane’s engines provide if it is to take off safely from its airport?
Answer in units of m/s2
Vf^2 = Vo^2 + 2ad,
a = (Vf^2 - Vo^2) / 2d,
a = ((40)^2 - 0) / 264 = 6.1m/s^2.
To find the average acceleration required for the small plane to take off safely, we can use the following equation:
v^2 = u^2 + 2as
Where:
v = final velocity (ground speed) = 40 m/s
u = initial velocity (0 m/s as the plane is initially at rest)
a = average acceleration (what we need to find)
s = distance traveled (runway length) = 132 m
Rearranging the equation to solve for acceleration (a):
a = (v^2 - u^2) / (2s)
Substituting the known values:
a = (40^2 - 0^2) / (2 * 132)
a = 1600 / 264
a ≈ 6.06 m/s^2
Therefore, the average acceleration required for the plane to take off safely is approximately 6.06 m/s^2.
To find the average acceleration required for the plane to take off safely, we can use the following kinematic equation:
v^2 = u^2 + 2as
where:
v is the final velocity (ground speed) of the plane, which is 40 m/s
u is the initial velocity of the plane, which is 0 m/s (as it starts from rest)
a is the average acceleration of the plane during takeoff
s is the displacement, which is the length of the runway, 132 m
Rearranging the equation, we get:
a = (v^2 - u^2) / (2s)
Substituting the given values, we have:
a = (40^2 - 0^2) / (2 * 132)
Simplifying further:
a = (1600 - 0) / 264
a ≈ 6.06 m/s^2
Therefore, the average acceleration the plane's engines must provide in order to take off safely from the airport is approximately 6.06 m/s^2.