An airport has runways only 196 m long. A
small plane must reach a ground speed of
38 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/s
2
To find the average acceleration needed for the plane to take off safely, we can use the following equation:
v^2 = u^2 + 2as
where:
v = final velocity (ground speed) = 38 m/s
u = initial velocity = 0 m/s (since the plane starts from rest)
a = average acceleration (what we need to find)
s = distance covered = 196 m
Rearranging the equation, we have:
a = (v^2 - u^2) / (2s)
Plugging in the values, we get:
a = (38^2 - 0^2) / (2 * 196)
a = (1444) / (392)
a ≈ 3.68 m/s^2
Therefore, the average acceleration the plane's engines must provide for it to take off safely from the airport is approximately 3.68 m/s^2.