An airport has runways only 136m long. A small plane must reach a ground speed of 39m/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.
Highway safety engineers build soft barriers so that cars hitting them will slow down at a soft rate. A person wearing a safety belt can withstand an acceleration of 300m/s^2.
How thick should barriers be to safely stop a car that hits the barriers at 92.8km/h?
Answer in units of m.
V = sqrt(2 a X)
V is the speed that must be attained
a is the acceleration
X is the distance required
Solve for a.
75m/s^2
To find the average acceleration the plane's engines must provide, we can use the equation:
v^2 = u^2 + 2as
Where:
v = final velocity (ground speed) = 39 m/s
u = initial velocity = 0 m/s (since the plane starts from rest)
a = average acceleration
s = distance covered = 136 m
Rearranging the equation, we get:
a = (v^2 - u^2) / (2s)
Substituting the given values, we have:
a = (39^2 - 0) / (2 * 136)
a = 1521 / 272
a ≈ 5.59 m/s^2
Therefore, the average acceleration the plane's engines must provide is approximately 5.59 m/s^2.