An airplane must reach a speed of 185 mi/h to take off. If the runway is 500 m long, what is the minimum value of the acceleration that will allow the airplane to take off successfully?
First of all, convert 185 mph to 82.7 m/s
Then use the equation
V(final)= sqrt (2 a X)
and solve for a.
X is the runway length in meters.
The answer will be in m/s^2.
To find the minimum value of acceleration that will allow the airplane to take off successfully, we need to apply the kinematic equation:
(v^2 - u^2) = 2as
where:
- v is the final velocity (which is 185 mi/h)
- u is the initial velocity (which is 0 mi/h since the airplane is at rest before taking off)
- a is the acceleration
- s is the distance (which is 500 m)
First, we need to convert the final velocity from mi/h to m/s since the distance is given in meters.
1 mi/h = 0.44704 m/s
So, the final velocity would be:
185 mi/h * 0.44704 m/s = 82.8808 m/s
Plugging in the values into the equation, we have:
(82.8808^2 - 0^2) = 2a * 500
Simplifying the equation:
(82.8808^2) = 1000a
Rearranging to solve for acceleration:
a = (82.8808^2) / 1000
Using a calculator, we can calculate a:
a ≈ 6.921 m/s^2
So, the minimum value of the acceleration that will allow the airplane to take off successfully is approximately 6.921 m/s^2.