An airplane must achieve a velocity of 71 m/s to takeoff. If the runway is 1000 m long, at what speed must the plane constantly accelerate?
Can you show the formula for how you found the answer and explain how to do it.
The velocity achieved after going a distance X while accelerating at a rate a, from a standing start, is
V = sqrt (2 a X)
Rearrange that equaiton to arrive at the acceleration "a" needed to achieve velocity V at runway distance X:
2aX = V^2
a = V^2/(2X)
2.52 m/s^2 is what I got...its a choice on the paper
Thanks
12,800
To find the speed at which the plane must constantly accelerate in order to take off, we can use the equations of motion.
Let's assume the initial velocity of the plane is 0 m/s and the final velocity is 71 m/s. The total distance it needs to cover is 1000 m.
Using the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity
u = initial velocity
a = acceleration
s = distance
We know the final velocity (v) is 71 m/s, the initial velocity (u) is 0 m/s, and the distance (s) is 1000 m.
Plugging these values into the equation, we get:
71^2 = 0^2 + 2a * 1000
Expanding the equation, we have:
5041 = 2000a
Now, we can solve for the acceleration (a):
a = 5041 / 2000
a = 2.52 m/s^2
Therefore, the plane must constantly accelerate at a rate of 2.52 m/s^2 to achieve a velocity of 71 m/s and take off on a runway that is 1000 m long.
The formula used in this calculation is a basic equation of motion, derived from Newton's laws of motion. By plugging in the given values and solving for the unknown variable, we can determine the required acceleration.