A Cessna 350 must reach a speed of 35 m/s for takeoff. If the plane accelerates at a steady 3.8 m/s2, how long must the runway be?
the time to accelerate to takeoff speed is __ 35 / 3.8
the average speed during acceleration is __ (0 + 35) / 2
the distance (length of runway) is speed times time
26.7
To find the required length of the runway, we can use the equations of motion. The formula we'll use is:
v^2 = u^2 + 2as
Where:
- v is the final velocity (in this case, 35 m/s)
- u is the initial velocity (0 m/s since the plane starts from rest)
- a is the acceleration (3.8 m/s^2)
- s is the distance traveled
Rearranging the equation, we get:
s = (v^2 - u^2) / (2a)
Plugging in the values, we have:
s = (35^2 - 0^2) / (2 * 3.8)
s = (1225 - 0) / 7.6
s = 161 meters
Therefore, the runway must be at least 161 meters long for the Cessna 350 to reach a speed of 35 m/s for takeoff.