What is the minimum runway length required for a Cessna 150 to take off with a speed of 34.8 m/s and an acceleration of 2.29 m/s²?

To determine the minimum runway length required for the Cessna 150 to take off, we can use the kinematic equation that relates acceleration, initial velocity, final velocity, and distance:

d = (v_f² – v_i²) / (2a)

Where:

• d = distance (runway length in meters)
• v_f = final velocity (takeoff speed)
• v_i = initial velocity (0 m/s for takeoff)
• a = acceleration

In this case:

• v_f = 34.8 m/s
• v_i = 0 m/s
• a = 2.29 m/s²

Plugging in the values:

d = (34.8² – 0²) / (2 * 2.29)

Calculating this gives:

d = (1218.24) / 4.58

d ≈ 265.5 m

This means that the minimum runway length required for the Cessna 150 to take off at the specified conditions is approximately 265.5 meters.