Final answer:
The constant acceleration for an airplane to reach a final takeoff velocity of 67 m/s on a 2 km runway is 1.1225 m/s².
Step-by-step explanation:
The constant acceleration of an airplane moving down the runway can be found using the kinematic equation ∙2as = v² - u², where 'v' is the final velocity, 'u' is the initial velocity, 'a' is the acceleration, and 's' is the distance covered. Here, we know the final velocity (v) is 67 m/s, the initial velocity (u) equals 0 m/s (since the airplane starts from rest), and the length of the runway (s) is 2 km, which is 2000 m. So, acceleration (a) can be calculated as follows:
a = (v² - u²) / (2s)
Substituting the values in:
a = (67² - 0²) / (2 * 2000 m)
After calculating, we obtain an acceleration of 1.1225 m/s², which is the constant acceleration needed to achieve a final takeoff velocity of 67 m/s over a 2 km runway.