Question

To take off from the ground, an airplane must reach a sufficiently high speed. The velocity required for the takeoff, the takeoff velocity, depends on several factors, including the weight of the aircraft and the wind velocity. Part A A plane accelerates from rest at a constant rate of 5.00 m/s2 along a runway that is 1800 m long. Assume that the plane reaches the required takeoff velocity at the end of the runway. What is the time tTO needed to take off

Answer 1

Answer:

t = 26.8 s

Explanation:

Here, we can use the second equation of motion to calculate the required time:

$s = v_it + \frac{1}{2}at^2$

where,

s = distance = 1800 m

vi = initial speed = 0 m/s

t = time needed = ?

a = acceleration = 5 m/s²

Therefore,

$1800\ m = (0\ m/s)t+\frac{1}{2}(5\ m/s^2)t^2\\\\t^2 = \frac{(1800\ m)(2)}{5\ m/s^2}\\\\t = \sqrt{720\ s^2}$

t = 26.8 s