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.A plane accelerates from rest at a constant rate of 5.00m/s2along a runway that is 1800mlong. Assume that the plane reaches the required takeoff velocity at the end of the runway. What is the timetTOneeded to take off?Express your answer in seconds using three significant figures.

Answer 1

Answer: 26,8 s

Explanation:

If we are talking about an acceleration at a constant rate , we are dealing with constant acceleration, hence we can use the following equations:

${V_{f}}^{2}={V_{o}}^{2}+2ad$ (1)

$V_{f}=V_{o}+at$ (2)

Where:

$V_{f}$ is the final velocity of the plane (the takeoff velocity in this case)

$V_{o}=0$ is the initial velocity of the plane (we know it is zero because it starts from rest)

$a=5m/s^{2}$ is the constant acceleration of the plane to reach the takeoff velocity

$d=1800m$ is the distance of the runway

$t$ is the time

Knowing this, let's begin with (1):

${V_{f}}^{2}=0+2(5m/s^{2})(1800m)$ (3)

${V_{f}}^{2}=18000m^{2}/s^{2}$ (4)

$V_{f}=134.164 m/s$ (5)

Substituting (5) in (2):

$134.164 m/s=0+(5m/s^{2})t$ (6)

Finding $t$:

$t=26.8 s$ This is the time needed to take off