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 5.00 m/s^2 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 t TO needed to take off?Express your answer in seconds using three significant figures.

Answer 1

Answer:

The time needed to take off is $t=26.8s$.

Explanation: Let's order the information.

$x_i=0.00m$

$v_i=0.00\frac{m}{s}$

$a=5.00\frac{m}{s^2}$

$x_f=1800.00m$

From Kinematics, the law for position as a function of time is:

$x(t)=x_i+v_it+\frac{1}{2}at^2$

So the time needed to take off will fallow this rule:

$x_f=x_i+v_it+\frac{1}{2}at^2$ ⇒ $(x_i-x_f)+v_it+\frac{1}{2}at^2=0$

⇒ $-x_f+\frac{1}{2}at^2=0$ ⇒ $t=\sqrt{\frac{2x_f}{a} }$

∴ $t=\sqrt{\frac{2x_f}{a} }=26.83s$.

This is the time needed to take off.

Written with only three significant figures:

$t=26.8s$.

Where the 8 stays the same since 3<5.