Question

1. planes take - off

this video is an etihad airbus a380 flight from abu dhabi to paris piloted by 2 women
the center line in the runways are 30m long and 20m apart. write down the position and times for the plane take - off.
airbus engineers have modeled the take off with the following equations:
position: s(t)=\frac{1}{2}1.7t^{2}
velocity: v(t)=1.7t
are these equations a good mathematical model of the take off?
questions

1. what is the planes initial position?
2. what is the planes initial velocity?
3. what is the planes final velocity before take off?
4. what is the acceleration?

motion graphs
sketch the position - time, velocity - time and acceleration - time graph for this motion
motion
describe this motion:

Explanation:

Step1: Find initial position

Initial time $t = 0$. Substitute $t = 0$ into position - function $s(t)=\frac{1}{2}1.7t^{2}$.
$s(0)=\frac{1}{2}\times1.7\times0^{2}=0$ m

Step2: Find initial velocity

Initial time $t = 0$. Substitute $t = 0$ into velocity - function $v(t)=1.7t$.
$v(0)=1.7\times0 = 0$ m/s

Step3: Assume take - off when plane reaches end of runway

Let's assume the runway length is used for take - off. If we assume the runway length is $500$ m (from the table values). We first find the time when $s(t)=500$ m. Given $s(t)=\frac{1}{2}1.7t^{2}$, then $500=\frac{1}{2}1.7t^{2}$. Solving for $t$:
$t^{2}=\frac{1000}{1.7}\approx588.24$, so $t=\sqrt{\frac{1000}{1.7}}\approx24.25$ s. Then substitute $t$ into $v(t)$: $v(t)=1.7t$, so $v(24.25)=1.7\times24.25 = 41.225$ m/s

Step4: Find acceleration

Acceleration $a(t)$ is the derivative of velocity function. Since $v(t)=1.7t$, then $a(t)=\frac{d}{dt}(1.7t)=1.7$ m/s²

Answer:

1. The plane's initial position is $0$ m.
2. The plane's initial velocity is $0$ m/s.
3. The plane's final velocity before take - off is approximately $41.225$ m/s.
4. The acceleration is $1.7$ m/s².

For the motion graphs:

• Position - time graph: The function is $s(t)=\frac{1}{2}1.7t^{2}$, which is a parabola opening upwards with the vertex at the origin $(0,0)$.
• Velocity - time graph: The function is $v(t)=1.7t$, which is a straight - line passing through the origin with a slope of $1.7$.
• Acceleration - time graph: The function is $a(t) = 1.7$, which is a horizontal line at $y = 1.7$ on the acceleration - time plane.

The motion is a uniformly accelerated linear motion, where the position of the plane increases quadratically with time, the velocity increases linearly with time, and the acceleration is constant.