Answer:
A) The minimum constant acceleration is [tex]1.9148\frac{m}{s^2}[/tex]
B) The airplane takes 17.553 seconds to become airborne.
Explanation:
First, consider the basic equations for the motion with constant acceleration:
[tex]v=v_0+at\\x=x_0+v_0t+\frac{1}{2}at^2[/tex]
Where [tex]x_0[/tex] and [tex]v_0[/tex] are the initial position and velocity respectively; a is the acceleration and t the time.
Let's imagine that the aircraft had become airborne after the 295 m run, and let's model this movement by the last equations. The airplane is considered to be at position zero ([tex]x_0=0[/tex] and start from repose (it's start velocity is zero: [tex]v_0=0[/tex])
Let's convert the velocity to m/s:
[tex]v=121\frac{km}{h} *\frac{1000m}{km} *\frac{h}{3600s} =33.611\frac{m}{s}[/tex]
The airplane has a velocity of 33.611 m/s and has run 295 meters when it becomes airborne, so at that time:
[tex]v=v_0+at[/tex]
[tex]33.611=at[/tex] (The initial velocity is zero)
[tex]x=x_0+v_0t+\frac{1}{2}at^2\\295=\frac{1}{2}at^2[/tex] (The initial position is zero)
At this point we have a two-variable-two-equation system; if we could solve it we would know the time and the acceleration.
Let's isolate the time from the first equation and replace it in the second one.
[tex]t=\frac{33.611}{a}\\t^2=\frac{1129.7}{a^2} \\295=\frac{1}{2}at^2\\295=\frac{1}{2}a*(\frac{1129.7}{a^2} )\\295=\frac{564.85}{a}[/tex]
Then, find the acceleration:
[tex]a=\frac{564.85}{295} =1.9148\frac{m}{s^2}[/tex]
Now, it is possible to find the time by means of any of the initial equations we wrote. Let's take, for example, the first one:
[tex]t=\frac{33.611}{a}\\t=\frac{33.611}{1.9148}=17.553s[/tex]
