Answer:
[tex]ax = 2.60m/s^{2}[/tex], [tex]t = 26.92s[/tex]
Explanation:
The acceleration of the plane can be determined by means of the kinematic equation that correspond to a Uniformly Accelerated Rectilinear Motion.
[tex](vx)f^{2} = (vx)i^{2} + 2ax \Lambda x[/tex] (1)
Where [tex](vx)f^{2}[/tex] is the final velocity, [tex](vx)i^{2}[/tex] is the initial velocity, [tex]ax[/tex] is the acceleration and [tex]\Lambda x[/tex] is the distance traveled.
Equation (1) can be rewritten in terms of ax:
[tex](vx)f^{2} - (vx)i^{2} = 2ax \Lambda x[/tex]
[tex]2ax \Lambda x = (vx)f^{2} - (vx)i^{2}[/tex]
[tex]ax = \frac{(vx)f^{2} - (vx)i^{2}}{2 \Lambda x}[/tex] (2)
Since the plane starts from rest, its initial velocity will be zero ([tex](vx) = 0[/tex]):
Replacing the values given in equation 2, it is gotten:
[tex]ax = \frac{(70m/s)^{2} - (0m/s)^{2}}{2(940m)}[/tex]
[tex]ax = \frac{4900m/s}{2(940m)}[/tex]
[tex]ax = \frac{4900m/s}{1880m}[/tex]
[tex]ax = 2.60m/s^{2}[/tex]
So, The acceleration of the plane is [tex]2.60m/s^{2}[/tex]
Now that the acceleration is known, the next equation can be used to find out the time:
[tex](vx)f = (vx)i + axt [/tex] (3)
Rewritten equation (3) in terms of t:
[tex]t = \frac{(vx)f - (vx)i}{ax}[/tex]
[tex]t = \frac{70m/s - 0m/s}{2.60m/s^{2}}[/tex]
[tex]t = 26.92s[/tex]
Hence, the plane takes 26.92 seconds to reach its take-off speed.
