Answer:
132.337 miles per hour
Explanation:
We are given that an aircraft with amss 43581 kg
The acceleration of an aircraft =[tex]1.7 m/s^2[/tex]
Initial velocity of an aircraft =0 (Because it start moving from rest)
Time taken by aircraft to reaches take-off speed=34.8 s
Using newton's second law of motion
[tex]s=ut+\frac{1}{2}at^2[/tex]
Substitute all values then we get
[tex]s=0\times 34.8+\frac{1}{2}1.7(34.8)^2[/tex]
[tex]s=1029.384 m[/tex]
Using Newton's third law of motion
[tex]v^2-u^2=2as[/tex]
Substitute all given values then we get
[tex]v^2-0=2\times 1.7\times 1029.384[/tex]
[tex]v=\sqrt{3499.9056}=59.16 m/s[/tex]
1 meter=0.000621371 miles
1 hour =3600 seconds
Therefore, v=[tex]\frac{59.16\times 0.000621371}{\frac{1}{3600}}[/tex]=132.337 miles per hour
