Answer:
The average acceleration is 8.06 m/s².
Explanation:
It is given that,
Initial speed of the jet, u = 120 mph = 176 ft/s
Final velocity of the jet, v = 30 mph = 44 ft/s
Distance, d = 1800 ft
We need to find the average acceleration required of the airplane during braking. It can be calculated using third law of motion as :
[tex]v^2-u^2=2ad[/tex]
a = acceleration
[tex]a=\dfrac{v^2-u^2}{2d}[/tex]
[tex]a=\dfrac{(44\ ft/s)^2-(176\ ft/s)^2}{2\times 1800\ ft}[/tex]
[tex]a=-8.06\ ft/s^2[/tex]
So, the average acceleration required of the airplane during braking is -8.06 ft/s². Hence, this is the required solution.
