Question

A jet airplane lands with a speed of 120 mph. It has 1800 ft of runway after touch- down to reduce its speed to 30 mph. Compute the average acceleration required of the airplane during braking A: a -8.1 ft/s2

Answer 1

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 :

$v^2-u^2=2ad$

a = acceleration

$a=\dfrac{v^2-u^2}{2d}$

$a=\dfrac{(44\ ft/s)^2-(176\ ft/s)^2}{2\times 1800\ ft}$

$a=-8.06\ ft/s^2$

So, the average acceleration required of the airplane during braking is -8.06 ft/s². Hence, this is the required solution.