A jet must reach a velocity of 75 m/s for takeoff. If the runway is 2100 meters long, what must the constant acceleration be?

Explanation:

Given that,

Initialy, the jet is at rest, u = 0

Final velocity of the jet, v = 75 m/s

Distance, d = 2100 m

We need to find the acceleration of the jet. It is based on the concept of equation of kinematics. Using third equation of motion, w get :

a is acceleration

So, the acceleration of the jet is .

the answer is kevlar

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures