Question

in order for am f-18 fighter jet to take-off from an aircraft carrier, it needs to reach a speed of 75m/s by quickly accelerating down a 90 m long runway. If an f-18 starts from rest, what acceleration does the fighter jet undergo during takeoff.

Answer 1

The F-18 fighter jet undergoes an acceleration of approximately 31.25 m/s^2 in order to reach a speed of 75 m/s when taking off from a 90 m long runway of an aircraft carrier.

Step-by-step explanation:

To determine the acceleration that an F-18 fighter jet undergoes during takeoff from an aircraft carrier, we can use the kinematic equation:

v2 = u2 + 2as

where v is the final velocity (75 m/s), u is the initial velocity (0 m/s as the jet starts from rest), a is the acceleration, and s is the displacement (90 m, the length of the runway).

By substituting the known values into the equation, we get:

752 = 0 + 2 * a * 90

5625 = 180a

a = 5625 / 180

a ≈ 31.25 m/s2

Therefore, the acceleration of the F-18 during takeoff is approximately 31.25 m/s2.

Answer 2

The acceleration of the F-18 fighter jet during takeoff is
`\(1.25 \, \text{m/s}^2\).`

Step-by-step explanation:

To determine the acceleration, we can use the kinematic equation
`\(v^2 = u^2 + 2as\)`, where v is the final velocity, u is the initial velocity (which is 0 m/s as the jet starts from rest), a is the acceleration, and s is the displacement. Rearranging the equation to solve for acceleration, we get
`\(a = (v^2 - u^2)/(2s)\)`. Plugging in the given values, we find the acceleration.

The positive acceleration indicates that the fighter jet is undergoing a uniform acceleration along the runway, gradually reaching the required takeoff speed.

The physics of motion, acceleration, and the principles behind the takeoff of aircraft from carriers.