Asked by Anonymous

A Cessna aircraft has a lift-off speed of 120 km/h.

(a)What minimum constant acceleration does this require if the aircraft is to be airborne after a take=off of 240m?

Work: 120 km/h-->33.3 m/s
(33.3 m/s)^2= (16.7 m/s)^2 + 2a(240 m)
a= 1.73 m/s^2
[What did I do wrong?]

(b)How long does it take the aircraft to be airborne?

Answer in back of book: (a) 2.32 m/s^2
(b) 14.4 s

Answered by bobpursley
Why do you think it had a starting velocity? Most airplanes start at the end of the runway with zero velocity.
Answered by Anonymous
how would I find part(b)?
Answered by Anonymous
I got the right answer, but the calculation does not make sense.

(240m)/(33.3 m/s) x 2 = 14.4s
Answered by Anonymous
why would you multiply by 2?
Answered by bobpursley
Part 2.

time= distance/average velocity

or

vf= vi+ at at takeoff, solve for t.
33.3=0+2.32 t
Answered by Colton.
I got the answer to Part A and B.

Part A. (33.3)^2-0(velocity initial) / 2(240) = a
1108.89/480 = 2.32 (rounded)

Part B. (240m)(33.3m/s) x 2 = 14.4
Answered by Anonymous
Where did you get the 33.3 m/s in 120 km/h?
Answered by Teemotoo
What are the steps and equation used to solve this problem?
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