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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

  • physics -

    Why do you think it had a starting velocity? Most airplanes start at the end of the runway with zero velocity.

  • physics -

    how would I find part(b)?

  • physics -

    I got the right answer, but the calculation does not make sense.

    (240m)/(33.3 m/s) x 2 = 14.4s

  • physics (EXPLAIN) -

    why would you multiply by 2?

  • physics -

    Part 2.

    time= distance/average velocity


    vf= vi+ at at takeoff, solve for t.
    33.3=0+2.32 t

  • physics -

    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

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