What is the minimum constant acceleration does this require if the aircraft is to be airborne after a take-off?

let's find acceleration plane goes @ 125 km/h after running 159m. higher acceleration plane airborne earlier, lower 1 not let reach desired speed before running 159 m of runway (i don't think asked demonstrate that, although not difficult).

let's call v speed @ take-off, t time elapsed between start , takeoff.

formula linking speed v acceleration @ time t, in case of constant acceleration , initial speed of vi,
v = * t + vi

in our case, @ start, plane immobile, vi = 0.

if apply formula @ time of take-off, get:

v = * t

formula linking position x after movement @ acceleration a, initial speed of vi, initial position of xi, @ time t is:

x = 1/2 * t^2 + vi * t + xi

let's mesure distances starting point (which means xi = 0). said, vi 0, in our case:

x = 1/2 * t^2

if apply @ time of take-off, when time t , distance start x,

x = 1/2 t^2

= 2*x / t^2

know x, that's 159 m. don't know t.

have 2 equations:

= 2*x / t^2
v = * t

know variables except t , a.

second equation can written t = v/a

can replace t in first equation v/a, get:

x = 1/2 * (v/a)^2
= / 2 * (v^2 /a^2 )
= (a * v^2) / (2 * a^2)
= v^2 / (2 * a)

same
= (v^2) / (2*x)

way, can check possible mistakes looking if units make sense.

v m/s
v^2 m^2/s^2
v m

(v^2) / (2*x) ( (m^2/s^2) / m) gives m/s^2, correct unit accelerations. seems ok.

know x = 159 m
v 125 km/h, need convert m/s

1km 1000 meters
1 hour 3600 seconds

so, 1 km/s same (1000) / (3600) = 1 / 3.6 m/s

let's compute acceleration now:

= (v^2) / (2*x) = ( (125/3.6)^2 ) / (2 * 159)

= 3.79 m/s^2

sound plausible? it's bit less third of g (9.8 m/s^2, acceleration of earth attraction), means feel inside plane force pushing of seat of bit less third of weight. seems reasonable.

a cessna 150 aircraft has lift-off speed of approximately 125 km/h. minimum constant acceleration require if aircraft airborne after take-off run of 159m? answer in units of m/s^2

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