The takeoff speed for an airbus a320 jetliner is 82 m/s . velocity data measured during takeoff are as follows: t(s) vx(m/s) 0 0 10 23 20 46 30 69 part a what is the jetliner's acceleration during takeoff, in m/s2? express your answer using two significant figures. a = m/s2 submitmy answersgive up part b what is the jetliner's acceleration during takeoff, in g's? express your answer using two significant figures. a = g submitmy answersgive up part c at what time do the wheels leave the ground? express your answer using two significant figures. tf = s submitmy answersgive up part d for safety reasons, in case of an aborted takeoff, the length of the runway must be three times the takeoff distance. what is the

a) from the data in the table, we observe that the acceleration is constant (because the rate of change in velocity is the same for each time interval of 10 seconds), so we can choose just one interval and calculate the acceleration as the ratio between the change in velocity and the change in time. Taking the first interval, we find

[tex]a=\frac{\Delta v_x}{\Delta t}=\frac{23 m/s-0}{10s -0}=2.3 m/s^2[/tex]

b) To find the jet's acceleration in g's, we just need to divide the acceleration in m/s^2 by the value of g, the acceleration of gravity (9.81 m/s^2), so we find

[tex]a_g=\frac{a}{g}=\frac{2.3 m/s^2}{9.8 m/s^2}=0.23 g[/tex]

c) the wheels leave the ground when the jet reaches its take-off velocity, which is 82 m/s.

At t=0s, the velocity of the jet is 0. We know that the acceleration is constant (a=2.3 m/s^2), so we can find the time t at which the jet reaches a velocity vf=82 m/s by using the equation

[tex]v_f = v_i +at[/tex]

Re-arranging and substituting numbers, we find

[tex]t=\frac{v_f}{a}=\frac{82 m/s}{2.3 m/s^2}=35.65 s[/tex]