formulas for acceleration, velocity and position of plane are
k²=2as, k=30a, s=450a and the value of k is 230 ft/sec
rate of changes of distances with respect to time.
rate of changes of velocity with respect to time.
Given, final velocity, v= k ft/s
time needed to attain speed k is 30 s
time, t= 30 sec
acceleration, a = constant
initially at time t=o, velocity and distance travel by aircraft was also o
initial velocity u =0, distance s = 0
as per the formula regarding velocity, distance, time and acceleration
v² = u²+2as
as per the question final velocity v= k, initial velocity u=0
k² = 2as
again, the formula v= u+at
the velocity of v after 30 second
k= a×30
acceleration, a= k/30
from the other formula, s=ut+1/2×at²
distance travel by aircraft after 30 second with initial velocity 0.
s= u×0+1/2×a(30) ²
s= 450a
previously we got a=k/30
now, s =450×k/30= 15k
if the runway is 3450 feet long and the plane needs to use the entire runway to take off then the value of k will be
s=15k
k=230 ft/sec
hence, the value of k is 230 ft/sec
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