Initial velocity of the plane is Vo = 0.
acceleration a = 1.3 m/s2
total distance = 2.5 km = 2500m
time taken to reach 2.5 km with 1.3m/s^2 acceleration = t
S = Vo t + 0.5 a t^2
2500 = 0 + (0.5*1.3* t^2)
t^2 = 3846.15
t = 62 s
the maximum velocity plan can reach within 62 s is Vt
Vt = Vo + a t
Vt = 0 + (1.3*62)
Vt = 80.6 m/s
Since 80.6 m/s is greater than 75 m/s, plane can use this runway to takeoff with required speed.
Answer:
808+808=1616 340+340=680 then 1616-680=976
Step-by-step explanation:
you need to add 808 plus 808 which is 1616 then add 340 plus 340 which is 680, then subtract 1616-680 which is 976.
The answer to your question is Y=12
C because 180-230 would be negative b as well as three quarters of seven
#b. The equation y = 50x + 200 represents the total cost that the contractor will charge for x hours of work.
#c. (1, 250); (2, 300); (3, 350)
