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:
C
Step-by-step explanation:
yeyeyeye
Answer:
I just learned this about a few weeks ago I hope this helps.
Step-by-step explanation:
Since it's cosine you're only looking at Adjacent and the hypotenuse which the adj is 20 and the hyp is 25 so since cosine is adj/hyp you do 20/25 then simplify it to 20/25 divide both by 5 so it's simplified to 4/5.
Then you do Cos-1(4/5)=36.869 which is rounded to the tenth 36.9 but rounded to the hundredth 36.87.
Hope this helps
