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:
(4,-2)
Step-by-step explanation:
plug in and simplify;
y+2=-3(x-4)
-2+2=-3(x-4)
0=-3x+12
3x=12
x=4
f( x ) = - 1/3 x + 5
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The answer is B because I just took the quiz.
Answer:
All real numbers.
Step-by-step explanation:
Since this is a quadratic function, that means the domain is all real numbers.
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