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.
Range is the value of the output
basically
y=f(x)
since it is -4
y=-4 is the range
not listed
Answer:
you can do it i belive
Step-by-step explanation:
Answer:
6a+150+18c
Step-by-step explanation:
6(a+25+3c)
6a +6 x 25 +6 x 3c
6a + 150+6 x 3c
6a +150 +18c
