A) A=500(1+0.015)^t
b)800=500(1.015)^t
800/500=1.015^t
t=log(800/500)/log(1.015)
t=31.6 years
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.
Can you post a picture please!
Answer:
a.) -20p - 2
Step-by-step explanation:
Answer:
1.02272727... mi per hr
Step-by-step explanation: