Given,
the initial velocity = 0 m /s.
acceleration = 3.20 m / s^2
time = 32.8 s
According to laws of motion.
s = ut + 1/2 at ^2
s = 1/2 at²
s=1/2(3.20)(32.8)²
s= 1721.344 m
the distance traveled before takeoff is 1731.3m
Answer:
12.14 cm
Explanation:
mass, m = 15.5 kg
frequency, f = 9.73 Hz
maximum amplitude, A = 14.6 cm
t = 1.25 s
The equation of the simple harmonic motion
y = A Sin ωt
y = A Sin (2 x π x f x t)
put, t = 1.25 s, A = 14.6 cm, f = 9.73 Hz
y = 14.6 Sin ( 2 x 3.14 x 9.73 x 1.25)
y = 14.6 Sin 76.38
y = 12.14 cm
Thus, the displacement of the particle from the equilibrium position is 12.14 cm.
