Answer:
the question is incomplete, below is the complete question
"An object is undergoing simple harmonic motion along the x-axis. Its position is described as a function of time by x(t) = 5.5 cos(4.4t - 1.8), where x is in meters, the time, t, is in seconds, and the argument of the cosine is in radians.
a.What is the object's velocity, in meters per second, at time t = 2.9?
b.Calculate the object's acceleration, in meters per second squared, at time t = 2.9.
c. What is the magnitude of the object's maximum acceleration, in meters per second squared?
d.What is the magnitude of the object's maximum velocity, in meters per second?"
a.![v(t)==24.1m/s](https://tex.z-dn.net/?f=v%28t%29%3D%3D24.1m%2Fs)
b.![a(t)=3.79m/s^{2}](https://tex.z-dn.net/?f=a%28t%29%3D3.79m%2Fs%5E%7B2%7D)
c.![a_{max}=106.48m/s^{2}](https://tex.z-dn.net/?f=a_%7Bmax%7D%3D106.48m%2Fs%5E%7B2%7D)
d.![v_{max}=24.2m/s](https://tex.z-dn.net/?f=v_%7Bmax%7D%3D24.2m%2Fs)
Explanation:
the gneral expression for the displacement of object in simple harmonic motion is represented by
![x(t)=Acos(wt- \alpha)\\](https://tex.z-dn.net/?f=x%28t%29%3DAcos%28wt-%20%5Calpha%29%5C%5C)
while the velocity is express as
![v(t)=-Awcos(4.4t-1.8)\\](https://tex.z-dn.net/?f=v%28t%29%3D-Awcos%284.4t-1.8%29%5C%5C)
and the acceleration is
![a(t)=-aw^{2}cos(wt- \alpha )\\](https://tex.z-dn.net/?f=a%28t%29%3D-aw%5E%7B2%7Dcos%28wt-%20%5Calpha%20%29%5C%5C)
Note: the angle is in radians
The expression for the displacement from the question is ![x(t)=5.5cos(4.4t-1.8)\\](https://tex.z-dn.net/?f=x%28t%29%3D5.5cos%284.4t-1.8%29%5C%5C)
comparing, A=5.5, <em>w=4.4,α=1.8</em>
a.To determine the object velocity at t=2.9secs,
we substitute for t in the velocity equation
![v(t)=-5.5*4.4sin(4.4*2.9-1.8)\\v(t)=-24.2sin(10.96)\\](https://tex.z-dn.net/?f=v%28t%29%3D-5.5%2A4.4sin%284.4%2A2.9-1.8%29%5C%5Cv%28t%29%3D-24.2sin%2810.96%29%5C%5C)
![v(t)=-24.2*(-0.9993)\\v(t)==24.1m/s](https://tex.z-dn.net/?f=v%28t%29%3D-24.2%2A%28-0.9993%29%5C%5Cv%28t%29%3D%3D24.1m%2Fs)
b.To determine the object acceleration at t=2.9secs,
we substitute for t in the acceleration equation
![a(t)=-5.5*4.4^{2} cos(4.4*2.9-1.8)\\a(t)=-106.48cos(10.96)\\](https://tex.z-dn.net/?f=a%28t%29%3D-5.5%2A4.4%5E%7B2%7D%20cos%284.4%2A2.9-1.8%29%5C%5Ca%28t%29%3D-106.48cos%2810.96%29%5C%5C)
![a(t)=-106.48*0.0356\\a(t)=3.79m/s^{2}](https://tex.z-dn.net/?f=a%28t%29%3D-106.48%2A0.0356%5C%5Ca%28t%29%3D3.79m%2Fs%5E%7B2%7D)
c. The acceleration is maximum when the displacement equals the amplitude. hence magnitude of the object acceleration is
![a_{max}=-w^{2}A\\ a_{max}=-4.4^{2}*5.5\\ a_{max}=106.48m/s^{2}](https://tex.z-dn.net/?f=a_%7Bmax%7D%3D-w%5E%7B2%7DA%5C%5C%20a_%7Bmax%7D%3D-4.4%5E%7B2%7D%2A5.5%5C%5C%20a_%7Bmax%7D%3D106.48m%2Fs%5E%7B2%7D)
d.The maximum velocity is expressed as
![v_{max}=wA\\v_{max}=4.4*5.5\\v_{max}=24.2m/s](https://tex.z-dn.net/?f=v_%7Bmax%7D%3DwA%5C%5Cv_%7Bmax%7D%3D4.4%2A5.5%5C%5Cv_%7Bmax%7D%3D24.2m%2Fs)