Question

What is the object’s velocity, in meters per second, at time t = 2.9? Calculate the object’s acceleration, in meters per second squared, at time t = 2.9. What is the magnitude of the object’s maximum acceleration, in meters per second squared? What is the magnitude of the object’s maximum velocity, in meters per second?

Answer 1

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`

b.
`a(t)=3.79m/s^(2)`

c.
`a_(max)=106.48m/s^(2)`

d.
`v_(max)=24.2m/s`

Step-by-step explanation:

the gneral expression for the displacement of object in simple harmonic motion is represented by

`x(t)=Acos(wt- \alpha)\\`

while the velocity is express as

`v(t)=-Awcos(4.4t-1.8)\\`

and the acceleration is

`a(t)=-aw^(2)cos(wt- \alpha )\\`

Note: the angle is in radians

The expression for the displacement from the question is
`x(t)=5.5cos(4.4t-1.8)\\`

comparing, A=5.5, w=4.4,α=1.8

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)\\`

`v(t)=-24.2*(-0.9993)\\v(t)==24.1m/s`

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)\\`

`a(t)=-106.48*0.0356\\a(t)=3.79m/s^(2)`

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)`

d.The maximum velocity is expressed as

`v_(max)=wA\\v_(max)=4.4*5.5\\v_(max)=24.2m/s`