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

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.$v(t)==24.1m/s$

b.$a(t)=3.79m/s^{2}$

c.$a_{max}=106.48m/s^{2}$

d.$v_{max}=24.2m/s$

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

$v_{max}=wA\\v_{max}=4.4*5.5\\v_{max}=24.2m/s$