Answer :
The vertical component of acceleration of the suspension system at [tex]t = 325\,s[/tex] is [tex]-2.157\times 10^{-133}[/tex] meters per square second.
Determination of the vertical component of acceleration of a suspension system
First, we obtain the function acceleration ([tex]a(t)[/tex]), in meters per square second, by deriving the function position ([tex]y(t)[/tex]), in meters, twice:
Position
[tex]y(t) = 0.75\cdot e^{-0.95\cdot t}\cdot \cos 6.3t[/tex] (1)
Velocity
[tex]v(t) = 0.75\cdot [-0.95\cdot e^{-0.95\cdot t}\cdot \cos 6.3t-6.3\cdot e^{-0.95\cdot t}\cdot \sin 6.3t ][/tex]
[tex]v(t) = (-0.713\cdot \cos 6.3t -4.725\cdot \sin 6.3t)\cdot e^{-0.95\cdot t}[/tex] (2)
Acceleration
[tex]a(t) = (4.492\cdot \sin 6.3t-29.768\cdot \cos 6.3t )\cdot e^{-0.95\cdot t} -0.95\cdot (-0.713\cdot \cos 6.3t - 4.725\cdot \sin 6.3t)\cdot e^{-0.95\cdot t}[/tex]
[tex]a(t) = (8.981\cdot \sin 6.3t -29.091\cdot \cos 6.3t) \cdot e^{-0.95\cdot t}[/tex] (3)
Now we evaluate the function acceleration for [tex]t = 325\,s[/tex]:
[tex]a(325) = [8.981\cdot \sin (6.3\cdot 325) - 29.091\cdot \cos (6.3\cdot 325)]\cdot e^{-0.95\cdot (325)}[/tex]
[tex]a(325) = -26.437\cdot e^{-308.75}[/tex]
[tex]a(325) = -26.437\cdot (8.158\times 10^{-135})[/tex]
[tex]a(325) = -(2.644\times 10^{1})\cdot (8.158\times 10^{-135})[/tex]
[tex]a(325) = -21.570\times 10^{-134}[/tex]
[tex]a(325) = -2.157\times 10^{-133}[/tex]
The vertical component of acceleration of the suspension system at [tex]t = 325\,s[/tex] is [tex]-2.157\times 10^{-133}[/tex] meters per square second. [tex]\blacksquare[/tex]
Remark
The statement is incomplete and incorrectly formatted, correct form is presented below:
A design team for an electric car company finds that under some conditions the suspension system of the car performs in a way that produces unsatisfactory bouncing of the car [tex]y[/tex] as a function of time [tex]t[/tex] under these conditions, they find that it is described by the relationship: [tex]y(t) = y_{o}\cdot e^{-\alpha\cdot t}\cdot \cos \omega t[/tex], where [tex]y_{o} = 0.75\,m[/tex], [tex]\alpha = 0.95\,s^{-1}[/tex], [tex]\omega = 6.3\,s^{-1}[/tex]. In order to find the vertical velocity of the car as a function of time we will need to evaluate the derivative of the vertical position with respect to time, [tex]\frac{dy}{dt}[/tex]. For this trajectory, what would the vertical component of acceleration for the module be at time [tex]t = 325\,s[/tex]? Recall that acceleration is the derivative of velocity with respect to time.
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