Question

1. (6pts total) for the position function

$x(t) = a\cos(\omega t)$
$a(t) = -a\omega^2 \cos \omega$
calculate the velocity $v(t)$ and acceleration $a(t)$ at $t = 2\pi$ s for the
constants $a = 1$ m and $\omega = 2$ rad/s.

Explanation:

Step1: Find Velocity Function

Velocity is the derivative of position. For \( x(t) = A\cos(\omega t) \), using the chain rule, \( v(t)=\frac{dx}{dt}=-A\omega\sin(\omega t) \).

Step2: Substitute Values for Velocity

Given \( A = 1\space m \), \( \omega = 2\space rad/s \), \( t = 2\pi\space s \). Substitute into \( v(t) \):
\( v(2\pi)=-1\times2\times\sin(2\times2\pi)= - 2\sin(4\pi) \). Since \( \sin(4\pi)=0 \), \( v(2\pi)=0\space m/s \).

Step3: Find Acceleration Function

Acceleration is the derivative of velocity. For \( v(t)=-A\omega\sin(\omega t) \), derivative is \( a(t)=\frac{dv}{dt}=-A\omega^{2}\cos(\omega t) \).

Step4: Substitute Values for Acceleration

Substitute \( A = 1 \), \( \omega = 2 \), \( t = 2\pi \) into \( a(t) \):
\( a(2\pi)=-1\times(2)^{2}\times\cos(2\times2\pi)= - 4\cos(4\pi) \). Since \( \cos(4\pi)=1 \), \( a(2\pi)=-4\space m/s^{2} \).

Answer:

Velocity \( v(2\pi) = \boldsymbol{0\space m/s} \), Acceleration \( a(2\pi) = \boldsymbol{-4\space m/s^{2}} \)