Answer
a) $v(t)=-2\sin t+3\cos t$
$a(t)=-2\cos t-3\sin t$
b) See image
c) $t=2.55359$
d) $3.606$
e) $t=-0.588+2\pi k$
$t=2.55+2\pi k$
Work Step by Step
a) The velocity is change in distance per time, so we can find the velocity equation by taking the derivative of the motion equation.
$v(t)=s'(t)=-2\sin t+3\cos t$
Acceleration is the change in speed per time, so we can find the acceleration equation by taking the derivative of the velocity equation.
$a(t)=v'(t)=-2\cos t-3\sin t$
b) See image
c) When the mass is pulled downward, it gains elastic potential energy. Here, the mass will have maximum acceleration. When it passes through the equilibrium, it will have no elastic potential energy, since it will all be converted to kinetic energy. Therefore, the velocity will be maximum at equilibrium. However, acceleration will be 0 because the mass is at equilibrium position.
$a(t)=0$
From the graph, we can see this value is at position $t=2.55359$
d) The position, velocity, and acceleration functions are all the same function shifted horizontally. Therefore, the distance the mass travels from the equilibrium position can be found by finding the amplitude of the velocity and acceleration functions, which is $3.606$
e) The speed is greatest for
$t=-0.588+2\pi k$
$t=2.55+2\pi k$
For any integer k.
