a particle moves along a line....?

Question

at any give time t's position is given by f(t) = 4πt + sin(4πt)
1. find the velocity
2. find the acceleration
3. what are all of the values from 0 to 3 for which the particle is at rest?
4. what is the maximum velocity?

Answer 1

velocity = df(t)/dt = 4π + 4πcos(4πt)
acceleration = dv(t)/dt = -16π^2sin(4πt)

Particle is at rest when velocity = 0, or 4π+4πcos(4πt) = 0
1+cos(4πt) = 0
cos(4πt) = -1 The values for which cosø = -1 are ø=n(odd)π, so when
4πt=n(odd)*π, t = 1/4*n(odd); thus for t = 1/4, 3/4, 5/4, 7/4, 9/4, 11/4 the velocity is zero

The maximum velocity occurs when cosø = 1, so the max velocity is 8π

Answer 2

1. is easy v = dv/dt= 4pi + 4pi cos4pit

2. Again a = dv/dt = -16pi^2sin4pit

3. V = 0 impkies 4pi = -4picos4pit, when 1 = 1cos4pit

when 4pit = cos-1(1), when t = (1/4pi) cos-1(1)..you should choose the plus signus.
4. Vmax when cos4pit=1

Answer 3

velocity=df/dt=4π + 4πcos(4πt)
a=d^f/dt^2=0-16π^2sin(4πt)
v=0=>cos(4πt)=-1=>t=t/4,3t/4 etc
vmax=4π(1+cos(4πt))=8π
when cos(4πt)=1