Calculus Question?

Question

A particle moves along a line so that at any time t, its position is given by x(t) = 2pi + cos 2pi t

1. what is the velocity?
2. what is the acceleration?
3. What are the values of t is the particle at rest?
4. what is the maximum velocity?

Answer 1

1. what is the velocity?
The first derivative of x(t) is the velocity:
x'(t) = -2pisin(2pi t)

2. what is the acceleration?
The second derivative of x(t) is the acceleration:
x''(t) = -(2pi)^2cos(2pi t)

3. What are the values of t is the particle at rest?
When at rest, the velocity = 0.
Thus, x'(t) = 0, => sin(2pi t) = 0, => t = n/2, where n can be any positive integer.

4. what is the maximum velocity? x'(t) max = 2pi
When at the maximum velocity, sin(2pit) = 1.
Thus, x'(t) max = 2pi

Answer 2

1. Taking the derivative gives -2(pi)(sin(2pit)).

2. Taking the derivative again gives -4(pi)^2(cost(2pit)).

3. Setting t=0 gives x(t) = 2pi-1.

4. From 1, above we have the derivative. It's value is maximized when sin(2pit) is -11, giving 2pi.

Answer 3

x(t) = 2π + cos (2π t)

Velocity is the derivative of position with respect to time...

v(t) = -2π sin ( 2π t )

Acceleration is the derivative of velocity with respect to time...

a(t) = -4π² cos (2π t)

The particle is at rest when v(t) = 0.
The particle has maximum velocity when sin ( 2π t ) is equal to one.