Calculus ( A Bit oh Physics)?

Question

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

a) find the velocity at time t

b) find the acceleration at time t

c) what are the value of t, 0 ≤ t ≤ 3, for which the particle is at rest?

d? what is the maximum velocity?

Answer 1

All math, really - the "particle" word probably threw you!

I'm not SURE the equation is phrased correctly, so I'm going to take a guess:
x(t) = 2*pi*t + cos(2*pi*t)

Then the velocity is the FIRST derivative,
x'(t) = 2*pi + sin(2*pi*t)*2*pi (use the chain rule on the second part)

The acceleration is the second derivative,
x''(t) = 2*pi*(-cos(2*pi*t))*2*pi (the first term drops out and the second uses the chain rule again)

The particle is at rest when the first derivative = 0, which happens when 0 = 2*pi + sin(2*pi*t)*2*pi
or
sin(2*pi*t)*2*pi = -2*pi
or
sin(2*pi*t) = -1
Sin(anything) = -1 when anything is -pi/2 plus or minus any multiple of 2*pi
Since the question does not allow you to provide a negative answer, let's look at the smallest positive answer:
2*pi*t = -pi/2+2*pi = pi*1.5
then 2*t = 1.5, and t = .75
Now, what if we add another 2*pi?
then 2*pi*t = -pi/2 + 4*pi = pi*3.5
Then 2*t = 3.5, t = 1.75
Clearly, we can also fit t = 2.75 into this as well, so the answer to c is .75. 1.75, and 2.75

For 4, it is easiest to assume that at some point sin(2*pi*t) will be 1 and this will be the
x'(t) (max) = 2*pi + 2*pi = 4*pi