If the position of a particle is given by x = 35t - 7t^3, where x is in meters and t is in seconds, answer the following questions.

(a) When, if ever, is the particle's velocity zero? (Enter the value of t or 'never'.)
(b) When is its acceleration a zero?

I wonder if you have had calculus...

velocity=dx/dt=35-21t^2.
so velocity is zero when t^2=35/32 solve for t.

acceleration= dv/dt=-42t so acceleation is zero only when t=0

To find the answers to these questions, we need to differentiate the equation for position with respect to time.

(a) To find the velocity, we differentiate x with respect to t:
v = dx/dt = d(35t - 7t^3)/dt

Differentiating each term separately, we get:
v = d(35t)/dt - d(7t^3)/dt
v = 35 - 21t^2

Now, we need to find when the velocity is zero by setting v = 0:
35 - 21t^2 = 0

Simplifying the equation, we have:
21t^2 = 35

Dividing by 21, we get:
t^2 = 35/21

Taking the square root of both sides, we obtain:
t = ±√(35/21)

So, the particle's velocity is zero at t = ±√(35/21).

(b) To find the acceleration, we differentiate the velocity equation with respect to time:
a = dv/dt = d(35 - 21t^2)/dt

Differentiating each term separately, we get:
a = d(35)/dt - d(21t^2)/dt
a = 0 - 42t
a = -42t

To find when the acceleration is zero, we set a = 0:
-42t = 0

This equation is satisfied when t = 0. Therefore, the particle's acceleration is zero at t = 0.