Let s(t)=6t^3−18t^2−54t be the equation of motion for a particle. Find a function for the velocity.
v(t)=

Where does the velocity equal zero? [Hint: factor out the GCF.]
t= and t=

Find a function for the acceleration of the particle.
a(t)=

The function for the velocity is .

The velocity is zero at t = 3 and t = -1

The function for acceleration is .

a) To find the function for velocity we'll find the first derivative of the distance function s(t).

b) We set the velocity equal to 0 and solve for t.

We set each factor equal to 0 and solve for t.

and

c) To find acceleration we differentiate v(t).

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Velocity:

The velocity equals 0 at

Acceleration:

Step-by-step explanation:

The velocity is the derivative of the position

The acceleration if the derivative of the velocity

The position is given by the following equation

So the equation for the velocity is:

Where does the velocity equal zero? [Hint: factor out the GCF.]

This is when .

So

The greatest common factor between 18,36 and 54 is 18. So, factoring by 18

How we find the roots of a quadratic function?

Given a second order polynomial expressed by the following equation:

.

This polynomial has roots such that , given by the following formulas:

So

We do not have a negative instant of time, so the velocity equals 0 at

Find a function for the acceleration of the particle.

if im undersanding the question correctly, its organized from 10

step-by-step explanation:

hope it

step-by-step explanation:

x^(1/6) · x^(1/6) · x^(1/6)   ·x^(1/6) = x^(4/6)   since we add the exponents

4/6 can be simplified to 2/3

the simplified form is x^(2/3)