#### To determine

**To find:** The velocity and acceleration as functions of *t*.

#### Answer

The velocity of the particle is v(t)=3t2−3 m/s.

The acceleration of the particle is a(t)=6t m/s2.

#### Explanation

**Given:**

The equation of motion of a particle is s=t3−3t .

Here, *s* is a meters and *t* is a seconds.

**Derivative rules:**

(1) Derivative of Constant Function: ddt(c)=0

(2) Constant Multiple Rule**:** ddt[c⋅f(t)]=c⋅ddtf(t)

(3) Power Rule: ddt(tn)=ntn−1

(4) Sum Rule: ddt[f(t)+g(t)]=ddt(f(t))+ddt(g(t))

(5) Difference Rule: ddt[f(t)−g(t)]=ddt(f(t))−ddt(g(t))

**Recall:**

If *s* is a displacement of a particle and *t* is in seconds, then the velocity of the particle is v=dsdt.

If *v* is a velocity of the particle and *t* is in seconds, then the acceleration of the particle is

a=dvdt.

**Calculation:**

Obtain the velocity of the particle.

v=ddt(s) =ddt(t3−3t)

Apply the difference rule (5) and the constant multiple rule (2),

v=ddt(t3)−ddt(3t)=ddt(t3)−3ddt(t)

Use power rule (3) and simplify the terms.

v=(3t3−1)−3(1t1−1) =3t2−3

Therefore, the velocity of the particle is v(t)=3t2−3 m/s .

Obtain the acceleration of the particle.

a=ddt(v)=ddt(3t2−3)

Apply the difference rule (5),

a=ddt(3t2)−ddt(3)

Apply the derivative rule (1) and (2),

a=ddt(3t2)−0=ddt(3t2)=3⋅ddt(t2)

Apply power rule (3) and to obtain the acceleration.

a=3⋅(2t2−1)=3⋅(2t1)=6t

Therefore, the acceleration of the particle is a(t)=6t m/s2 .

#### To determine

**To find:** The acceleration of particle after 2 seconds.

#### Answer

The acceleration of the particle after 2 seconds is 12 m/s2_.

#### Explanation

**Given:**

The acceleration of the particle is a(t)=6t m/s2. (From part(a))

**Calculation:**

Obtain the acceleration of the particle after 2 seconds.

That is, to find the value of a(2) .

Substitute 2 for *t* in a(t)=6t,

a(2)=6(2)=12

Therefore, the acceleration of the particle after 2 seconds is 12 m/s2_.

#### To determine

**To find:** The acceleration of the particle when velocity is zero.

#### Answer

The acceleration of the particle is 6 m/s2_ when velocity is zero.

#### Explanation

From part (a), the velocity of the particle is v(t) is 3t2−3 m/s and the acceleration of the particle is a(t) is 6t m/s2.

**Calculation:**

Obtain the time variable *t* when the velocity is zero,

That is, to solve the equation 3t2−3=0 for *t*.

3(t2−1)=0t2−1=0t2=1t=±1

Here, t=−1 an absurd, since time cannot be negative.

Thus, the value of t=1.

Therefore, the velocity is zero when time is one second.

Obtain the acceleration when time t=1.

That is, to find the value of a(1).

Substitute 1 for *t* in a(t)=6t,

a(1)=6(1)=6

Therefore, the acceleration is 6 m/s2_ when velocity is zero.