#### To determine

**(a)**

**To find: **The velocity and acceleration function

#### Answer

The velocity = 3t2-12 and the acceleration = 6t

#### Explanation

**Concept:**

By differentiating position function obtain velocity function and further by differentiating velocity function obtain acceleration function

**Formula:**

i.

velocity=v=dydt

ii.

Acceleration=a=dvdt

iii. Power rule:

ddxxn=nxn-1

iv. Sum rule:

ddxfx+gx=ddxfx+ddx(gx)

v. Constant multiple rule:

ddxCf(x)=Cddx(fx)

vi. Difference rule:

ddxfx-gx=ddxfx-ddx(gx)

vii. Constant function rule:

ddxC=0

**Given:**

Position of particle =y=t3-12t+3, t≥0

**Calculation:**

velocity=v=dydt=ddt(t3-12t+3)

By using sum and difference rule,

v=ddt(t3)-ddt(12t)+ddt(3)

By using power rule, constant multiple rule and constant rule,

v=3t2-12ddt(t)+0

v=3t2-12

Now,

Acceleration=a=dvdt=ddt(3t2-12)

By using difference rule,

a=ddt(3t2)-ddt(12)

By using constant function and constant multiple rule,

a=3ddt(t2)-0

By using power rule,

a=6t

**Final statement:**

Velocity = 3t2-12 and Acceleration = 6t

#### To determine

**(b)**

**To find:** When the particle is moving upward and moving downward

#### Answer

Particle is moving in upward direction in the interval (2, ∞)

And moving downward in the interval [0, 2)

#### Explanation

**Concept:**

Find value of t when velocity is positive and velocity is negative. When velocity is positive particle shall move upward and when negative particle shall move downward

**Given:**

v=3t2-12

**Calculation:**

The particle is moving in upward direction when the velocity is positive

That is, v=3t2-12≥0⇒3t2≥12⇒t2≥4⇒t>2 or t< -2

But since, t≥0

We get,

t>2

Therefore, particle is moving in upward direction in the interval (2, ∞)

And moving downward in the interval [0, 2)

**Final statement:**

Particle is moving in upward direction in the interval (2, ∞)

And moving downward in the interval [0, 2)

#### To determine

**(c)**

**To find:** The distance of particle in the time interval 0≤t≤3

#### Answer

The distance of particle in the interval 0≤t≤3 = 23

#### Explanation

By using distance formula and critical points calculate the distance in the given interval

**Formula: **

Distance of particle in the time interval a≤t≤b

|yb-ya|

**Given:**

Position of particle =y=t3-12t+3, t≥0

Velocity of particle = v=3t2-12

**Calculation:**

Consider the velocity function,

v=3t2-12

Equate it with 0 to find critical point

v=3t2-12=0⇒3t2=12⇒t2=4⇒t=2 or t -2

But since, t≥0

Therefore, t=2

The distance of particle in the interval 0≤t≤3 is,

y2-y0+|y3-y2|

First calculate y0, y2 and y(3)

Position of particle =y=t3-12t+3

For t = 0, y0=0-0+3=3

For t = 2, y2=8-24+3=-13

For t = 3, y3=27-36+3=-6

Therefore,

y2-y0+y3-y2=-13-3+-6+13=16+7=23

**Final statement:**

The distance of particle in the interval 0≤t≤3 = 23

#### To determine

**(d)**

**To sketch:** Graph of position, velocity and acceleration function for time interval 0≤t≤3

#### Answer

The graph of position, velocity and acceleration function for time interval 0≤t≤3 is,

#### Explanation

By using equations for position, velocity and acceleration functions plot a graph

**Given:**

Position of particle =y=t3-12t+3

Velocity of particle = v=3t2-12

Acceleration of particle = a=6t

**Graph:**

Here red is position curve, blue is velocity and green is acceleration **Conclusion:**

The graph of position, velocity and acceleration function for time interval 0≤t≤3 is,

#### To determine

**(e)**

**To find:** When is the particle speeding up and speeding down

#### Answer

Particle is speeding up in the interval (2, 3) and speeding down in the interval (0, 2)

#### Explanation

Find when the velocity and acceleration have the same and opposite signs Since, when velocity and acceleration have the same sign then particle is speeding up and when velocity and acceleration have the opposite sign then particle is speeding down

**Given:**

Velocity of particle = v=3t2-12

Acceleration of particle = a=6t

**Calculation:**

The particle is speeding up when velocity and acceleration have the same sign

And speeding down when velocity and acceleration have the opposite sign

**Graph**:

From the graph,

velocity and acceleration have the same sign in the interval (2, 3)

velocity and acceleration have the opposite sign in the interval (0, 2)

Therefore, Particle is speeding up in the interval (2, 3) and speeding down in the interval (0, 2)

**Final statement:**

Particle is speeding up in the interval (2, 3) and speeding down in the interval (0, 2).