#### To determine

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

#### Answer

The velocity of the particle is v(t)=4t3−6t2+2t−1 m/s.

The acceleration of the particle is a(t)=12t2−12t+2 m/s2.

#### Explanation

**Given:**

The equation of motion of a particle is s=t4−2t3+t2−t, where *s* is in meters and *t* is in seconds.

**Derivative rules:**

(1) Constant Function: ddx(c)=0

(2) Constant Multiple Rule: ddx[c⋅f(x)]=c⋅ddxf(x)

(3) Power Rule: ddx(xn)=nxn−1

(4) Sum Rule: ddx[f(x)+g(x)]=ddx(f(x))+ddx(g(x))

(5) Difference Rule: ddx[f(x)−g(x)]=ddx(f(x))−ddx(g(x))

**Recall:**

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

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

**Calculation:**

Obtain the velocity of the particle.

v=ddt(s) =ddt(t4−2t3+t2−t)

Apply the sum rule (4) and the difference rule (5),

v=ddt(t4)−ddt(2t3)+ddt(t2)−ddt(t)

Apply the constant multiple rule (2) and the power rule (3),

v=ddt(t4)−2ddt(t3)+ddt(t2)−ddt(t)=(4t4−1)−2(3t3−1)+(2t2−1)−(1t1−1)=(4t3)−2(3t2)+(2t1)−(1t0)=4t3−6t2+2t−1

Therefore, the velocity of the particle is v(t)=4t3−6t2+2t−1 m/s.

Obtain the acceleration of the particle.

a=ddt(v)=ddt(4t3−6t2+2t−1)

Apply the sum rule (4) and difference rule (5),

a=ddt(4t3)−ddt(6t2)+ddt(2t)−ddt(1)

Apply the derivative rules (1), (2) and (3),

a=4ddt(t3)−6ddt(t2)+2ddt(t)−ddt(1)=4(3t3−1)−6(2t2−1)+2(1t1−1)−0=4(3t2)−6(2t1)+2(1t0)=12t2−12t+2

Therefore, the acceleration of the particle is a(t)=12t2−12t+2 m/s2.

#### To determine

**To find:** The acceleration of the particle after 1 second.

#### Answer

The acceleration of the particle after 1 second is 2 m/s2

#### Explanation

**Given:**

The acceleration of the particle is a(t)=12t2−12t+2 . (from part (a))

**Calculation:**

Obtain the acceleration of the particle after 1 second.

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

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

a(1)=12(1)−12+2=2

Therefore, the acceleration of the particle after 1 seconds is 2 m/s2.

#### To determine

**To sketch:** The graph the position, velocity and the acceleration of the functions.

#### Explanation

Use the online graphing calculator to draw the graph of the position, velocity, and the acceleration functions as shown in Figure 1.

From Figure 1, it is observed that v(t)=4t3−6t2+2t−1 m/s is the derivative of the function s(t)=t4−2t3+t2−t and a(t)=12t2−12t+2 m/s2 is the derivative of the function v(t)=4t3−6t2+2t−1 m/s.