#### To determine

**To find:** The velocity at time *t*.

#### Answer

The velocity at time is
f′(t)=0.04t3−0.12t2 ft/sec.

#### Explanation

**Given:**

The given equation is as below.

f(t)=0.01t4−0.04t3 (1)

**Calculation:**

Calculate the velocity at time
t.

Differentiate the equation (1) with respect to time.

v(t)=f′(t)=(4)0.01t3−(3)0.04t2=0.04t3−0.12t2

Therefore, the velocity at time
t is
f′(t)=0.04t3−0.12t2 ft/sec.

#### To determine

**To find:** The velocity after 1 second.

#### Answer

The velocity after 1 second is
v(1)=0.3678 ft/sec_.

#### Explanation

Calculate the velocity after 1 second.

Substitute 1 for
t in the equation (2).

f′(t)=0.04t3−0.12t2f′(1)=0.04(1)3−0.12(1)2ft/sec=−0.08ft/s

Therefore, the velocity after 1 second is
v(1)=−0.08 ft/sec.

#### To determine

**To find:** The time when particle at rest.

#### Answer

The particle is at rest when
t=0sec and
t=3sec.

#### Explanation

Calculate the time when particle will be at rest.

The velocity will be zero, when the particle is at rest.

Substitute 0 for
v(t) in the equation (2).

v(t)=0.04t3−0.12t20=t2(0.04t−0.12)t=0 sec

When the second factor is equated to zero then,

0.04t−0.12=00.04t=0.12t=0.120.04t=3sec

Therefore, the particle at rest when
t=0sec and
t=3sec.

#### To determine

**To find:** The particle moving in the positive direction.

#### Answer

The particle will be moving in the positive direction within the time limits are
t>3 sec.

#### Explanation

Calculate the time during which the particle will be moving in the positive direction.

If the particle moves in positive direction, the velocity at any time *t* will be greater than zero which is
v(t)=0.04t3−0.12t2>0.

There arises three cases as
t<0; 0<t<3; t>3.

When
t<0 the velocity becomes,

v(−1)=0.04(−1)3−0.12(−1)2=−0.04−0.12=−0.16<0

When
0<t<3 the velocity becomes,

v(1)=0.04(1)3−0.12(1)2=0.04−0.12=−0.08<0

When
t>3 the velocity becomes,

v(4)=0.04(4)3−0.12(4)2=2.56−1.92=0.64>0

The particle is moving in the negative direction for the intervals
t<0; 0<t<3 and is in the positive direction for the time intervals
t>3 sec.

Therefore, the particle will be moving in the positive direction within the time limits are
t>3 sec.

#### To determine

**To find:** The total distance traveled during the first 6 seconds.

#### Answer

The total distance travelled during first 6 second is
f(6)=4.86 ft .

#### Explanation

Calculate the total distance traveled during first 6 seconds.

Since the particle is at rest at time
t=0,3, it is required to find the position function at points where *t* takes the values 0, 3 and 6.

Substitute 0 for
t in the equation (1).

f(t)=0.01t4−0.04t3f(0)=0.01(0)4−0.04(0)3f(0)=0

Substitute 3 for
t in the equation (1).

f(t)=0.01t4−0.04t3f(3)=0.01(3)4−0.04(3)3f(3)=−0.27

The change from the time interval
t=0 to
t=3 is
|f(3)−f(0)|=|−0.27−0|=0.27.

Substitute 6 for
t in the equation (1).

f(t)=0.01t4−0.04t3f(6)=0.01(6)4−0.04(6)3f(6)=4.32

The change from the time interval
t=3 to
t=6 is
|f(6)−f(3)|=|4.32−(−0.27)|=4.59.

Hence, the total distance travelled can be calculated as,

|f(3)−f(0)|+|f(6)−f(3)|=0.27+4.59=4.86

Therefore, total distance travelled during first 6 seconds is
f(6)=4.86 ft .

#### To determine

**To draw:** The diagram to illustrate the motion of the particle.

#### Explanation

Show the diagram to illustrate the motion of the particle as shown below in figure (1).

From the above Figure 1, it can be notices that
t=0 and
t=3 are the points where the particles is at rest.

#### To determine

**To find:** The acceleration at time *t* and after 1 second.

#### Answer

The acceleration at time is
0.12t2−0.24t and after one second is
−0.12 ft/s2.

#### Explanation

Calculate the acceleration at time *t.*

Differentiate the equation (2) with respect to *t.*

f″(t)=(12)0.01t2−(6)0.04tf″(t)=0.12t2−0.24t

Therefore, the acceleration at time is
0.12t2−0.24t.

Calculate the acceleration after 1 second.

Substitute 1 for
t in the above equation.

f″(t)=0.12t2−0.24tf″(1)=0.12(1)2−0.24(1)=0.12−0.24=−0.12

f″(1)=−0.12 ft/s2

Therefore, the acceleration after 1 second is
−0.12 ft/s2.

#### To determine

**To sketch:** The graph of the position, velocity, and acceleration function for
0≤t≤6.

#### Explanation

The position function of the particle is
f(t)=0.01t4−0.04t3, the velocity function is
f′(t)=0.04t3−0.12t2 and the acceleration function is
f″(t)=0.12t2−0.24t.

Use online graphing calculator and sketch the graph of the position, velocity and acceleration function as shown below in Figure 2.

From the Figure 2, observe the position function
f(t)=0.01t4−0.04t3, velocity function
f′(t)=0.04t3−0.12t2 and acceleration function
f″(t)=0.12t2−0.24t of the particle.

#### To determine

**To find:** The time when the particle is speeding up and slowing down.

#### Answer

The particle is speeding up in the time interval
0≤t<2 and for
t>3.

The particle slows down in the time interval
2<t<3.

#### Explanation

Calculate the time when particle is speeding up and slowing down.

Substitute 0 for
f″(t) in the equation
f″(t)=0.12t2−0.24t.

0=0.12t2−0.24t0=t(0.12t−0.24)t=0; t=2

From the graph of part (h), it can be observed that the acceleration and velocity function have the same sign in the interval
0≤t<2.

The acceleration and velocity function have the same sign in the interval
t>3.

The particle speeds up when the acceleration function and the velocity function have the same sign and hence the particle speeds up in the intervals
0≤t<2 and
t>3.

Notice that the acceleration function and the velocity function have different signs in the interval
2<t<3.

The particle slows down when the acceleration function and the velocity function have the different sign and hence the particle speeds up in the interval
2<t<3.