#### To determine

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

#### Answer

The velocity at time is
3t2−18t+24 ft/sec.

#### Explanation

**Given:**

The given equation is as below.

s=f(t)=t3−9t2+24t (1)

**Calculation:**

Calculate the velocity at time
t.

Differentiate the equation (1) with respect to time.

v(t)=f'(t)=3t2−18t+24 (2)

Therefore, the velocity at time
t is
3t2−18t+24 ft/sec.

#### To determine

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

#### Answer

The velocity after 1 second is
v(1)=9ft/sec.

#### Explanation

Calculate the velocity after 1 second.

Substitute 1 for
t in the equation (2).

v(t)=f'(t)=3t2−18t+24v(1)=f'(1)=3(1)2−18(1)+24v(1)=9 ft/sec

Therefore, the velocity after 1 second is
v(1)=9ft/sec.

#### To determine

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

#### Answer

The particle is at rest at
t=2,4.

#### 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)=3t2−18t+240=3t2−18t+24t=18t±(18)2−4(3)(24)2(3)=18±366

The above equation can be simplified as
t=18+366 and
t=18−366.

Hence the particle at rest can be found out as,

t=18±366=18±66=246,126=4,2

Therefore, the particle is at rest at time
t=2,4.

#### To determine

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

#### Answer

The particle will move in positive direction in the time interval
t>4 or
0≤t<2.

#### Explanation

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

If speed is positive, the particle moves in positive direction whereas the speed is negative, the particle moves in negative direction.

v(t)=3t2−18t+24=3(t−2)(t−4)

The velocity is positive when
v(t)>0 and hence the cases wherein the particles moves in positive direction are
3(t−2)(t−4)>0 and this is possible when both the factors are positive or both the factors are negative

The factors are positive when
t>4 and both are negative
t<2.

Hence, the particle will move in positive direction in the time interval
t>4 or
0≤t<2.

#### To determine

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

#### Answer

The total distance travelled during first 6 seconds is
f(6)=44 ft .

#### Explanation

Calculate the total distance traveled during first 6 seconds.

Since the particle is at rest at time
t=2,4, it is required to find the position function at points where *t* takes the values 0, 2, 4 and 6.

Substitute 0 for
t in the equation (1).

f(t)=t3−8t2+24tf(0)=(03)−9(0)2+24(0)f(0)=0

Substitute 2 for
t in the equation (1).

f(t)=t3−8t2+24tf(2)=(23)−9(2)2+24(2)f(2)=20

The change from the time interval
t=0 to
t=2 is
|f(2)−f(0)|=20.

Substitute 4 for
t in the equation (1).

f(t)=t3−8t2+24tf(4)=(43)−9(4)2+24(4)f(4)=14

The change from the time interval
t=2 to
t=4 is
|f(4)−f(2)|=|16−20|=4.

Substitute 6 for
t in the equation (1).

f(t)=t3−9t2+24tf(6)=(63)−9(6)2+24(6)f(6)=36

The change from the time interval
t=4 to
t=6 is
|f(6)−f(4)|=|36−16|=20.

Hence, the total distance travelled can be calculated as,

|f(2)−f(0)|+|f(4)−f(2)|+|f(6)−f(4)|=20+4+20=44

Therefore, total distance travelled during first 6 seconds is
f(6)=44 ft .

#### To determine

**To find:** 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, observe that the particle is rest at the time when *t* equals to the value 2 and 4.

#### To determine

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

#### Answer

The acceleration at time is
f"(t)=6t−18 and after one second is
−12 ft/s2.

#### Explanation

Calculate the acceleration at time *t.*

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

f'(t)=3t2−18t+24f"(t)=6t−18

Therefore, the acceleration at time is
6t−18.

Calculate the acceleration after 1 second.

Substitute 1 for
t in the above equation.

f"(t)=6t−18f"(1)=6(1)−18=−12fts2

Therefore, the acceleration after 1 second is
−12 ft/s2.

#### To determine

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

#### Explanation

The position function of the particle is
f(t)=t3−9t2+24t, the velocity of the particle is
v(t)=3t2−18t+24 and acceleration is given by
a(t)=6t−18.

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

From the above Figure 2, observe the position, velocity and acceleration function in the same graph.

#### To determine

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

#### Answer

The time when the particle is speeding up is when time *t* is
2<t<3 and 4<t≤6 and the time when the particle is slowing down is when time is
0≤t<2 and 3<t<4.

#### Explanation

From the graph of part (h) it is clear that the particle has same sign for velocity and acceleration in the two intervals
2<t<3 and
4<t≤6 respectively.

Hence, the particle speeds up in the interval
2<t<3 and
4<t≤6.

Similarly from the graph the particle has different sings for the velocity and acceleration in the two intervals
0≤t<2 and 3<t<4.

Hence, the particle slows up in the interval
0≤t<2 and 3<t<4.