#### To determine

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

#### Answer

The velocity at time *t* is f'(t)=−9(t2−9)(t2+9)2 ft/sec_ .

#### Explanation

**Given:**

The given equation is as below.

s=f(t)=9tt2+9 (1)

**Calculation:**

Calculate the velocity at time t.

Differentiate the equation (1) with respect to time by applying uv method.

f'(t)=(t2+9)(9)−(9t×(2t))(t2+9)2=9t2+81−18t2(t2+9)2=−9t2+81(t2+9)2

f'(t)=−9(t2−9)(t2+9)2 (2)

Therefore, the velocity at time *t* is f'(t)=−9(t2−9)(t2+9)2ft/sec_.

#### To determine

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

#### Answer

The velocity after 1 second is v(1)=0.72 ft/sec_.

#### Explanation

Calculate the velocity after 1 second.

Substitute 1 for t in the equation (2).

v(t)=f'(t)=−9(t2−9)(t2+9)2v(1)=−9(1−9)(1+9)2=0.72 ft/sec

Therefore, the velocity after 1 second is v(1)=0.72 ft/sec_.

#### To determine

To find: The time when particle at rest.

#### Answer

The time when particle at rest is t=3sec_.

#### Explanation

Calculate the time when particle at rest.

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

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

v(t)=−9(t2−9)(t2+9)20=−9(t2−9)(t2+9)2−9(t2−9)=0t2−9=0t=3sec

Therefore, the time when particle at rest is t=3sec_.

#### To determine

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

#### Answer

The particle always moves in positive direction when time is within the range 0≤t<3_.

#### 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.

v(t)=−9(t2−9)(t2+9)2>0−9(t2−9)(t2+9)2>0−9(t2−9)>0t2−9>0t<3

Therefore, the particle moving in the positive direction when 0≤t<3_.

#### To determine

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

#### Answer

The total distance travelled during first 6 second is 1.8ft_.

#### Explanation

Calculate the total distance traveled during first 6 seconds.

Here, the particle moves in positive and negative direction, the total distance traveled should be calculated between the intervals of [0,3] and [3,6].

Substitute 3 and 0 for t between interval of [0,3] in the equation (1) and subtract them.

|f(3)−f(0)|=|9(3)(3)2+9−9(0)(0)2+9|=|32−0|=32

Substitute 6 and 3 for *t* between intervals of [3,6] for t in the equation (1) and subtract them.

|f(6)−f(3)|=|9(6)(6)2+9−9(3)(3)2+9|=|65−32|=310

The total distance travelled is as given below.

f(6)=32+310=1.8ft

Therefore, the total distance travelled during first 6 second is f(6)=1.8ft_.

#### 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.

Figure 1 shows the movement of particle at different times.

#### To determine

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

#### Answer

The acceleration at time *t* is f"(t)=18t(t2−27)(t3+9)3_ and after one second is −0.468 ft/s2_.

#### Explanation

Calculate the acceleration at time *t.*

Differentiate the equation (2) with respect to *t* by applying uv method.

f'(t)=−9(t2−9)(t2+9)2f"(t)=−9(t2+9)2(2t)−[(t2−9)(2(t2+9)×2t)][(t2+9)2]2=−92t(t2+9)[(t2+9)−2(t2−9)](t2+9)4

On further simplification,

f"(t)=−18t[t2+9−2t2+18](t2+9)3=−18t(−t2+27)(t2+9)3=18t(t2−27)(t3+9)3

Therefore, the acceleration at time is f"(t)=18t(t2−27)(t3+9)3_.

Calculate the acceleration after 1 second.

Substitute 1 for t in the above equation.

f"(t)=18t(t2−27)(t3+9)3f"(1)=18(1)((1)2−27)(13+9)3=−0.468fts2

Therefore, the acceleration after 1 second is f"(1)=−0.468fts2_.

#### To determine

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

#### Explanation

Calculate the position of the particle with respect to time using the expression.

s(t)=9tt2+9

Substitute 0 for t in the above equation.

s(0)=9(0)02+9=0

Similarly, calculate the remaining values.

Calculate the value of t and s(t) as shown in the table (1).

t | s(t)=9tt2+9 |

0 | 0 |

0.5 | 0.48649 |

1 | 0.9 |

1.5 | 1.2 |

2 | 1.38462 |

2.5 | 1.47541 |

3 | 1.5 |

3.5 | 1.48235 |

4 | 1.44 |

4.5 | 1.38462 |

5 | 1.32353 |

5.5 | 1.26115 |

6 | 1.2 |

Calculate the velocity using the formula.

v(t)=−9(t2−9)(t2+9)2

Substitute 0 for t in the above equation.

v(0)=−9(02−9)(02+9)2=1

Similarly, calculate the remaining values.

Calculate the value of t and v(t) as shown in the table (1).

t | v(t)=−9(t2−9)(t2+9)2 |

0 | 1 |

0.5 | 0.92038 |

1 | 0.72 |

1.5 | 0.48 |

2 | 0.26627 |

2.5 | 0.10642 |

3 | 0 |

3.5 | -0.0648 |

4 | -0.1008 |

4.5 | -0.1183 |

5 | -0.1246 |

5.5 | -0.1241 |

6 | -0.12 |

Calculate the acceleration using the formula.

a(t)=18t(t2−27)(t3+9)3

Substitute 0 for t in the above equation.

a(0)=18(0)(02−27)(03+9)3=0

Similarly, calculate the remaining values.

Calculate the value of t and a(t) as shown in the table (1).

t | a(t)=18t(t2−27)(t3+9)3 |

0 | 0.00 |

0.5 | -0.32 |

1 | -0.47 |

1.5 | -0.35 |

2 | -0.17 |

2.5 | -0.06 |

3 | -0.02 |

3.5 | -0.01 |

4 | 0.00 |

4.5 | 0.00 |

5 | 0.00 |

5.5 | 0.00 |

6 | 0.00 |

Draw the graph of the position, velocity and acceleration functions as shown in the Figure 2.

#### To determine

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

#### Answer

the acceleration is zero when time *t* is greater than 33 sec and the acceleration is negative when time *t* is less than 33 sec.

#### Explanation

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

Substitute 0 for f"(t) in the equation f"(t)=18t(t2−27)(t3+9)3.

0=18t(t2−27)(t3+9)3t2−27=0t=27t=33 sec

Substitute 1 sec for t in the equation f"(t)=18t(t2−27)(t3+9)3.

f"(1)=18×(1)((1)2−27)((1)3+9)3=−0.468<0

Substitute 4 sec for t in the equation f"(t)=18t(t2−27)(t3+9)3.

f"(4)=18×(4)((4)2−27)((4)3+9)3=0

Therefore, the acceleration is zero when time *t* is greater than 33 sec and the acceleration is negative when time *t* is less than 33 sec.