#### To determine

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

#### Answer

The velocity at time is f'(t)=π2cos(πt2)ft/sec_.

#### Explanation

**Given:**

The given equation is as below.

s=f(t)=sin(πt2) (1)

**Calculation:**

Calculate the velocity at time t.

Differentiate the equation (1) with respect to time.

v(t)=f'(t)=cos(πt2).π2

v(t)=f'(t)=π2cos(πt2)ftsec (2)

Therefore, the velocity at time t is f'(t)=π2cos(πt2)ft/sec_.

#### To determine

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

#### Answer

The velocity after 1 second is v(1)=0 ft/sec_.

#### Explanation

Calculate the velocity after 1 second.

Substitute 1 for t in the equation (2).

v(t)=f'(t)=π2cos(πt2)v(1)=f'(1)=π2cos(π(1)2)v(1)=0 ft/sec

Therefore, the velocity after 1 second is v(1)=0 ft/sec_.

#### To determine

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

#### Answer

The particle never is at rest when t=1+2n_.

#### 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)=π2cos(πt2)0=π2cos(πt2)cos(πt2)=0(πt2)=π2+nπt=(π2+nπ)2πt=1+2n

Here, n is the non-negative integer where t≥0.

#### To determine

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

#### Answer

The velocity of particle always moves in positive direction when 0<t<1_,3<t<5_, 7<t<9_.

#### Explanation

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

The particle will move in positive direction when v(t)>0, and the equation from part (c), the velocity changes its sign at every positive odd integer. Therefore, the intervals are 0<t<1, 3<t<5, 7<t<9 and so on.

#### 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)=6ft_ .

#### Explanation

Calculate the total distance traveled during first 6 seconds.

Here, the velocity changes from 1, 3 and 5 which appears in the interval of [0,6].

Substitute 1 and 0 for t in the equation (1).

|f(1)−f(0)|=|sinπ(1)2−sinπ(0)2|=|1−0|=1|f(3)−f(1)|=|sinπ(3)2−sinπ(1)2|=|−1−1|=2|f(5)−f(3)|=|sinπ(5)2−sinπ(3)2|=|1−(−1)|=2|f(6)−f(5)|=|sinπ(6)2−sinπ(5)2|=|0−1|=1

The total distance travelled is as below.

f(t)=1+2+2+1=6 ft

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

#### To determine

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

#### Answer

The diagram to illustrate the motion of particle is shown in the figure (1).

#### Explanation

Calculate the distance *s* using the expression.

s=sinπt2

Substitute 0 for t in the above equation.

s=sinπt2s=sinπ(0)2s=0

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

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

#### To determine

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

#### Answer

The acceleration at time is −π24sinπt2 ft/s2_ and after one second is −π24 ft/s2_.

#### Explanation

Calculate the acceleration at any time *t.*

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

a(t)=π2(−sinπt2×π2)=−π24sinπt2ft/s2

Therefore, the acceleration at any time is −π24sinπt2 ft/s2_.

Calculate the acceleration after 1 second.

Substitute 1 for t in the above equation a(t)=π2(−sinπt2×π2).

a(1)=−π24sinπ(1)2ft/s2=−π24ft/s2

Therefore, the acceleration after 1 second is −π24 ft/s2_.

#### To determine

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

#### Answer

The position, velocity, and acceleration functions are plotted for time limits 0≤t≤6 in figures (2), (3), and (4).

#### Explanation

Calculate the position with respect to time using the formula.

f(t)=sinπt2

Substitute 0 for t in the above equation.

f(t)=sinπt2f(0)=sinπ(0)2=0

Similarly, calculate the remaining values.

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

t | f(t)=sinπt2 |

0.00 | 0.00 |

0.50 | 0.71 |

1.00 | 1.00 |

1.50 | 0.71 |

2.00 | 0.00 |

2.50 | -0.71 |

3.00 | -1.00 |

3.50 | -0.71 |

4.00 | 0.00 |

4.50 | 0.71 |

5.00 | 1.00 |

5.50 | 0.71 |

6.00 | 0.00 |

Calculate the velocity using the expression.

v(t)=π2cosπt2

Substitute 0 for t in the above equation.

v(0)=π2cosπ(0)2=1.57

Similarly, calculate the remaining values.

Calculate the value of t and v(t) as shown in the table (2).

t | v(t)=π2cosπt2 |

0.00 | 1.57 |

0.50 | 1.11 |

1.00 | 0.00 |

1.50 | -1.11 |

2.00 | -1.57 |

2.50 | -1.11 |

3.00 | 0.00 |

3.50 | 1.11 |

4.00 | 1.57 |

4.50 | 1.11 |

5.00 | 0.00 |

5.50 | -1.11 |

6.00 | -1.57 |

Calculate the acceleration using the formula.

a(t)=−π24sinπt2

Substitute 0 for t in the above equation.

a(t)=−π24sinπt2a(0)=−π24sinπ(0)2=0

Similarly, calculate the remaining values.

Calculate the value of t and a(t) as shown in the table (3).

t | a(t)=−π24sinπt2 |

0.00 | 0.00 |

0.50 | -1.74 |

1.00 | -2.47 |

1.50 | -1.74 |

2.00 | 0.00 |

2.50 | 1.74 |

3.00 | 2.47 |

3.50 | 1.74 |

4.00 | 0.00 |

4.50 | -1.74 |

5.00 | -2.47 |

5.50 | -1.74 |

6.00 | 0.00 |

Draw the position as a function of time curve as shown in the Figure (2).

Draw the speed as a function of time curve as shown in the figure (3).

Draw the acceleration as a function of time curve as shown in the figure (4).

#### To determine

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

#### Answer

The acceleration is negative when the value of time is 0<t<2 seconds and this will be slowing down and the acceleration is positive when the value of time is 2<t<4 sec and this will be speeding up. This cycle will repeat in the next 4 seconds, and go on.

#### Explanation

Calculate the time when particle speeding up and slowing down.

Substitute 0 for f"(t) in the equation f"(t)=−π24sinπt2.

0=−π24sinπt2sinπt2=0t=0,2,4, ...2n sec

Here, variable n is given as natural numbers starting from 0.

Therefore, the acceleration is negative when the value of time is 0<t<2 seconds and this will be slowing down and the acceleration is positive when the value of time is 2<t<4 sec and this will be speeding up. This cycle will repeat in the next 4 seconds, and go on.