# math portfolio type 2 (eleator)

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Introduction

Hl type 2

Math portfolio 2

Designing a freight elevator

Candidate name: lokesh baid

Candidate Number:2279-014

Session: 2007-2009

Supervisor: Mrs. Jyoti Uppal

Math’s Portfolio-Designing a Freight Elevator

In this portfolio I will attempt to design a freight Elevator model. Primarily I will study the given model to assess the usefulness and the strengths and weaknesses This will allow me to then form conditions for designing an effective and usefulelevator model.

Therefore I will now study the motion of the elevator by the given displacement equation :

Displacement

Checking to interpret straight line motion for time (t) =0,1,2,3,4,5,6

Substituting t=0

Substituting t=1

Substituting t=2

Substituting t=3

Substituting t=4

Substituting t=5

Substituting t=6

Displacement/Time Graph

Velocity

As velocity is the change in displacement at the given time, we differentiate the displacement equation to obtain an equation to give us the velocity of the elevator at a given time.

Therefore :

We now substitute r values of t = 0,1,2,3,4,5,6 to obtain the velocity at the given times

Substituting t=0

Substituting t=1

Substituting t=2

Substituting t=3

Substituting t=4

Substituting t=5

Substituting t=6

Velocity/Time graph

Acceleration

As acceleration is the change in velocity in a given time, we differentiate the velocity equation to obtain an equation to give us the acceleration of the elevator at a given time.Therefore:

We now substitute r values of t = 0,1,2,3,4,5,6 to obtain the velocity at the given times

Substituting t=0

Substituting t=1

Substituting t=2

Substituting t=3

Substituting t=4

Substituting t=5

Substituting t=6

Accerleration/Time graph

Interpretation

Middle

as f=0 and g=0

Substituting condition 4. t=3, v=0

as f=0

Substituting condition 5. t=6, s=0

As f=0

Substituting condition 6. t=6, v=0

As f=0

Substituting condition 7. t=6, a=0

Analysis

As there are 5 variables: a, b, c, d, e and 5 equations with the variables. I have simultaneously equated them using technology.

Variables | Values |

a | |

b | |

c | |

d | |

e, f, g |

After substituting the values found above for a, b, c, d, e, f and g in the general equation. The model that I have final obtained is:

For Displacement (s) as a function of time

For Velocity (v) as a function of time

For Acceleration (a) as a function of time

I will now substitute the values of the variables in the equations below for t=0,1,2,3,4,5,6 to find the desired values for:

- Displacement
- Velocity
- Acceleration

Displacement (s)

When t=0

When t=1

When t=2

When t=3

When t=4

When t=5

When t=6

Velocity (v)

When t=0

When t=1

When t=2

When t=3

When t=4

When t=5

When t=6

Acceleration (a)

When t=0

When t=1

Conclusion

Therefore:

Ideal Table

Time | Displacement (s) | Velocity (v) | Acceleration (a) |

1 | 0 | 0 | |

2 | 50 | 0 | |

3 | 0 | 0 | 0 |

Therefore we will use 6 degree equation as there are 7 variables needed for the 7 equations.

For

Displacement (s):

For

Velocity (v):

For

Acceleration (a):

I will now use these equations to find out the desired horizontal movement of the elevator.

Substituting condition 1. t=0, x=0

Substituting condition 2. t=2, x=50

Substituting condition 3. t=4, x=0

Substituting condition 4. t=0,v=0

Substituting condition 5. t=2, v=0

Substituting condition 6. t=4, v=0

Substituting condition 6. t=4, a=0

As f and g are 0 all the equations become 5 variables and I have solved them using technology.

Therefore:

Variables | Values |

a | |

b | |

c | |

d | |

e |

Therefore the general equations of the horizontal model are:

For

Displacement (s):

For Velocity (v):

For Acceleration (a):

By substituting the t=0, 1,2,3,4 into displacement, velocity and acceleration general equations, the value of the displacement, velocity and acceleration is found at different times.

Time | Displacement (s) | Velocity (v) | Acceleration (a) |

0 | 0 | 0 | 0 |

1 | |||

2 | 0 | ||

3 | |||

4 | 0 | 0 | 0 |

Now the elevator is designed to have vertical motion and horizontal motion. For Dual motion in x and y , the equation will be represented in parametric form as independent equations.

Therefore horizontal motion:

Therefore vertical motion:

Through this model we see that the lift can move horizontally as well as vertically. The 2 equations have been kept independent as this allows more flexibility in the movement of the Elevator. The horizontal motion of the lift can be used for unloading at 50 m displacement or for picking up people. This model has now become useful for the intended purpose and has also allowed a horizontal motion to be incorporated when needed. Therefore the elevator model is successfully formed.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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