One Dimensional Motion

Slides:



Advertisements
Similar presentations
Derivation of Kinematic Equations
Advertisements

The Basics of Physics with Calculus – Part II
Slopes and Areas Frequently we will want to know the slope of a curve at some point. Or an area under a curve. We calculate slope as the change in height.
Linear Kinematics Chapter 3. Definition of Kinematics Kinematics is the description of motion. Motion is described using position, velocity and acceleration.
Integral as Accumulation What information is given to us by the area underneath a curve?
3.4 Velocity, Speed, and Rates of Change
Slopes and Areas Frequently we will want to know the slope of a curve at some point. Or an area under a curve. We calculate area under a curve as the sum.
Constant velocity Average velocity equals the slope of a position vs time graph when an object travels at constant velocity.
Denver & Rio Grande Railroad Gunnison River, Colorado
3.1 Derivative of a Function
Derivation of Kinematic Equations
A Mathematical Model of Motion
Describing Motion. the study of motion motion is a change in position two branches Mechanics.
Every slope is a derivative. Velocity = slope of the tangent line to a position vs. time graph Acceleration = slope of the velocity vs. time graph How.
3.4 Velocity, Speed, and Rates of Change. downward , 8.
Derivation of Kinematic Equations
10.1 Parametric Functions. In chapter 1, we talked about parametric equations. Parametric equations can be used to describe motion that is not a function.
Derivative Examples 2 Example 3
Graphical Model of Motion. We will combine our Kinematics Equations with our Graphical Relationships to describe One Dimensional motion! We will be looking.
Motion graphs Position (displacement) vs. time Distance vs. time
DAY 1 Motion A Puzzler – You ride your bike from Ossining to NYC, 30 miles away at 15 mph. How fast must you return to Ossining to average 30 mph?
Graphs of Motion Physics Notes. Introduction to Graphs  Graphs are mathematical pictures.  They are the best way to convey a description of real world.
Derivation of Kinematic Equations
3.4 Velocity, Speed, and Rates of Change
Graphical Analysis of Motion
Motion Graphs Position-Time (also called Distance-Time or Displacement-Time) d t At rest.
Derivation of Kinematic Equations
AP Physics C.
Graphical Analysis of Motion
Accelerated Motion Chapter 3.
Motion with Constant Acceleration
3.4 Velocity and Other Rates of Change, p. 127
Derivative of a Function
9.2 Calculating Acceleration
Acceleration Changing velocity (non-uniform) means an acceleration is present Acceleration is the rate of change of the velocity Units are m/s² (SI)
9.2 Calculating Acceleration
Derivation of Kinematic Equations
Non-constant Forces Up until this time, we have mainly dealt with forces that are constant. These produce a uniform, constant acceleration. Kinematic.
3.2 Motion With Constant Acceleration
Graphs of Linear Motion
9.2 Calculating Acceleration
Pictures worth even MORE words now!
Kinematics Formulae & Problems Day #1
Kinematics: The Mathematics of Motion
Derivation of Kinematic Equations
Denver & Rio Grande Railroad Gunnison River, Colorado
9.2 Calculating Acceleration
The Basics of Physics with Calculus – Part II
Graphical Analysis of Motion
Graphical Analysis of Motion
Derivation of Kinematic Equations
Graphical Analysis of Motion
3.4 Velocity and Other Rates of Change
Graphical Analysis of Motion
3.3 Velocity, Speed, and Rates of Change
Graphical Analysis of Motion
Graphical Analysis of Motion
The integral represents the area between the curve and the x-axis.
Derivation of Kinematic Equations
Graphical Analysis of Motion
Graphical Analysis of Motion
Velocity-Time Graphs for Acceleration
Putting all the Graphs Together
Motion Graphs AP Physics C.
Derivation of Kinematic Equations
Velocity vs Time Graphs
Journal Entry 5 Accelerated Motion, Pt I
We have talked about graphical analysis that includes finding slope of a graph… Tells us acceleration Tells us velocity t t.
Straight Line Motion (continued)
Motion in One Dimension
Presentation transcript:

One Dimensional Motion The Area Under the Curve

The Last Lecture During the last lecture, we discussed: The use of integration and algebra to derive the “BIG FIVE” Today we will discuss the significance of the “Area Under the Curve”

Let us recall the motion of the elevator. The graph shows that the position vs. time follows a non-linear (polynomial) relationship from 1-3 and 8-11 seconds, and a constant change from 3-8 seconds. y = -0.65t2 + 14.31t - 48.82 y = 4.00t - 8.00 With the exception of the first and last second of the dataset, the motion of the elevator can be broken up into three distinct segments: Positive acceleration from 1 – 3 s Constant velocity from 3 – 8 s Negative acceleration from 8 – 11s y = t2 – 2.00t + 1.00

If you calculate the area under the curve from 1 – 11 seconds, what do you get? It should be clear by now that the area under a velocity vs. time curve equals the displacement of the object. 𝑨𝟏=½ 𝒃∗𝒉 𝒅=½ 𝒗∗𝒕 𝒅=𝒗𝒂𝒗𝒈∗𝒕 𝒅=½ (𝟒𝒎/𝒔)(𝟐𝒔) 𝒅=𝟒 𝒎𝒆𝒕𝒆𝒓𝒔 𝑨𝟑=½ 𝒃∗𝒉 𝒅=½ (𝟒𝒎/𝒔)(𝟑𝒔) 𝒅=𝟔 𝒎𝒆𝒕𝒆𝒓𝒔 𝑨𝟐= 𝒃∗𝒉 𝒅=𝒗∗𝒕 𝒅=(𝟒𝒎/𝒔)(𝟓𝒔) 𝒅=𝟐𝟎 𝒎𝒆𝒕𝒆𝒓𝒔 The total vertical displacement is equal to the sum of the areas: 𝑑𝑡𝑜𝑡𝑎𝑙=𝐴1+𝐴2+𝐴3=4𝑚+20𝑚+6𝑚=30𝑚

If you calculate the area under the curve from 1 – 3 and 8 – 11 seconds, what do you get? As should be clear once again, the area under an acceleration vs. time graphs equates to the change in velocity of the object. 𝑨=𝒃∗𝒉 𝒗=𝒂∗𝒕 𝒗=(𝟑𝒔)(−𝟏.𝟑𝒎/𝒔𝟐) 𝒗=~−𝟒𝒎/𝒔 𝑨=𝒃∗𝒉 𝒗=𝒂∗𝒕 𝒗=(𝟐𝒔)(𝟐𝒎/𝒔𝟐) 𝒗=𝟒𝒎/𝒔 Note: If you add the two areas together, you get zero, which tells you that the initial and final velocities should be the same. In this case, it is zero.

When the acceleration is uniform or constant, the area under a velocity vs. time curve, and an acceleration vs. time curve are easy to calculate. When the acceleration is zero or constant, the position of the particle will follow a first or second order polynomial, respectively. Constant Velocity 𝑥−𝑥𝑜=𝑣𝑜𝑡 First Order Constant Acceleration 𝑥−𝑥𝑜=𝑣𝑜𝑡+ 1 2 𝑎𝑡2 Second Order 𝑥=2.0+2.0𝑡 𝑣=2 𝑎=0 𝑥=5.0+2.0𝑡+1.5𝑡2 𝑣=2.0+3.0𝑡 𝑎=3 𝐴= 1 2 𝑏×ℎ 𝐴=𝑏×ℎ 𝐴=𝑏×ℎ 𝐴=𝑏×ℎ

However, what if the position vs However, what if the position vs. time of a particle does not follow a first or second order polynomial relationship? Acceleration not constant 𝑥=3𝑡−4𝑡2+𝑡3 Third Order

Now that we know what the area under the curve represents, how do you calculate it? We can break up the area into individual rectangles and add those areas up. 𝑨𝟐=𝒃∗𝒉 𝒙𝟐=𝒗∗𝒕 𝒙𝟐=(𝟓.𝟕𝟓𝒎/𝒔)(𝟏𝟎𝒔) 𝒙𝟐= 𝟓𝟕.𝟓𝒎 𝒙𝟑= 𝟓𝟕.𝟓𝒎 𝑨𝟏=𝒃∗𝒉 𝒙𝟏=𝒗∗𝒕 𝒙𝟏=(𝟐.𝟓𝒎/𝒔)(𝟏𝟎𝒔) 𝒙𝟏= 𝟐𝟓𝒎 𝒙𝟏= 𝟐𝟓𝒎 The total displacement is: 𝑥𝑡𝑜𝑡𝑎𝑙=𝑥1+𝑥2+𝑥3+𝑥4=25𝑚+57.5𝑚+57.5+25𝑚=165𝑚

Repeating the process using a smaller time interval We will now reduce the time interval by a factor of 2 by going from 10s to 5s. 𝟓𝟗.𝟕𝟓𝒎 𝟓𝟗.𝟕𝟓𝒎 𝟓𝟐.𝟎𝒎 𝟓𝟐.𝟎𝒎 𝟑𝟔.𝟕𝟓𝒎 𝟑𝟔.𝟕𝟓𝒎 𝟏𝟑.𝟓𝒎 𝟏𝟑.𝟓𝒎 Recalculating the area: 𝑥𝑡𝑜𝑡𝑎𝑙=𝑥1+𝑥2+…𝑥8=13.5𝑚+36.75𝑚+52.0+59.75𝑚+…=162𝑚

We can repeat this process again and again infinitely producing a smaller and smaller time interval each and every iteration.

Note that the area better approximates the displacement of the object as the time interval decreases. This is especially noticeable near the peak velocity. 𝑥= lim 𝑡𝑛→0 𝑛 𝑣𝑎𝑣𝑔 𝑡 𝑣𝑎𝑣𝑔 (for the interval) 𝑡𝑛

In calculus, if we know the formula, we can calculate the area under the curve through integration 𝑥 = 0 40 −0.0153𝑡2 + 0.6139𝑡 − 0.0697 𝑑𝑡 𝑣 = −0.0153𝑡2 + 0.6139𝑡 − 0.0697 𝑥 = 0.0051𝑡3 + 0.3070𝑡2 − 0.0697𝑡 𝑥 = −0.0051𝑚 𝑠3 40𝑠 3 + 0.3070𝑚 𝑠2 40𝑠 2 − 0.0697𝑚 𝑠 40𝑠 =162𝑚