Motion Graphs Part Motion Graphs Part TwoTwo

Review: Line of Best FitReview: Line of Best FitLine of best fit Line of best fit - A line on a graph drawn - A line on a graph drawn

so that there are the same number of data so that there are the same number of data points above as below the line.  points above as below the line. 

What does the slope or rate What does the slope or rate of change mean in this of change mean in this graph?graph?

SPEEDSPEED

What does the slope or rate What does the slope or rate of change mean in this of change mean in this graph?graph? VELOCITYVELOCITY

Acceleration: Acceleration: a = a = ΔΔvv ΔΔtt

AccelerationAcceleration is rate at which an is rate at which an object changes its velocity. object changes its velocity.

It is a It is a vector quantityvector quantity A person can be moving very fast, A person can be moving very fast,

and still not be accelerating.and still not be accelerating. Anytime an object's velocity is Anytime an object's velocity is

changing, that object is said to be changing, that object is said to be accelerating.accelerating.

Acceleration on position-time graphsAcceleration on position-time graphs

On a position-timeOn a position-time

graph, or distance-timegraph, or distance-time

graph, a curved linegraph, a curved line

represents a change in represents a change in

velocity. velocity.

Finding Instantaneous Finding Instantaneous Velocity on a GraphVelocity on a Graph

Find Instantaneous VelocityFind Instantaneous Velocity

1. Draw a tangent line to the slope1. Draw a tangent line to the slope

2. Pick two points2. Pick two points

(3, 2) & (0.5, 11)(3, 2) & (0.5, 11)

3. Calculate the slope3. Calculate the slope

A = ∆v / ∆tA = ∆v / ∆t

A = (2 – 11) / (3 – 0.5)A = (2 – 11) / (3 – 0.5)

A = -3.6 m/s2A = -3.6 m/s2

Graphing Acceleration- A car that moves at a uniform (constant) velocity to the right at + 10 m/s - A car with a constant velocity has an acceleration of ZERO (no change in the velocity).

- The slope of a horizontal line on a graph = 0

NOTE: Y-axis isVELOCITY andNot distance

- A car moves to the right and its velocity is increasing at a steady rate. This means that it has a positive acceleration.

- The graph would look like:

- Note that the slope is positive.

We can have 3 graph shapes for Velocity vs. Time.

1. No acceleration

v

t

2. Positive acceleration (speeding up)

v

t

3. Negative acceleration (slowing down, decelerating) v

t

Finding Acceleration on a Finding Acceleration on a Velocity Time GraphVelocity Time Graph

Acceleration = Acceleration = ∆v∆v ∆ ∆ttA = A = (v(vfinalfinal – v – vinitialinitial)) (t(tfinalfinal – t – tinitialinitial))A = A = (8m/s – 0m/s)(8m/s – 0m/s) (4s – 0s)(4s – 0s)A = A = 8m/s8m/s 4s4sA = 2m/sA = 2m/s22

Graphing Instantaneous Acceleration Graphing Instantaneous Acceleration

What is the instantaneousWhat is the instantaneous

acceleration at 3.5 s?acceleration at 3.5 s?

Draw tangentDraw tangent

Pick two points on linePick two points on line

(1.5, 8) & (4.5, 10)(1.5, 8) & (4.5, 10)

Calculate the slopeCalculate the slope

(10 – 2) / (4.5 – 1.5)(10 – 2) / (4.5 – 1.5)

2.67 m/s2.67 m/s22

Completing TablesCompleting Tables

∆∆t = tf – ti t = tf – ti ∆∆t = 55s – 5s = 50s t = 55s – 5s = 50s ∆∆v = vf – vi v = vf – vi ∆∆v = -15m/s - +10m/s = -25 m/s v = -15m/s - +10m/s = -25 m/s Aav = ∆v ÷ ∆t Aav = ∆v ÷ ∆t Aav = -25m/s / 50s = -0.5 m/sAav = -25m/s / 50s = -0.5 m/s22

Tf = 25sTf = 25s∆∆v = +65m/sv = +65m/sa = +2.6m/sa = +2.6m/s22

tti i (s)(s) ttf f (s)(s) ∆∆t (s)t (s) vvii (m/s) (m/s) vvff (m/s) (m/s) ∆∆v (m/s)v (m/s) aaavav (m/s (m/s22))

55 5555 +10+10 -15-15

3535 6060 +25+25 +95+95