3024 Rectilinear Motion

AP Calculus

On a line

Position

Defn: Rectilinear Motion: Movement of object in either direction along a

coordinate line (x-axis , or y-axis)

 s(t) = position function - position versus time graph (historical note:

  x(t) = horizontal axis y(t) = vertical axis

  a directed distance (a vector quantity) of the particle from some point,

p, at instant t . negative time = time before

 

s(t) positive – the particle is_______________________________________ 

negative – the particle is _______________________________________ 

= 0 – the particle is _______________________________________ _

physi

cs

math

SpotiaLatin

Direction and quantity

Located to the right

Located to the left

At the origin

Velocity

v(t) = velocity function - the rate of change of position

Velocity gives both quantity of change and direction of change (again a vector quantity) Speed finds quantity only. - absolute value of velocity (a scalar quantity)

Rem: Average Velocity = change in position over change in time =

  = Instantaneous Velocity the derivative

st

0t

sLimt

( ) ( )v t s t

Magnitude only

𝑥 (𝑏) −𝑥(𝑎)𝑏−𝑎

Average speed

𝑥 ′ (𝑡 ) Ticketed speed

Velocity

v(t) = velocity function - the rate of change of position

  = Instantaneous Velocity the derivative

v(t) positive – the particle’s position is ____________________ < velocity in a positive direction - _______________________

negative – the particle’s position is _____________________< velocity in a negative direction - _______________________

= 0 - the particle is _____________________________

{This is the 1st Derivative Test for increasing /decreasing!}

0t

sLimt

( ) ( )v t s t

First derivative is positive y is increasing

increasing

Moving to the rightdecreasing

Moving to the leftstationary not moving

*

Acceleration

a(t) = acceleration function - rate of change of velocity

a(t) positive - velocity is __________________________________ < acc. in a positive direction – _________________________

  negative - velocity is __________________________________ < acc. in neg. direction – ______________________________

  = 0 - velocity is __________________________________

{This is the 2nd derivative test for concavity}

CAREFUL: This is not SPEEDING UP or SLOWING DOWN!

( ) ( )( ) ( ) ( )

v t a ts t v t a t

increasing

Pushed to the rightdecreasing

Pushed to the left

constant cruise control

Speed and Direction

Determining changes in Speed

speed increasing if v(t) and a(t) have same sign -

also for v(t) = 0 and a(t) 0

  speed decreasing if v(t) and a(t) have opposite signs -

  

Determining changes in Direction

direction changes if v(t) = 0 and a(t) 0

no change if both v(t) = 0 and a(t) = 0

Moving Pushed

Moving Pushed OrMoving Pushed

Ball bouncing

Sitting still

Method (General):

1) Find the Critical Numbers in First and Second Derivatives.

1) Answer any questions at specific locations.

2) Do the Number Line Analysis (Brick Wall).

1) Find direction - moving , pushed , and speed

3) Identify the Change of Direction locations

1) Find values at beginning, ending, and change of direction times.

4) Sketch the Schematic graph.

5) Find the Displacement and Total Distance Traveled.

Set = 0

Set = 0

Example: A particle’s position on the y –axis is given by:

4 2( ) 6 6 [ 3,3]y t t t t

1) Find y(t), v(t) and a(t) at t = 2. Interpret each value.

𝑚𝑒𝑡𝑒𝑟𝑠

𝑡𝑖𝑚𝑒𝑖𝑛𝑠𝑒

𝑐𝑜𝑛𝑑𝑠

𝑦 (𝑡 )=𝑡 4 −6 𝑡 2− 6 𝑦 (2 )=16 − 24 −6=− 14

𝑣 (𝑡 )=𝑦 ′ (𝑡 )=4 𝑡 3 −12𝑡 𝑦 ′ (2 )=32− 24=8

𝑎 (𝑡 )=𝑦 ′ ′ (𝑡 )=12 𝑡2 −12 𝑦 ′ ′ (2 )=48−12=36

Located 14 units left

Moving right 8 units/sec

Pushed 36 units2/sec

Speed is increasing

Example: A particle’s position on the y –axis is given by:4 2

3 2

( ) 6 6 [ 3,3]

( ) 4 12 ( ) 12 12

y t t t t

v t t t a t t

2) Determine the motion within each interval: location, direction moving

and direction pushed.

3) Find the values at t = -3, t = 3 and where the particle changes directions.

4) Find the Displacement and Total Distance Traveled.

v(t)

a(t)

m

P

s

-3 −√3 √3 30

neg pospos neg

-2 -1 1 2

Moving left

4 𝑡 (𝑡2− 3) 𝑡=0 𝑡=±√3

𝑦 (− 3 )=21𝑦 (−√3 )=−15

𝑦 (3 )=21

𝑦 (0 )=6𝑦 (√3 )=−15

>36> 9>36>9

Original equation y values

36+9+9+36Ending – beginning 21-21=0

Number line analysis

21

6

-15

Example 2 : A particle’s position on the x –axis is given by:3 2( ) 12 36 24 [ 1,9]x t t t t t

Find and interpret x(t), v(t), and a(t) at t = 5

Example: A particle’s position on the x –axis is given by:

Sketch:

v

a

mps

3 2( ) 12 36 24 [ 1,9]x t t t t t

v

a

mps

𝑠 (𝑡 )=3 𝑡2 −12 𝑡+1 ; [0,5 ]

v

a

mps

𝑠 (𝑡 )=𝑡2+3 𝑡− 6 ; [− 2,2]

v

a

mps

𝑠 (𝑡 )=𝑡3 −9 𝑡+1 ;[−3,3 ]

v

a

mps

𝑠 (𝑡 )=−2 𝑡3+15 𝑡 2− 24 𝑡− 6 ;[0,5 ]

v

a

mps

𝑠 (𝑡 )=24=6 𝑡− 𝑡3

v

a

mps

v

a

mps

Last Update

• 11/22/10

• Assignment: work sheet - Swokowski