Calculus

6.3 Rectilinear Motion

6.3 Rectilinear Motion8. Apply derivatives

4 Solve a problem that involves applications and combinations of these skills.

3 Use implicit differentiation to find the derivative of an inverse function

Use derivatives to solve problems involving rates of change with velocity, speed and acceleration Use derivatives to find absolute and relative extrema in real-life context (Unit 4 Ch 5-6)Use derivatives to model rates of change (Unit 4 Ch 5-6)

2 Uses implicit differentiation to find the derivative of an inverse but makes errors in writing the inverses or in computing the derivativeCan state the relationship between position, velocity, speed and acceleration, and can write the velocity function given position, or acceleration function given the velocity functionShows understanding of using the derivatives to find extrema, but makes mistakes, or finds local extrema when absolute are required.Writes an equation to describe a situation but does not write a differential equation, or makes mistakes in evaluating the function.

1 With help, partial success at 2.0 & 3.0 content

0 Even with help, no success

Vocabulary

• Day 1:– Rectilinear Motion

• Position function– Velocity function

• Instantaneous velocity

• Day 2:– Speed function

• Instantaneous speed

• Day 3– Acceleration Function

• Speeding up• Slowing down

Rectilinear Motion

• Motion on a line

Moving in a positive direction from the origin

Moving in a negative direction from the origin

Position Function

• Horizontal axis:– time

• Vertical Axis:– position on a line

Moving in a positive direction from the origin

time

position

Moving in a negative direction from the origin

Position function: s(t)s = position (sposition duh!)t = times(t)= position changes as time changes

Sketchpad Example

Example

• Use the position and time graph to describe how the puppy was moving

time

position

Velocity• Rate

– position change vs time change

– Velocity can be positive or negative

• positive: going in a positive direction

• negative: going in a negative direction

18

16

14

12

10

8

6

4

2

-2

-4

-6

-8

-10

p

1 2 3 4 5 6 7 8 9 10 11 12

t

position

time

A A

18

16

14

12

10

8

6

4

2

-2

-4

-6

-8

-10

-12

p

-1 1 2 3 4 5 6 7 8 9 10 11

t

v(t) x = 3x2+-34x+76

4

time

Animate Points

Vel

ocity

Pos

ition

Velocity function

• Velocity is the slope of the position function (change in position /change in time)

• velocity =

– Technically this is instantaneous velocity

dt

dstv )( )(ts

Position Velocity Meaning

Positive Slope Positive y’s moving in a positive direction

Negative slope Negative y’s

Moving in a negative direction

Practice

• Let s(t)= t3-6t2 be the position function of a particle moving along an s-axis were s is in meters and t is in seconds. – Graph the position function– On a number line, trace the path that the particle

took. – Where will the velocity be positive? Negative?– Graph the instantaneous velocity. – Identify on the velocity function when the particle was

heading in a positive direction and when it was heading in a negative direction.

Velocity or Speed

• Speed change in position with respect to time in any direction

• Velocity is the change in position with respect to time in a particular direction– Thus – Speed cannot be negative – because

going backwards or forwards is just a distance– Thus – Velocity can be negative – because

we care if we go backwards

Speed

• Absolute Value of Velocity–

dt

dstv

)(

speed

ousinstantane

example: • if two particles are moving on the same coordinate line • with velocity of v=5 m/s and v=-5 m/s,• then they are going in opposite directions• but both have a speed of |v|=5 m/s

Example - s(t)= t3-6t2 position

time

23 6)( ttts

time

velocity

tttv 123)( 2

tttv 123)( 2

time

speed

Practice

• Graph the velocity function

• What will the speed function look like?

• At what time(s) was the particle heading in a negative direction? Positive direction?

19163)( 2 tttv

Acceleration

• the rate at which the velocity of a particle changes with respect to time.– If s(t) is the position function of a particle

moving on a coordinate line, then the instantaneous acceleration of the particle at time t is

•

• or

dt

dvta )(

2

2

)(")(dt

sdtsta

Example

• Let s(t) = t3 – 6t2 be the position function of a particle moving along an s-axis where s is in meters and t is in seconds. Find the instantaneous acceleration a(t) and shows the graph of acceleration verses time

tttstv 123)(')( 2 126)('')(')( ttstvta

Day 3: Speeding Up & Slowing down

• Speeding up when slope of speed is positive

• Slowing down when slope of speed is negative

Example

• When is s(t) speeding up and slowing down?

position

time

23 6)( ttts

time

velocity

tttv 123)( 2 speed

acceleration

Velocity & Acceleration function20

18

16

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12

10

8

6

4

2

-2

-4

-6

-8

-10

-12

p

-1 1 2 3 4 5 6 7 8 9 10 11

t

Animate Points

A AB

Slowing down

Velocity +

Acceleration -

Speeding up

Velocity -

Acceleration -

Slowing down

Velocity -

Acceleration +

Speeding up

Velocity +

Acceleration +

Analyzing MotionGraphically Algebraically Meaning

Position

Velocity

Acceleration

Positive “s” values Positive side of the number line

Negative side of the number line

Negative “s” values

s(t)=velocity.

Look for Critical Pts

Postive “v” values

0 “v” values (CP)

Negative “v” values

Moving in + direction

Turning/stopped

Moving in a – direction

v(t)=accelerationLook for Critical Pts

+ a, + v = speeding up- a, - v = speeding up+ a, - v = slowing down- a, + v = slowing down

Example

Suppose that the position function of a particle moving on a coordinate line is given by s(t) = 2t3-21t2+60t+3 Analyze the motion of the particle for t>0

Graphically Algebraically Meaning

Pos

ition

Vel

ocity

Acc

eler

atio

n

0360212)( 23 tttts

Never 0 (t>0), always postive

Always on postive side of number line

060426)()( 2 tttvts0)107(6 2 tt0)5)(2(6 tt

0 2 5

+ - +0 0

0<t<2 going pos direction

t=2 turning

2<t<5 going neg. directiont=5 turning

t>5 going pos. direction

t=0 t=2t=5

04212)()( ttatv4212 t 5.3t

+ - - +

0 2 53.5

va - - + +

0<t<2 slowing down

2<t<3.5 speeding up

3.5<t<5 slowing down

5<t speeding up

Day 4: Applications; Gravity•

• s = position (height)

• s0= initial height

• v0= initial velocity

• t = time• g= acceleration due to gravity

– g=9.8 m/s2 (meters and seconds)– g=32 ft/s2 (feet and seconds)

200 2

1gttvss

s0

Example• Nolan Ryan was capable of throwing a baseball at 150ft/s (more

than 102 miles/hour). Could Nolan Ryan have hit the 208 ft ceiling of the Houston Astrodome if he were capable of giving the baseball an upward velocity of 100 ft/s from a height of 7 ft?

2161007 tts tv 32100 the maximum height occurs when velocity = 0

t=100/32=25/8 seconds

s(25/8)=163.25 feet