Position,Position, Velocity,Velocity, andand AccelerationAcceleration

Mr.Mr. MiehlMiehl www.tesd.net/miehlwww.tesd.net/miehl [email protected]@tesd.net Position,Position, VelocityVelocity && AccelerationAcceleration

Velocity is the rate of change of position with respect to time. ΔD Velocity = ΔT Acceleration is the rate of change of velocity with respect to time. ΔV Acceleration = ΔT Position,Position, VelocityVelocity && AccelerationAcceleration

Warning: Professional driver, do not attempt!

When you’re driving your car… Position,Position, VelocityVelocity && AccelerationAcceleration

squeeeeek!

…and you jam on the brakes… Position,Position, VelocityVelocity && AccelerationAcceleration

…and you feel the car slowing down… Position,Position, VelocityVelocity && AccelerationAcceleration

…what you are really feeling… Position,Position, VelocityVelocity && AccelerationAcceleration

…is actually acceleration. Position,Position, VelocityVelocity && AccelerationAcceleration

I felt that acceleration. Position,Position, VelocityVelocity && AccelerationAcceleration

How do you find a function that describes a physical event?

Steps for Modeling Physical Data

1) Perform an experiment.

2) Collect and graph data.

3) Decide what type of curve fits the data.

4) Use statistics to determine the equation of the curve. Position,Position, VelocityVelocity && AccelerationAcceleration

A crab is crawling along the edge of your desk. Its location (in feet) at time t (in seconds) is given by P (t ) = t 2 + t.

a) Where is the crab after 2 seconds?

b) How fast is it moving at that instant (2 seconds)? Position,Position, VelocityVelocity && AccelerationAcceleration

A crab is crawling along the edge of your desk. Its location (in feet) at time t (in seconds) is given by P (t ) = t 2 + t.

a) Where is the crab after 2 seconds? 2 P()22=+ ()( 2)

P()26= feet Position,Position, VelocityVelocity && AccelerationAcceleration

A crab is crawling along the edge of your desk. Its location (in feet) at time t (in seconds) is given by P (t ) = t 2 + t.

b) How fast is it moving at that instant (2 seconds)? Velocity is the rate of change of position.

Pt()=+ t2 t Vt()== Pt'( ) 21t +

Velocity function P'2( )= 22( )+ 1

P'2( )= 5feet per second Position,Position, VelocityVelocity && AccelerationAcceleration

A disgruntled calculus student hurls his calculus book in the air. Position,Position, VelocityVelocity && AccelerationAcceleration

The position of the calculus book: pt( )= −+16 t2 96 t t is in seconds and p(t) is in feet

a) What is the maximum height attained by the book?

b) At what time does the book hit the ground?

c) How fast is the book moving when it hits the ground? Position,Position, VelocityVelocity && AccelerationAcceleration a) What is the maximum height attained by the book? The book attains its maximum height when its velocity is 0. pt( )= −+16 t2 96 t vt( )= p′(t)= −32t + 96 03296= −+t 32t = 96

t = 3 seconds 2 Velocity function p(3163963)()=− + ( ) p(3144288)= −+

p(3144)= feet Position,Position, VelocityVelocity && AccelerationAcceleration b) At what time does the book hit the ground? The book hits the ground when its position is 0.

pt( )= −+16 t2 96 t 01696= −+tt2 016(6)= −−tt −16t = 0 t −=60

t = 0 sec. t = 6 sec. Position,Position, VelocityVelocity && AccelerationAcceleration c) How fast is the book moving when it hits the ground? Good guess: 0 ft/sec This is incorrect.

vt( )= −+32 t 96 v(632696)= −+( ) v(6)= −+ 192 96

v(696)= − ft/sec

Downward direction Position,Position, VelocityVelocity && AccelerationAcceleration

Acceleration: the rate of change of velocity with respect to time.

Velocity function vt( )= −+32 t 96

Acceleration function at( )()= v′ t =−32 ft/sec2

How is the acceleration function related to the position function?

Acceleration is the second derivative of position. at( )= p′′( t)

Jerk is the rate of change of acceleration with respect to time. Position,Position, VelocityVelocity && AccelerationAcceleration

A red sports car is traveling, and its position P (in miles) at time t (in hours) is given by P (t ) = t 2 –7t.

a) When is the car 30 miles from where it started? b) What is the velocity at the very moment the car is 30 miles away? c) What is the acceleration at the very moment the car is 30 miles away? d) When does the car stop? Position,Position, VelocityVelocity && AccelerationAcceleration

A red sports car is traveling, and its position P (in miles) at time t (in hours) is given by P (t ) = t 2 –7t. a) When is the car 30 miles from where it started?

30=−tt2 7

0730=−−tt2

0103=−()tt( +) t −=10 0 t + 30=

t =10 hours t = −3 Position,Position, VelocityVelocity && AccelerationAcceleration

A red sports car is traveling, and its position P (in miles) at time t (in hours) is given by P (t ) = t 2 –7t. b) What is the velocity at the very moment the car is 30 miles away? Vt()==− Pt'27 () t Vt()==− Pt'27 () t P'10()= 210( )− 7

P'10()= 13Miles per hour Position,Position, VelocityVelocity && AccelerationAcceleration

A red sports car is traveling, and its position P (in miles) at time t (in hours) is given by P (t ) = t 2 –7t. c) What is the acceleration at the very moment the car is 30 miles away? Vt()==− Pt'27 () t

At()== P''( t) 2 Miles per hour2 Position,Position, VelocityVelocity && AccelerationAcceleration

A red sports car is traveling, and its position P (in miles) at time t (in hours) is given by P (t ) = t 2 –7t. d) When does the car stop?

Vt()==− Pt'27 () t 02=−t 7 72= t

t = 3.5 hours ConclusionConclusion z The velocity function is found by taking the derivative of the position function. z In order for an object traveling upward to obtain maximum position, its instantaneous velocity must equal 0. z As an object hits the ground, its velocity is not 0, its height is 0. z The acceleration function is found by taking the derivative of the velocity function.