KINEMATICS – Motion with constant acceleration

Level 1 Physics

Objectives and Essential Questions

Objectives Define and apply

definitions of displacement, avg. velocity, instantaneous velocity, and average acceleration

Demonstrate proficiency in problem solving using kinematic equations including problems involving free-fall

Analyze motion of graphs qualitatively and quantitatively.

Essential Questions What is displacement?

How does it differ from distance?

How is displacement affected by time, velocity, and acceleration?

How is scientific data displayed?

SYMBOLS

x, y Displacement

t Time

vo Initial velocity

v Finial velocity

a Acceleration

g Acceleration due to gravity

Equation #1

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a =Δv

Δt=v − vot

= v − vo = at

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v = vo − at

EXAMPLE #1A speedboat has a constant acceleration of +2.0 m/s2. If the initial velocity of the boat is 6.0 m/s, how fast is the boat moving after 8 seconds?

What do Iknow?

What do I need?

vo = 6 m/s v = ?

a = +2.0 m/s2

t = 8 sec

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v = vo +at

v = 6 ms( ) + +2.0 ms2( ) 8sec( )[ ]

v = 22m

s

Equation #2

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x = vot +1

2at 2

b. Find the displacement of the boat after the 8 second.

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x = vot +1

2at 2

x = 6.0 ms( ) 8sec( )[ ] +1

2+2.0 m

s2( ) 8sec( )2

[ ]

x =112m only 2 significant figures therefore

x =110m

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Ar = bh

Ar = 8( ) 6( )

Ar = 48m

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At =1

2bh

At =1

28( ) 16( )

At = 64m

Graphical RepresentationArea of triangle

Area of rectangle

Total Displacement = 64 m + 48 m = 112 m 110 m

22 m/s

Equation/Example #3

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v2 = vo2 + 2ax

Example: You are driving through town at 12 m/s when suddenly a ball rollsout in front of your car. You apply the brakes and begin decelerating at 3.5 m/s2.

What do I know?

What do I want?

vo = 12 m/s x = ?

a = - 3.5 m/s2

v = 0 m/s

How far do you travel before coming to a complete stop?

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v2 = vo2 + 2ax

x =v2 − vo

2( )

2a( )

x =0( )2− 12( )

2

[ ]

2( ) −3.5( )[ ]

x = 20.57m

Confusion??How do I know what equation to use and what variable do I need to solve for?

Equation Missing quantity

x

v

t

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v = vo + at

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x = vo t +12 at

2

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v2 = vo2 + 2ax

VERTICAL KinematicsAll of the 3 previous equations can be applied to the verticaldirection simply by adding a few changes.

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v = vo + at vy = voy − gt

x = vot +12 at

2 y = voyt −12 gt

2

v 2 = vo2 + 2aΔx vy

2 = voy2 − 2gΔy

Notice: Since g has a value of 9.8 m/s2 and is directed downward, the positivesign has been replaced with a negative sign.