derivation of kinematic equations

22
Derivation of Kinematic Equations

Upload: omaregypt

Post on 30-Jun-2015

583 views

Category:

Education


2 download

DESCRIPTION

Derivation of Kinematic Equations

TRANSCRIPT

Page 1: Derivation of Kinematic Equations

Derivation of Kinematic Equations

Page 2: Derivation of Kinematic Equations

Constant velocityAverage velocity equals the slope of a position vs time graph when an object travels at constant velocity.

v x

t

Page 3: Derivation of Kinematic Equations

Displacement when object moves with constant velocity

The displacement is the area under a velocity vs time graph

v

t

x

x v t

Page 4: Derivation of Kinematic Equations

Uniform acceleration

v

t

v0

v

t

This is the equation of the line of the velocity vs time graph when an object is undergoing uniform acceleration.

The slope is the acceleration

The intercept is the initial velocity

v f at v 0

Page 5: Derivation of Kinematic Equations

Displacement when object accelerates from restDisplacement is still the area under the

velocity vs time graph. Use the formula for the area of a triangle.

From slope of v-t graph:

Now, substitute for ∆v:

x 12vt

v

t

v

v a t

x 12a t t

x 12a (t)2

Page 6: Derivation of Kinematic Equations

Displacement when object accelerates with initial velocityBreak the area up into two parts:

the rectangle representingdisplacement due to initial velocity

and the triangle representingdisplacement due to acceleration

v

t

v0

x v 0t

x 12a(t)2

Page 7: Derivation of Kinematic Equations

Displacement when object accelerates with initial velocity

Sum the two areas:

Or, if starting time = 0, the equation can be written:

x v 0t 12

a(t )2

x v0t 12at2

v

t

v0

v

Page 8: Derivation of Kinematic Equations

Time-independent relationship between ∆x, v and a

Another way to express average velocity is:

v x

t

v v f v 0

2

Page 9: Derivation of Kinematic Equations

Time-independent relationship between ∆x, v and a

We have defined acceleration as:

This can be rearranged to:

and then expanded to yield :

a v

t

t v

a

t vf v0

a

Page 10: Derivation of Kinematic Equations

Time-independent relationship between ∆x, v and a

Now, take the equation for displacement

and make substitutions for average velocity and ∆t

x v t

Page 11: Derivation of Kinematic Equations

Time-independent relationship between ∆x, v and a

x v t

v v f v 0

2

t vf v0

a

Page 12: Derivation of Kinematic Equations

Time-independent relationship between ∆x, v and a

x v t

x vf v0

2vf v0

a

Page 13: Derivation of Kinematic Equations

Time-independent relationship between ∆x, v and a

Simplify

x vf v0

2vf v0

a

x vf

2 v02

2a

Page 14: Derivation of Kinematic Equations

Time-independent relationship between ∆x, v and a

Rearrange

x vf

2 v02

2a

2a x v f2 v 0

2

Page 15: Derivation of Kinematic Equations

Time-independent relationship between ∆x, v and a

Rearrange again to obtain the more common form:

2a x v f2 v 0

2

vf2 v0

2 2a x

Page 16: Derivation of Kinematic Equations

Which equation do I use?First, decide what model is appropriate

Is the object moving at constant velocity? Or, is it accelerating uniformly?

Next, decide whether it’s easier to use an algebraic or a graphical representation.

Page 17: Derivation of Kinematic Equations

Constant velocityIf you are looking for the velocity,use the algebraic form

or find the slope of the graph (actually the same thing)

v x

t

Page 18: Derivation of Kinematic Equations

Constant velocityIf you are looking for the displacement,

use the algebraic form

or find the area under the curve

x v t

v

t

x

Page 19: Derivation of Kinematic Equations

Uniform accelerationIf you want to find the final velocity,

use the algebraic form

If you are looking for the accelerationrearrange the equation above

which is the same as finding the slope of a velocity-time graph

vf at v0

a vt

Page 20: Derivation of Kinematic Equations

Uniform accelerationIf you want to find the displacement,use the algebraic formeliminate initial velocity if the object

starts from rest

Or, find the area under the curve

x v0t 12at2

v

t

v0

v

Page 21: Derivation of Kinematic Equations

If you don’t know the time…You can solve for ∆t using one of the earlier equations, and then solve for the desired quantity, orYou can use the equation

rearranging it to suit your needs

vf2 v0

2 2a x

Page 22: Derivation of Kinematic Equations

All the equations in one placeconstant velocity uniform acceleration

vf2 v0

2 2ax

x v0t 12at2

a vt

v f at v 0

v x

t

x v t