Derivation Kinematic Eqns

Derivation of Kinematic Equations

View this afterMotion on an Incline Lab

Constant velocity

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

v = Δx

Δt

Displacement when object moves with constant velocity

The displacement is the area under a velocity vs time graph

v

t

Δx

Δx = v Δt

Uniform accelerationThis 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 0v

t

v0

Δv

Δt

Displacement when object accelerates from rest

Displacement is still the area under the velocity vs time graph. However, velocity is constantly changing.

v

t

Displacement when object accelerates from rest

Displacement is still the area under the velocity vs time graph. Use the formula for the area of a triangle.

Δx = 12ΔvΔt

v

t

Δv

Displacement when object accelerates from rest

From slope of v-t graphRearrange to get

Now, substitute for ∆v

a = Δv

Δt Δv = a Δt

Δx = 12 a Δt ⋅ Δt

v

t

Δv

Displacement when object accelerates from rest

Simplify

Assuming uniform acceleration and a starting time = 0, the equation can be written:

v

t

Δv

Δx = 12 a (Δt)2

Δx = 12

at 2

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

the rectangle representingdisplacement due to initial velocity

v

t

v0 Δx = v 0Δt

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

and the triangle representingdisplacement due to acceleration

Δx = 12 a(Δt)2

v

t

v0

Δv

Displacement when object accelerates with initial velocitySum the two areas:

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

Δx = v 0Δt + 12

a(Δt )2

Δx = v0t + 12 at2

v

t

v0

Δv

Time-independent relationship between ∆x, v and a

Sometimes you are asked to find the final velocity or displacement when the length of time is not given.To derive this equation, we must start with the definition of average velocity:

v = Δx

Δt

Time-independent relationship between ∆x, v and a

Another way to express average velocity is: v = Δx

Δt

v = v f +v 0

2

Time-independent relationship between ∆x, v and a

We have defined acceleration as:

This can be rearranged to:

and then expanded to yield :

a = ΔvΔt

Δt = Δv

a

Δt = vf −v0

a

Time-independent relationship between ∆x, v and a

Now, take the equation for displacement

and make substitutions for average velocity and ∆t

Δx = v Δt

Time-independent relationship between ∆x, v and a

Δx = v Δt

v = v f +v 0

2 Δt = vf −v0

a

Time-independent relationship between ∆x, v and a

Δx = v Δt

Δx =

vf + v0

2⋅vf −v0

a

Time-independent relationship between ∆x, v and a

Simplify

Δx =

vf + v0

2⋅vf −v0

a

Δx =

vf2 −v0

2

2a

Time-independent relationship between ∆x, v and a

Rearrange Δx =

vf2 −v0

2

2a

2a Δx = v f2 − v 0

2

Time-independent relationship between ∆x, v and a

Rearrange again to obtain the more common form:

2a Δx = v f2 − v 0

2

vf2 = v0

2 + 2a Δx

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.

Constant velocity

If you are looking for the velocity,– use the algebraic form

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

v = Δx

Δt

Constant velocity

• If you are looking for the displacement,– use the algebraic form

– or find the area under the curve

Δx = v Δt

v

t

Δx

Uniform acceleration

• If you want to find the final velocity,– use the algebraic form

• If you are looking for the acceleration– rearrange the equation above

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

vf = at + v0

a =

ΔvΔt

Uniform acceleration

If you want to find the displacement,– use the algebraic form– eliminate initial velocity if the object

starts from rest

– Or, find the area under the curve

Δx = v0t + 12 at2

v

t

v0

Δv

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 vf

2 = v02 + 2a Δx

All the equations in one place

constant velocity uniform acceleration

vf2 = v0

2 + 2aΔx Δx = v0t + 1

2 at2

a =

ΔvΔt

v f = at +v 0

v = Δx

Δt

Δx = v Δt