STARTER

Rearrange the equation for acceleration, solving for vf. Start with a = (vf - vi )/t and solve for vf .

vf = vi + at

Kinematic Equations for Constant Acceleration

If the acceleration is constant,

a = Dv/Dt = (vf - vi )/t

or 1. vf = vi + at

If a = constant, v vs. t is a straight line.

The average value of v is halfway on the line, so:

Vavg = ½(vi +vf) = (xf – xi )/ t ,or

xf = xi + (t/2)(vi +vf)

So far:

1. vf = vi + at

2. xf = xi + (t/2)(vi +vf)

Solving 1. for t and inserting into 2 gives you :

xf = xi + (vf -vi )(vi +vf)(a/2)

Or vf2 = vi

2 +2a(xf –xi)

So far:

1. vf = vi + at

2. xf = xi + (t/2)(vi +vf)

3. vf2 = vi

2 +2a(xf –xi)

Substituting 1. into 2. gives: xf = xi + t/2(vi +vi +at), or xf = xi + vit + 1/2at2

Finally, the Four Kinematic Equations

1. vf = vi + at

2. xf = xi + (t/2)(vi +vf)

3. vf2 = vi

2 +2a(xf –xi)

4. xf = xi + vit + (1/2)at2

How To Use Them

1st List the possible unknowns

xf = vf = a = xi = vi = t =

2nd Read the problem and fill in all you can ( usually 4 ).

3rd Choose a kinematic equation with just one unknown in it.

ExampleA car starts from rest and accelerates to 40 m/s in 10 seconds.

1. What is the acceleration of the car?

2. How far does the car move?

1st List the possible unknowns – fill in.xf = vf = a =

xi = vi = t = xf = ? vf = 40 a = ?xi = 0 vi = 0 t = 10

To get a, you need an equation with a in it, but without xf. Which

one is it? vf = vi + at

40 = 0 + 10a or

a = 4.00 m/s2

xf = ? vf = 40 a = ?xi = 0 vi = 0 t =10

To get xf you have a choice.Using 2.,

xf = xi + (t/2)(vi +vf)

xf = 0 + (10/2)(0 +40) = 5(40) = 200m

xf = ? vf = 40 a = ?xi = 0 vi = 0 t =10

ExampleA car moving at 50m/s sees a dog in the road 100m ahead. If the drivers stops just in time, what acceleration must the brakes provide? How long does it take to stop?

1. What is the acceleration of the car?2. What is t?

xf = vf = a = xi = vi = t =

xf = 100 vf = 0 a = ?xi = 0 vi = 50 t = ?

To get a, you need an equation with a in it, but without t. Which

one is it?

vf2 = vi

2 +2a(xf –xi)

0 =502 + 200a or

a = -502 / 200 = -12.5 m/s2

xf = 100 vf = 0 a = ?xi = 0 vi = 50 t = ?

To get t, you have a choice.Let’s use:

vf = vi + at

0 =50 -12.5t or t = -50 /-12.5 = 4.00 seconds

xf = 100 vf = 0 a = ?xi = 0 vi = 50 t = ?

Summary

1. vf = vi + at

2. xf = xi + (t/2)(vi +vf)

3. vf2 = vi

2 +2a(xf –xi)

4. xf = xi + vit + (1/2)at2

EXIT

A relay runner moving at 2 m/s, speeds up to 6m/s in 4 seconds. What is her acceleration?