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

2. Kinematic Equations for Constant Acceleration
If the acceleration is constant,
a = Dv/Dt = (vf - vi )/t
or vf = vi + at

3. 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)

4. 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)

5. So far: 1. vf = vi + at 2. xf = xi + (t/2)(vi +vf)
3. vf2 = vi 2 +2a(xf –xi)
2. xf = xi + (t/2)(vi +vf)
Substituting 1. into 2. gives:
xf = xi + t/2(vi +vi +at), or
xf = xi + vit + 1/2at2

6. Finally, the Four Kinematic Equations
1. vf = vi + at
3. vf2 = vi 2 +2a(xf –xi)
2. xf = xi + (t/2)(vi +vf)
4. xf = xi + vit + (1/2)at2

7. How To Use Them xf = vf = a = xi = vi = t =
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.

8. Example xf = vf = a = xf = ? vf = 40 a = ? xi = vi = t =
A car starts from rest and accelerates to 40 m/s in 10 seconds.
1st List the possible unknowns – fill in.
xf = vf = a =
xi = vi = t =
xf = ? vf = a = ?
xi = vi = t = 10
What is the acceleration of the car?
How far does the car move?

9. To get a, you need an equation with a in it, but without xf
To get a, you need an equation with a in it, but without xf. Which one is it?
vf = vi + at
xf = ? vf = a = ?
xi = vi = t =10
40 = a or
a = m/s2

10. To get xf you have a choice. Using 2.,
xf = xi + (t/2)(vi +vf)
xf = ? vf = a = ?
xi = vi = t =10
xf = (10/2)(0 +40)
= 5(40) = 200m

11. Example xf = 100 vf = 0 a = ? xf = vf = a = xi = 0 vi = 50 t = ?
A 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?
xf = 100 vf = a = ?
xi = vi = t = ?
xf = vf = a =
xi = vi = t =
What is the acceleration of the car?
What is t?

12. vf2 = vi 2 +2a(xf –xi) 0 =502 + 200a or a = -502 / 200 = -12.5 m/s2
To get a, you need an equation with a in it, but without t. Which one is it?
vf2 = vi 2 +2a(xf –xi)
xf = 100 vf = a = ?
xi = vi = t = ?
0 = a or
a = -502 / 200 =
-12.5 m/s2

13. To get t, you have a choice. Let’s use:
vf = vi + at
xf = 100 vf = a = ?
xi = vi = t = ?
0 = t or
t = -50 /-12.5 =
4.00 seconds

14. Summary 1. vf = vi + at 2. xf = xi + (t/2)(vi +vf)
3. vf2 = vi 2 +2a(xf –xi)
2. xf = xi + (t/2)(vi +vf)
4. xf = xi + vit + (1/2)at2

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