1. UNIT 1: 1-D KINEMATICS Lesson 5: Kinematic Equations
CENTRE HIGH: PHYSICS 20
UNIT 1: 1-D KINEMATICS
Lesson 5:
Kinematic Equations

2. Reading Segment:
Kinematic Equations
To prepare for this section, please read:
Unit 1: p.19

3. Kinematic Equations
When an object has a constant acceleration,
there are many equations which describe the motion.
A complete list of equations is on the next page
(and on your formula sheet)

4. Here is a list of all the equations:
a = vf - vi vf2 = vi a d t
d = vi t a t2 d = vf t a t2
d = vi + vf t 2
where vavg = vi + vf

5. To figure out which equation to use, you make a list of
the 5 kinematic variables:
vi, vf, a, d, t
If you can identify the variable that is NOT involved
in the question, you can choose the proper equation.
This will be illustrated in the next pages.

6. Variables involved in question:
vi, vf, a, t
Variable NOT involved in the question: d
Equation to use: a = vf - vi t
This is the only kinematic equation that
does not have displacement (d) in it.

7. Variables involved in question:
vi, vf, a, d
Variable NOT involved in the question: t
Equation to use: vf2 = vi a d
This is the only kinematic equation that
does not have time (t) in it.

8. The process is similar for the other equations:
Equations Variables
vi vf a d t
d = vi t a t2    
d = vf t a t2    
d = vi + vf t
2    

9. Note:
 If the object has a constant velocity:
- the acceleration is zero
- the final velocity is equal to the initial velocity (vi = vf)
Use the equation: v = d t

10. Ex. 1 A ball is rolled up a constant incline at a speed of 9.0 m/s.
How much time is required to reach a maximum
displacement of 6.45 m ?

11. Diagram: Rest
9.0 m/s m
If it has reached its maximum displacement,
then it cannot go any higher.
Thus, it has come to rest (vf = 0)

12. List: Ref: Up slope is positive
vi = m/s Down is negative
vf = 0
a 
d = m
t ?
Make a list of the kinematic variables.
Be certain you get the right sign for the vector quantities.

13. List: Ref: Up slope is positive
vi = m/s Down is negative
vf = 0
a 
d = m
t ?
d = vi + vf t 2
This is the only equation that does not
have acceleration (a) in it.

14. d = vi + vf t 2
d = vi t
Since vf = 0, then remove this variable from the equation.

15. d = vi t 2
2 d = vi t
t = 2 d = (6.45 m)
vi m/s
= s

16. Ex. 2 A particle experiences an acceleration of 470 m/s2 North
for a total time of 960 milliseconds. If its displacement
is km South, then find the initial velocity.

17. List: Ref: North is positive
vi = ? South is negative
vf 
a = m/s2
d =  103 m = m
t = 960  10-3 s = s
Make a list of the kinematic variables.
Be certain you get the right sign for the vector quantities
and all quantities are in standard units.

18. List: Ref: North is positive
vi = ? South is negative
vf 
a = m/s2
d =  103 m = m
t = 960  10-3 s = s
Equation: d = vi t a t2
This is the only equation that does not have
the final velocity (vf) in it.

19. d = vi t a t2
d a t2 = vi t
vi = d a t2 t
= (-123 m) (470 m/s2) (0.960 s)2
0.960 s
= m/s
vi = m/s South

20. Practice Problems
Try these problems in the Physics 20 Workbook:
Unit 1 p. 10 #1 - 10