KINEMATIC EQUATIONS
New equations and how to use them!
KINEMATIC EQUATIONS
 Kinematic
Equations are considered to be
“equations of motion” and are based on the
fundamental definitions of average velocity
and acceleration:
d
v
t
v1  v2
v
2
v2  v1
a
t
OUR VARIABLES
 There
are 5 basic variables that are used in
any motion-related calculation:





Initial Velocity = v0 or vi or v1
Final Velocity = v or vf or v2
Acceleration = a
Displacement = d (sometimes also s )
Time = t
 Bold
face indicates a vector
 Each of the kinematic equations will use 4 of
these 5 variables
DERIVING THE EQUATIONS
 Each
of the kinematic equations starts with a
rearranged version of the equation for average
velocity:
d  v t
And uses substitution, rearranging, and
simplifying the equations to get to the end
result.
For example…
KINEMATICS EQUATION #1
 Step
1:
 Step 2: Substitute
equation for v
 Step 3: Rearrange
acceleration equation
to solve for t, then
substitute
 Step 4: Simplify by
multiplying fractions
 Step 5: Rearrange
d  v t
v v
v 1 2
2
t
v2  v1
a
 v1  v2 
d


t
→
 2 
v2  v1   v2  v1 


 2   a 
→d  
 v2 2  v12 

d  

 2a 
2ad  v2  v1
2
2
v2  v1  2ad
2
2
KINEMATICS EQUATION #2
 Step
1:
 Step
2: Substitute
 Step
3: Rearrange
acceleration equation to
solve for v, then
substitute
 Step 4: Simplify
 Step 5: Distribute the t
through the equation
 Step
6: Simplify again
d  v t
v v
v 2 1
2
 v2  v1 
d


t
→
 2 
 (v  at )  v1 
v2  v1  at → d   1
t
2


 2v  at 
d  1
t
 2 
 2v1t  at 2 

d  
2


1 2
d  v1t  at
2
SUMMARY OF EQUATIONS
v2  v1  at
v2  v1  2ad
2
2
1 2
d  v1t  at
2
 You
will NOT be required to memorize these 
HOW DO THESE RELATE TO OUR
LABS?
 The
equation of the displacement-time graph is:
d  vt  d1
The slope of this graph = velocity
The y-intercept of this graph = initial position
(displacement)
HOW DO THESE RELATE TO OUR
LABS?
 The
equation of the velocity-time graph is:
v  at  v1
The slope of this graph = acceleration
The y-intercept of this graph = initial velocity
PROBLEM SOLVING STRATEGY
 When
given problems to solve, you will be
expected to “show your work” COMPLETELY!
 “Showing work” means that you will be expected
to include the following pieces in your full answer
(or you will not receive full credit for the
problem…)
List of variables – include units on this list
 Equation – in variable form (no numbers plugged in yet)
 If necessary, show algebra mid-steps (still no numbers)
 Plug in your value(s) for the variables
 Final answer – boxed/circled with appropriate units
and sig figs

PRACTICE PROBLEM #1
A
school bus is moving at 25 m/s when the
driver steps on the brakes and brings the bus
to a stop in 3.0 s. What is the average
acceleration of the bus while braking?
v2 = 0 m/s
v2  v1  at
v1 = 25 m/s
v2  v1  at
t = 3.0 s
v2  v1
a= ?
a
t
0 m  25 m
s
s
a
3.0s
a = -8.3 m/s2
PRACTICE PROBLEM #2
 An
airplane starts from rest and accelerates at
a constant 3.00 m/s2 for 30.0 s before leaving
the ground.
(a) How far did it move?
(b) How fast was it going when it took off?
v2 = ?
v1 = 0 m/s
t = 30.0 s
a = 3.00 m/s2
d= ?
1 2
d  v1t  at
2
1
d  0  (3.00)(30.0) 2
2
d = 1350 m
v2  v1  at
v2  0  (3.00)(30.0)
v2 = 90.0 m/s