1. KKiinneemmaattiiccss 11DD
KKiinneemmaattiicc EEqquuaattiioonnss

2. Equation How Used
How altered when
objects start from
rest, v0 = 0
How altered when
v is constant,
a = 0
KKiinneemmaattiicc EEqquuaattiioonnss
1
2
v=v0+at
The kinematic equations can only be used in problems with
uniform (constant magnitude) acceleration.
For constant velocity we already have
v = Dx
t
Dx=v0t+
at2
v2=v0
2+2aDx
Dx=
1
2
at2
v=2aDx
v=at
Dx=v0t
v=v0
v=v0
Relates
displacement and
time
Relates velocity
and time
When time in not
mentioned

3. ZZeerroo’’ss,, PPlluusssseess,, aanndd MMiinnuusseess
These are often the trick to the problem and hidden in the text.
Zero Positives and Negatives
x0 Start at origin + Starts right of origin
− Starts left of origin
x Ends at origin + Ends right of origin
− Ends left of origin
Δx Returns to starting point + Final position is to the right of initial position
− Final position is to the left of initial position
v0 Initially at rest + Moving right
− Moving left
v Stops + Moving right
− Moving left
©
a Constant velocity
+ Moving right (+) and speeding up (+)
+ Moving left (−) and slowing down (−)
− Moving right (+) and slowing down (−)
− Moving left (−) and speeding up (+)
t N/A + Always

4. EExxaammppllee 11
An object initially at rest is displaced 150 m reaching a
final speed of 30 m/s.
a. Determine the object’s acceleration.
Making a variable list is a huge step in the right direction.
Look for hidden zero quantities, and picture a coordinate axis to assist in
determining the correct sign of each variable.
x0 = 0 m Never said where it started: Assume the origin.
x = Never given, but can be figured out from displacement.
Δx = 150 mGiven in the text of the problem.
v0 = 0 m/s Hidden in the words, “initially at rest.”
v = 30 m/s Given in the text of the problem.
a = ? The unknown.
t = Never given.
We can eliminate the first two variables since we already have displacement.
©

5. EExxaammppllee 11
An object initially at rest is displaced 150 m reaching a
final speed of 30 m/s.
a. Determine the object’s acceleration.
Δx = 150 m
v0 = 0 m/s
v = 30 m/s
a = ?
t =
Deciding on an equation
1. Type of motion?
If you answer constant velocity, use
If you answer acceleration continue.
2. Time mentioned?
If no, use
If yes continue.
3. Time and displacement mentioned?
©
If yes, use
4. Time and velocity mentioned?
If yes, use
v = Dx
t
v2 = v0
2 + 2aDx
Dx = v0t + 1
2
at2
v = v0 + at
Accelerating,
and time is not
mentioned.
v2 = v0
2 + 2aDx

6. EExxaammppllee 11
An object initially at rest is displaced 150 m reaching a
final speed of 30 m/s.
a. Determine the object’s acceleration.
2 + 2aDx
©
Δx = 150 m
v0 = 0 m/s
v = 30 m/s
a = ?
t =
Solve showing the following work
1. Write the original equation (or an equivalent).
v2 = v0
2. Rearrange to solve for requested variable.
3. Show substitution of given values.
4. Solve boxing answer and including correct
units.
Accelerating,
and time is not
mentioned.
v2 = v0
2 + 2aDx
a =
v2 - v0
2
2Dx
a =
(30)2 - (0)2
2(150)
a = 3 m s2

7. EExxaammppllee 11
An object initially at rest is displaced 150 m reaching a
final speed of 30 m/s.
©
b. Determine the time of the motion.
Δx = 150 m
v0 = 0 m/s
v = 30 m/s
a =
t =
Acceleration from part (a) can be added to the list.
1. I’m going to choice to easier (non-quadratic)
formula.
2. Rearrange to solve for requested variable.
3. Show substitution of given values.
4. Solve boxing answer and including correct units.
Now time is
mentioned, and
both velocity and
displacement are
known.
t =
v - v0
a
t =
(30)- (0)
(3)
t = 10 s
Dv=v0+at
1
2
x=v0t+
at2
v=v0+at
3 m/s2

8. EExxaammppllee 22
An object moving at 32.0 m/s decelerates to a stop in a
distance of 240 m. Determine the objects acceleration.
©
Δx = 240 m
v= 32.0 m/s
0 v = 0 m/s
a =
t =
Accelerating, and
time is not
mentioned.
v2 = v2 + 2aDx
0
v2 = v0
2 + 2aDx
a =
v2 - v0
2
2Dx
a =
(0)2 - (32.0)2
2(240)
a = -2.13 m s2
Acceleration has two signs.
It is a vector, so direction of
motion influences its sign.
Since no direction was given
I assumed the object was
moving in the +x direction.
It did not reverse. Direction
remained positive.
Acceleration is also a rate. If
velocity is increasing
(speeding up) acceleration is
positive, and if decreasing
(slowing) it is negative.
Here the object was
decelerating in the positive
direction. − + a = −a .

9. EExxaammppllee 33
An object moves at a constant 15 cm/s for 1 minute.
Determine the objects displacement.
©
v = Dx
t
Dx = 9.0 m
Ask the first question.
Type of motion?
Constant velocity
Don’t waste time with the
kinematic equations and
all those variables.
It’s the simple equation
from the summer
homework.
However, watch out for
conversions !!!!!!!!!!!!!
Dx = vt
Dx = (0.15)(60)

10. An object uniformly accelerates from rest reaching a
speed v while moving a distance x . Determine its speed
when it has moved a distance 2x . Answer in terms of v .
No values, but you can still use a variable list. Simply
check of any variable mentioned or implied.
An object uniformly accelerates check off aa
from rest vv00 == 00
reaching a speed v check off vv
while moving a distance x check of ΔΔxx
How would final speed change if x is doubled to 2x ?
Double x in the derived equation and determine the
multiplier for v that will maintain the equality.
Doubling x increases the right side by root two.
Therefore, the left side must also be multiplied by
root two.
EExxaammppllee 44
©
Δx =
v0 =
v =
a =
t =
Accelerating.
Time not
mentioned.
00 (rreesstt)
v2 = v0
2 + 2aDx
Substitute v0 = 0 ,
and rearrange in
terms of v .
(? v)= 2a(2x)
v = 2ax 2v v = 2ax

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1st Using the Quadratic Equation 2nd Avoiding the Quadratic Equation
©
2 + 2a x - x0 Solve for v : ( )
Result: Two velocities (parabolic).
Positive velocity is +x motion and the
negative velocity is −x motion. Decide
the direction of the final velocity and
choose the correct velocity.
2 + 2a x - x0 ( )= v0 + at
± v0
2 2 ( )
0 0 Time unknown: v = v + 2a x - x
v = ± v0
Dx = v0t + 1
Main equation: at2
2
t =
2 - 2a(-Dx)
a
-v0 ± v0
1
2
a
æ
è ç
ö
ø ÷
t2 + v0 ( )t + (-Dx)= 0
Quadratic:
t =
æ
- v0 ( )± v0 ( )2 - 4 1
2
a
è ç
ö
ø ÷(-Dx)
æ
2 1
2
a
è ç
ö
ø ÷
Result: Two times
Why? Quadratic equations are
parabolic with 2 x points for every
y .
Which time is correct? Choose
positive nonzero times. However, if
there are two positives it can be
tricky to pick.
0 Substitute into: v = v + at

12. EExxaammppllee 55
An object is moving in the +x direction at 25 m/s. At the
instant it passes through the origin it encounters an
acceleration of
3.0 m/s2 directed opposite its motion. The acceleration ceases to
act when the object is 6.0 m to the left of the origin. Determine
the elapsed time of the motion.
First we’ll try solving this using the quadratic method.
©
Δx =
v0 =
v =
a =
t =
6.0 m to the left
Begin by identifying key variables and their correct signs.
−6.0 m
s
+25 m/s
3.0 m/s2 directed opposite
−3.0 m/s2
? s
Determine
the elapsed time
Dx = v0t +
1
2
at2
(-6.0)= (25)t + 1
2
(-3.0)t2
-1.5t2 + 25t + 6 = 0
t =
-b ± b2 - 4ac
2a
t =
-(25)± (25)2 - 4(-1.5)(6)
2(-1.5)
t =
-25 ± 625 + 36
-3
t = -0.24s or 17s
Negative time
makes no sense in
real life.
Mathematically it is
a point on a
parabola if time
had run backwards.
Ignore the negative,
and choose the
positive value.

13. EExxaammppllee 55
An object is moving in the +x direction at 25 m/s. At the
instant it passes through the origin it encounters an
acceleration of
3.0 m/s2 directed opposite its motion. The acceleration ceases to
act when the object is 6.0 m to the left of the origin. Determine
the elapsed time of the motion.
Now let’s try the other two equations, avoiding the quadratic.
©
Δx =
v0 =
v =
a =
t =
6.0 m to the left
−6.0 m
s
+25 m/s
3.0 m/s2 directed opposite
−3.0 m/s2
? s
Determine
the elapsed time
v2 = v0
2 + 2aDx
2 + 2aDx
v = ± v0
The quadratic is technically
superior, but this method
has major advantages.
1.Same amount of work.
2.Velocity is calculated.
3.Choosing the correct sign
on velocity is just a matter
of direction.
4.Students mess up the
quadratic 99% of the time.
It’s harder to get all the
signs correct in the
quadratic.
v = ± (25)2 + 2(-3.0)(-6.0)
v = 25.7m/s or - 25.7m/s
If the object is to the left of
the origin it must be moving
left. Choose the negative
velocity.
v = v0 + at
t =
v - v0
a
t =
(-25.7)- (25)
(-3.0)
t = 17 s

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eevveennttuuaallllyy ttuurrnn uupp..””
©
v = v0 + at
v2 = v0
2 + 2aDx
v = v0 + at

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ffoolllloowwss..
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aanndd aadddd tthheemm ttooggeetthheerr..
©

16. EExxaammppllee 66
This problem has three distinct parts, with three different accelerations
“constant velocity”
“slows to a stop”
800 m 180 m
©
Finally the car slows
to a stop in a distance of 180 m.
Δx =
v0 =
v =
a =
t =
Δx =
v0 =
v =
a =
t =
“ 2.4 m/s2 ”
Δx =
v0 =
v =
a =
t =
A car is at rest at a traffic signal. When the light turns green
the car accelerates at 2.4 m/s2 for 15 s.
0 m/s
2.4 m/s2
Then the driver
maintains a constant velocity for 0.80 km.
0 m/s2
15 s
−? m/s2
What other variables can be discerned from the problem above?
Are there any important facts we should keep in mind?
0 m/s
The final velocity of one phase becomes the initial velocity for the next.
Constant velocity means that initial and final velocities are equal.

17. A car is at rest at a traffic signal. When the light turns green
the car accelerates at 2.4 m/s2 for 15 s. Then the driver
maintains a constant velocity for 0.80 km. Finally the car slows
to a stop in a distance of 180 m.
800 m 180 m
−? m/s2
EExxaammppllee 66
−3.6 m/s2
Δx =
v0 =
v =
a =
t =
22 s 10 s
Constant Velocity
Equation
©
0 m/s
36 m/s
v = v0 + at
v = (0)+ (2.4)(15)
v = 36m s
36 m/s
36 m/s
36 m/s
Dx = v0t + 1
2
at2
Dx = (0)(15)+ 1
2
(2.4)(15)2
Dx = 270m
270 m
v =
Dx
t
(36)=
(800)
t
t = 22m s
v2 = v0
2 + 2aDx
(0)2 = (36)2 + 2a(180)
a = -3.6m s2
v = v0 + at
(0)= (36)+ (-3.6)t
t = 10s
Δx =
v0 =
v =
a =
t =
Δx =
v0 =
v =
a =
t =
2.4 m/s2 0 m/s2
15 s
0 m/s

18. A car is at rest at a traffic signal. When the light turns green
the car accelerates at 2.4 m/s2 for 15 s. Then the driver
maintains a constant velocity for 0.80 km. Finally the car slows
to a stop in a distance of 180 m.
800 m 180 m
−? m/s2
EExxaammppllee 66
ttotal = t1 + t2 + t3 = (15)+ (22)+ (10)= 47 s
©
0 m/s
36 m/s
36 m/s
36 m/s
36 m/s
270 m
−3.6 m/s2
Δx =
v0 =
v =
a =
t =
22 s 10 s
Δx =
v0 =
v =
a =
t =
Δx =
v0 =
v =
a =
t =
2.4 m/s2 0 m/s2
15 s
0 m/s
To find the total time sum the times of all three phases.
To find the total displacement sum the displacements of all three phases.
Dxtotal = Dx1 + Dx2 + Dx3 = (270)+ (800)+ (180)= 1250m

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aannootthheerr aanndd ssuuppppoorrtt oonnee aannootthheerr..
©
Dx = v0t + 1
2
at2

20. DDeerriivviinngg tthhee KKiinneemmaattiicc EEqquuaattiioonnss
at2 æ
©
Dx = v0t + 1
2
at2
x = x0 + v0t + 1
2
at2
Dx = v0t + 1
2
at2
Constant Velocity
a = 0
Dx = v0t + 1
2
(0)t2
Dx = v0t
Constant Velocity
v = v0
Dx = vt
v =
Dx
t
Velocity is the
derivative of position
with respect to time.
Need to use position,
not displacement.
v = dx
dt
v =
d x0 + v0t + 1
2
è ç
ö
ø ÷
dt
v = v0 + at
Combine with
v = v0 + at
rearranged as
t =
v - v0
a
Dx = v0
v - v0
a
æ
è ç
ö
ø ÷
+ 1
2
a
v - v0
a
æ
è ç
ö
ø ÷
2
v2 = v0
2 + 2aDx