The Kinematics Equations
(1D Equations of Motion)
Day #1
Introduction to the Equation of Motion
What are the 5 key equations we have
learned thus far?
 A) ∆x = xf – xi
finding a change in displacement
 B) ∆v = vf – vi
finding a change in velocity
 C) s = d/∆t
finding speed (distance/time)
 D) v¯ = ∆x/∆t
finding velocity (displacement/time)
 E) a¯ = ∆v/∆t
finding acceleration (change in velocity
change in time)
There are 4 new equations for motion.
 We’ll start with a = ∆v/∆t
 Substitute in ∆v = vf – vi
¯
 The formula is now
a = vf-vi
t
 Rearrange to get a∙t = vf - vi
Remember…
slope of the line =
a¯ = vf-vi
t
Example #1
 A car starting from rest accelerates uniformly to a speed of 75
m/s in 12 seconds. What is the car’s acceleration?
 First, identify the known values, then plug and chug
•Vi = 0
• Vf = 75 m/s
•t = 12 s
•a = ?
*always make
a list of your
knowns
a = vf-vi or a∙t = vf - vi
t
= 75 m/s – 0
12s
=+6.25 m/s2
Example #2
 A spacecraft traveling at 1025 m/s is uniformly
accelerated at a rate of 105 m/s2 for 9 seconds. What is
the final velocity of the spacecraft?
 First, identify the known values, rearrange the formula,
then plug and chug
•Vi = 1025 m/s
• Vf = ?
•t = 9 s
•a = 105 m/s2
a = vf-vi or a∙t = vf - vi
t
Vf= Vi + at
= 1025 m/s + 9s(105m/s2)
= 1970 m/s
The second new formula
Recall: area under the curve
on a v/t chart gives us
displacement = ∆x (or d)
Consider a random, uniform
accl object.
Use the area of a trapezoid
area  h(b1  b2 )
x  12 t (v1  v2 )
1
2
Example #3
 A plane is moving at a speed of 500m/hr when it
lands on a runway. Accelerating uniformly, it comes
to a stop after covering 15m. How long did it take to
stop?
•Vi = 500 m/s
• Vf = 0
•t = ?
•a = ?
•∆x = 15 m
∆x =½∆t(vf+vi)
15 m = ½(t)(0+500 m/hr)
30 m = 500m/hr(t)
t = 0.06 hr
= 216 sec
Example #3
 A plane is moving at a speed of 500m/hr when it
lands on a runway. Accelerating uniformly, it comes
to a stop after covering 15m. How long did it take to
stop? What was its acceleration?
•Vi = 500 m/hr
• Vf = 0
•t = 0.03 hr or 108 s
(previous slide)
•a = ?
•∆x = 15 m
a = vf-vi
t
= 0-500m/hr
0.06 hr
=-8333.33 m/hr2
Or -0.0006 m/s2
HW
 In book,
 Complete pg 53,55, & 58 odds
Day #2
 RE-Introduction to the equation of motion
a
v f  vi
t
v f  vi  at
…. area under curve = x
area  Atriangle  Arectangle  bh  BH
1
2
x  t (v f  vi 
)  t (vi )
1
2
x  at  v1t
1
2
2
2.
a  2 sm2
vi  8 ms
x  12 at 2  vi t

100  (2)t  8t
1
2
2
0  t 2  8t 100


You have to use the
quadratic formula to
solve the problem. 
x  100m
The first answer doesn’t make
sense 
because you can’t have a
negative time.
t  14.77sec, 6.77sec
a = vf-vi (rearrange)
t
v f  vi  at
Substitute our newest formula into the
derivation below
v   v  at 
2
x  at  vi t
1
2
2
2
f
i

v f  vi  2vi at  a t
2
2
2 2
v f  vi  2a(vi t  at )
2
2
1
2
2
v f  vi  2a (x)
2
2
Make a decision as to which direction is POSITIVE & which is NEGATIVE
List what you have (using “COMPATIBLE” units & using proper SIGNS)
choose your equation
Solve the equation
Make sure your answer makes sense (both in MAGNITUDE & DIRECTION)
3.
vi  40 ms
v f  20 ms
v f  vi  2ax
2
2

202  402  2a(50)
a  12 sm2

x  12 t (vi  v f )

50  12 t(40  20)
t  1.67sec

x  50m
4.
vi  0 ms
v f  60mph  26.82 ms
v f  vi  at
26.82  0  a(3.5)
t  3.5sec

a  7.66 sm2

x01  12 at 2  vi t  12 (7.66)(1) 2  0(1)  3.83m

x0 2  12 at 2  vi t  12 (7.66)(2) 2  0(2)  15.32m
d01 3.83m

d12 15.32m  3.83m 11.49m

Review – Solve
 Solve the following:
 You roll a ball up an incline at a speed of 4.5 m/s. After 5
sec, the ball is on the way down travelling at a new speed
of 1.5 m/s. Find the ball’s acceleration. (Sketch)

-1.5-4.5
-6
5
5
rookie mistake…. 1.5-4.5
-1.2 m/s2
 Find the time to the peak.
 -1.2 = 0-4.5
-1.2 = -4.5
=3.75 sec
t
t
How far up does it go?
0 = (4.5)2 + 2(-1.2)d
=8.44 m
Day #3
 Free Fall
Free Fall
 Free fall problems are acceleration problems where we
know the acceleration.
 For all free fall problems (at the surface of the earth),
 a = 9.8 m/s2 (down)
the object speeds up by 9.8m/s2 after every second that
it falls
 When solving:
 Make a chart
 Use arrows next to variables to show the direction of the
vectors
 Fill in known variables
 Solve for unknowns
Free Fall Problems
 Initial Velocity (Vi) will be 0, so our formulas change
slightly. No need to memorize these..
x  at  vi t
x  at
v f  vi  2a(x)
v f  2a(x)
2
1
2
2
2
a = vf-vi
t
1
2
2
a = vf
t
2
9.8
2
m/s
0 m/s
9.8 m/s
19.6 m/s
29.4 m/s
39.2 m/s
49 m/s
Free Fall Problems
 1. A ball is dropped from the top of a bridge, and it takes 8
seconds to hit the water.
 A. How tall was the bridge?
 B. How fast was the ball going just before it hits the water?
a
9.8m
/s2
vi
0
a. height? So find ∆x.
∆x = vi ∆t +1/2(a)(∆t )2
= ½(9.8)(8) 2
= 313.6 m
b. a = (vf-vi)/t
9.8= vf/8
= 78.4 m/s
downward
vf
∆x
∆t
8s
*You could solve part B first and then use the
equation ∆x=1/2t(vi+vf )
vi= 0
a
9.8m/
s2
vi
0
a = -9.8 m/s2
x = -20 m
v f  vi  2ax
2
2
vf
∆x
20m
∆t
v f  02  2(9.8)( 20)
2
v f   392
v f  19.8 ms
Because the diver is falling
DOWNWARD, we can
also use a negative sign
 If all objects accelerate at the same rate when falling
near the earth’s surface, why do some objects actually
hit the ground faster than others? (assuming they are
dropped at the same time from the same height)?
 AIR RESISTANCE!!
 Some objects are narrower (like pencils) and therefore
the air resists their falling less.
 Some things are wider (like frisbees) and the air provides
more resistance and slows down their fall.
 NOTICE HOW THE MASS OF THE OBJECTS DOES
NOT MATTER!
5.
y
v
t
Sometimes the variable “y”
is used instead of “x” to
simply show that the object
is moving vertically
t
Notice that the object speeds up
Notice that the acceleration is
CONSTANT and NEGATIVE the ENTIRE
way
Day #4
 Throw-ups, Come-downs, Throw-downs
Free Fall Problem – Throw up
 “A ball is thrown up from the ground…”
 In this type of problem, the initial velocity is always up. Also, the primary
type of problem will ask “what is the maximum height”, which means vf = 0
m/s.
 2. A ball is thrown up from the ground at an initial speed of 20 m/s.
 A. what is the maximum height the ball reaches?
 B. How long before it gets to the max altitude?
A. ∆x =?
B. ∆t =?
Vf2=vi2 +2a ∆x
a=(vf-vi)/ ∆ t
0=2o2 +2(-9.8) ∆x
-9.8=(0-20) ∆ t
2
a -9.8m/s
∆x = 20.4m
∆ t=2.04s
vi
20 m/s
vf 0
∆x
∆t
*You can solve for B first
then use the formula
∆x =1/2t(vi+vf)
Problem #2
2
UP +, DOWN a = -9.8 m/s2
v2 = 0 (at the top)
a)
vi = 20 m/s
a = -9.8 m/s2
vf = 0 m/s
1
b)
v f  vi  2ax
2
2
0  20  2(9.8)x
x  20.41m
2
2
v f  vi  at
c)
x 
11

1
2
1
2
0  20  ( 9.8)t
t  2.04 sec
at 2  vi t
( 9.8)t
2
 20t
t= 0.66s and 3.42 sec
(going up)
(going down)
To solve for
“t”, use the
quadratic
formula 
Throw –up/Come down
 Assume:
3 scenarios
 Up+, Down –
 A =-9.8m/s2
 v2= 0 (at the top)
 We’ll start with scenario #3. V3=-V1
1.
2.
3.
Scenario 3 – object is thrown up and caught at the same height
2
b)
a)
vf = 0 m/s
a = -9.8 m/s2
x = 30 m
vf = 0 m/s
a = -9.8 m/s2
v i= 24.25 m/s
v f  vi  2ax
2
t  2.47 sec
2
3
1
t total  2(2.47sec)  4.94 sec
2
0  vi  2(9.8)(30)
m
vi   588   24.25
s
2
v f  vi  at
0  24.25  ( 9.8)t
c) Due to the fact that the ball is
thrown and caught at the same
height …
v 3  v1  24.25 ms
Scenario 2 and part d of our problem
2
vi= 24.25 m/s
3
1
4
a = -9.8 m/s2
x = -1 m (because down is -)
vf2=vi2 +2a∆x
vf2=24.252 +2(-9.8)(-1)
24.65 m/s down
Or-24.65m/s
What if we were solving for the time it was when the object
hits the ground?
vi= 24.25 m/s
a = -9.8 m/s2
x = -1 m (because down is -)
x  12 at 2  vi t
 1  12 (9.8)t 2  24.25t
t   .04 sec,4.99 sec
To solve for “t”, use the
quadratic formula 
Only positive times make
sense
Day #5
 Chase Problems
Chase Problems
 A chase problem is a scenario where two objects are
involved, and they have the same position at some
later time.
 They can
 Both start at the same place and same time
 Start at different places but at the same time
 Both start at the same place but at different times
 Start at different places at different times
 B or D but be moving in different directions
The Key Equation
 ∆x=1/2at2+vit
∆x=xf-xi
 Substitute the second formula into the first, creating:
x2=1/2at2+vit+xi
 This formula will give you the final position for both
objects. Since you don’t know that final position, set the
two formulas equal to each other.
Chase Problem #1
 Timmy is running at a constant speed of 6m/s. He sees Susie running
50m in front of him. She is moving at a constant speed of 4 m/s, in the
same direction. How long will it take for Timmy to catch Susie?
 Notice that while Timmy and Susie have different starting positions
(xiT=0m while xiS=50m), they have the same final position (xf ) after
Timmy catches Susie.
.
.
X
xiT
xiS
x2
Tim
1/
=
Susie
2+v t+x = 1/ at2+v t+x
at
2
i
i
2
i
i
½(0)t2+6t+0 = 1/2(0)t2+4t=50
6t = 4t=50
t = 25sec
Chase Problem #2
 A continuation of the last problem.
 Timmy passes Susie, running at his constant speed of 6m/s. Susie decides
to pick up the pace, very gradually. She begins to accelerate at a constant
rate of 0.1m/s2. How long will it take her to catch Timmy?
 Notice that Timmy and Susie have the same starting position (xiT=0m xiS=0m)
and ending position (xf ) after Susie catches Timmy.
Really, you could have
made the initial
position anything you
wanted – they cancel
out. I decided to
choose 0.
.
xit=xis
Tim
=
Susie
1/ at2+v t+x = 1/ at2+v t+x
2
i
i
2
i
i
½(0)t2+6t+0 = 1/2(0.1)t2+4t+0
6t=0.05t2+4t
6=0.05t+4
t=40sec
x
x2
Chase Problem #3
 A man drops a penny off the top of a 100m tall building. Exactly 1
+
0
second later another man throws a nickel downward from the same
place as the first man. What is the minimum speed the nickel must be
thrown at in order to catch the penny?
 Start with the time it takes the penny to hit the bottom using the
formula:
∆x=1/2at2+vit
-100=1/2(-9.8)t2+0t
t= 4.52 sec
-
penny
=
nickel (1 sec later)
1/ at2+v t+x = 1/ at2+v t+x
2
i
i
2
i
i
½(-9.8)(4.52)2+0(4.52)+0 = 1/2(-9.8)(3.52)2+vi(3.52)+0
Vi= -11.19 m/s (neg. because it’s going
down)
Chase Problem #4 really hard
 In a strange, yet exciting, crash-test-dummy crash, two cars start by
facing each other 1000m apart on a straight road. The first car
accelerates from rest with a constant accel of 4m/s2. The second car
accel, at a rate of 8 m/s2 for 5 sec but then settles into a constant speed.
Find the elapsed time before these two cars collide.
 Notice that the dummies have different starting positions (xiA=0,
xiB=1000m), but they have the same final position (x2) when they crash.
 Start by finding the final velocity of car B by using the following formula:
 a = vf-vi
-8=(vf-0)/5
so vf =-40m/s
t
1/
2+v t+x = 1/ at2+v t+x
at
2
i
i
2
i
i
Test Review
Step #1
Write Down What You
Have (Look for “Key” Words)
“Coming to a stop”
“Starting from rest”
“Coasting”
“Maximum Height”
“Dropped”
vf = 0
vi = 0
vi = vf = constant
vf = 0
a = -9.8 m/s2
“Slowing Down”
a = - __
“Braking”
“Speeding up”
“Accelerating from rest”
a = + __
Step #3
Solve the Equation
Step #4
Make sure your answer
makes sense
Helpful Tip #1
Choose your “Key
Points” in every
problem…and do so
wisely.
Vertical Problems
1
“Throwdowns”
“Free-fall”
vi = 0
2
1
vi 0
2
“Throw up”
2
vf = 0 (at top)
1
Vertical Problems
“Throw up / Come Down”
(throw and catch at same height)
2
vf = 0 (at top)
v1 = -v3
1
3
“Throw up / Come Down”
(throw and catch at different heights)
Use x  (9.8)t  vi t and solve
quadratically for “t”
1
2
2
2
2
x13 = 
3
1
1
x13 = 
3
Helpful Tip #2
Assign positive and
negative to different
directions.
Helpful Tip #3
When solving a
quadratic equation,
do so with minimal
effort.
Solving
a
quadratic
equation
Choice B
Choice A
Factoring
b  b  4ac
2a
2
Unlikely on a physics problem 
Choice C

Trace  Zero
(on graphing calculator)
nd
2
x  at  v1t
1
2
2
Can be used at constant
speeds (a=0) or when
accelerating. Awesome Dude!
Chase
Problems
Since the two objects
(A and B) end up at the
same position by the end of
the chase, use …
x 2A  x 2B
1
2
atA  v1A t A  x1A  at B  v1B t B  x1B
2
1
2
2
But
what
if…
The objects start at
different places?
x 2A  x 2B
1
2
atA  v1A t A  x1A  at
2
 v t  x1B
2
1
2
B
1B B
It’s already accounted for
here and here 
But
what
if…
The objects start at
different TIMES?
x 2A  x 2B
 v t  x  at B  v1B t B  x1B
2
1
1
2
A
1A A
1A
2
You’ll need to use an extra equation
relating the two times. Plug this new
equation into the long equation
above.
at
2
Example: tA = tB + 1
Helpful Tip #6
It is always important to
remember that when
something is thrown up or
down (or simply falls), the
acceleration at ALL times is
constant.
+
-
The acceleration of the ball at
EVERY point on this red path
(When it’s rising up, when it’s
stopped, when it’s falling
down) is always -9.8 m/s2.
An object thrown up has a constant
acceleration at ALL times….
….the acceleration due to gravity
x
Objects rise and fall in the
same amount of time
(assuming no parachute )
t
v
t
Constant slope = constant accel.
TONIGHTS HW
Complete Review Worksheet