Position-time and velocity-time graphs Uniform

Position-time and velocity-time graphs
Uniform motion problems – algebra
Acceleration and displacement
Topics:
 The kinematics of motion in
one dimension: graphing and
calculations
 Problem-solving strategies
 Free fall
Sample question:
Horses can run much much faster than humans, but if the length
of the course is right, a human can beat a horse in a race. When,
and why, can a man outrun a horse?
Speed, Velocity, & Acceleration
 Speed (v) – how fast you go (scalar)
 Velocity (v) – how fast and which way;
the rate at which position changes (vector)
 Average speed ( v ) – distance / time
 Acceleration (a) – how fast you speed
up, slow down, or change direction;
the rate at which velocity changes
 During your 8 mi. trip, which took 15 min., your
speedometer displays your instantaneous speed,
which varies throughout the trip.
 Your average speed is 32 mi/hr.
 Your average velocity is 32 mi/hr in a SE direction.
 At any point in time, your velocity vector points
tangent to your path.
 The faster you go, the longer your velocity vector.
Acceleration
Acceleration – how fast you speed up, slow down, or
change direction; it’s the rate at which velocity
changes. Two examples:
t (s)
v (mph)
t (s)
v (m/s)
0
55
0
34
1
57
1
31
2
59
2
28
3
61
3
25
a = +2 mph / s
a = -3 ms/ s = -3 m / s 2
Velocity & Acceleration Sign Chart
VELOCITY
A
C
C
E
L
E
R
A
T
I
O
N
+
-
+
-
Moving forward;
Moving backward;
Speeding up
Slowing down
Moving forward;
Moving backward;
Slowing down
Speeding up
Displacement is
a change of position
in a certain direction,
not
the total distance traveled
Motion
 Takes place over time
 Depends upon the frame of reference
 What is frame of reference?
 How do we choose one?
 Once we decide upon a frame of
reference, it remains fixed for the problem
Keep in mind…
 Displacement is not always equal to the
distance traveled!
 Displacement can be positive or negative!
 For motion along x or y axis, the displacement is
determined by the x or y coordinate of its final
position. Example: Consider a car that travels 8 m,
E then 12 m, W.
Net displacement Δx, is
from the origin to the final
position:
Δx
8 m,E
Δx = 4 m, W
What is the
distance traveled?
d = 20 m !!
x = -4
x = +8
12 m,W
x
Displacement
 When things start moving, the length of a
straight line drawn from the object’s initial
position to it’s final position is it’s
displacement
 In one dimension…
x  x f  xi
Change in position along x-axis = (final position on x-axis) – (initial position on x-axis)
Motion in One Dimension
 Same concepts apply to any axis!
y  y f  yi
Change in position along y-axis = (final position on y-axis) – (initial position on y-axis)
POSITIVE AND NEGATIVE
QUANTITIES
ARE JUST DIRECTIONS
IN MOTION PROBLEMS!
Positiontime
graphs
the relationship
between the
shape of a p-t
graph and the
motion of the
object.
consider a car moving with a constant,
rightward (+) velocity of +10 m/s.
If the position-time data for
the car were graphed, it
would look like the graph at
the right.
Now consider a car moving with a rightward (+),
changing velocity - that is, a car that is moving
rightward but speeding up or accelerating.
If the position-time data for such
a car were graphed, the graph
would look like the graph at the
right.
.
Position-time graphs for the two types of
motion - constant velocity and changing
velocity (acceleration).
Constant Velocity
Positive Velocity
Positive Velocity
Changing Velocity
(acceleration)
Motion diagram (student walking to school)
Provide a reasonable explanation for the motion diagram.
Draw a position-time graph.
(time on the x-axis)
Motion diagram (student walking to school)
Table of data
Graph
Interpreting position-time graphs
Starting at t = 0:
• Position away from the origin in + direction has + velocity
• Position away from the origin in – direction has a – velocity
• No change in position – 0 velocity
x
p-t graph
B
Describe this motion.
A
t
C
A … Starts at home (origin) and goes forward slowly
B … Not moving (position remains constant as time
progresses)
C … Turns around and goes in the other direction
quickly, passing up home
E
x
A
C
B
THS
D
W
Practice problems 1-3, pg. 85
Uniform motion (constant velocity)
Equal displacements occur during
successive equal time intervals.
the slope of the line on a position-time
graph reveals useful information about
the velocity of the object.
The importance of Slope
The slope of the line on a position-time graph reveals
useful information about the velocity of the object.
Whatever characteristics the velocity has, the
slope will exhibit the same (and vice versa). If the
velocity is constant, then the slope is constant (i.e.,
a straight line). If the velocity is changing, then the
slope is changing (i.e., a curved line). If the
velocity is positive, then the slope is positive (i.e.,
moving upwards and to the right). This very
principle can be extended to any motion
The object represented by the graph on the right is traveling
faster than the object represented by the graph on the left.
Slow, Rightward(+)
Constant Velocity
Fast, Rightward(+)
Constant Velocity
Consider the two graphs below. Both graphs show plotted points forming a curved line.
The curved line of changing slope is a sign of accelerated motion (i.e., changing velocity).
Applying the principle of slope to the graph on the left, the object depicted by the graph is
moving with a negative velocity (since the slope is negative ). The object is starting with a
small velocity (small slope) and finishes with a large velocity. This object is moving in the
negative direction and speeding up.
Negative (-) Velocity
Slow to Fast
Leftward (-) Velocity
Fast to Slow
The graph on the right also depicts negative velocity (since there is a negative
slope). The object begins with a high velocity (the slope is initially large) and
finishes with a small velocity (since the slope becomes smaller). So this object
is moving in the negative direction and slowing down.
Applying algebra to uniform motion
The slope of the line on a position-time graph
is equal to the velocity of the object.
If the object is moving with a velocity of +2 m/s, then the slope of the
line will be +2 m/s.
Slope = rise/run
For p-t graph
rise = ∆d run = ∆t
slope of the line = ave. velocity
V ave = ∆d = d1 – d0
∆t
t1 – t0
ave. velocity is ∆d
∆t
NOT
d*
t
*instantaneous speed
A va = 2 - (-2) = 4 m = 10 m/s
0.4 – 0
.4
A
4
2
m
0
-2
-4
B
0.2
book, pg.86
0.4
0.6
0.8
s
B va = -4 - (-2) = -2 m = -5 m/s
0.4 – 0
.4
For Uniform motion
the derivative:
Vave = ∆d = d1 – d0
∆t
t1 – t0
When t0 = 0, it can be eliminated from the problem
Vave = d1 – d0
t1
d1 = Vave (t1) + d0
Vave = v because constant velocity – so no need for ave.
d = vt + d0
d0 represents position at t = 0
An airplane is moving at a uniform velocity of 75 m/s from a
starting position of 40 m. Find the position of the airplane after 2.5
seconds.
d = vt + d0
d = 40 m + (75 m/s)(2.5 s)
d = 230 m
practice problems 9-12, pg. 89
The Cheetah: A cat that is built for speed. Its
strength and agility allow it to sustain a top speed of
over 100 km/h. Such speeds can only be maintained
for about ten seconds.
graphing velocity
Definition of Speed
 Speed is the distance traveled per unit of
time (a scalar quantity).
d = 20 m
A
Time t = 4 s
B
vs =
d
t
=
20 m
4s
s = 5 m/s
Not direction dependent!
Definition of Velocity
Velocity is the displacement per unit
of time. (A vector quantity.)
s = 20 m
B
Δx=12 m
A
20o
Time t = 4 s
= 3 m/s at 200 N of E
Direction required!
Average Speed and
Instantaneous Velocity
 The average speed depends ONLY on
the distance traveled and the time
required.
A
s = 20 m
C
Time t = 4 s
B
The instantaneous
velocity is the magnitude
and direction of the
velocity at a particular
instant. (v at point C)
Example: A runner runs 200 m, east, then changes
direction and runs 300 m, west. If the entire trip takes 60 s,
what is the average speed and what is the average velocity?
Recall that average speed is a
function only of total distance
and total time:
s2 = 300 m
s1 = 200 m
start
total path 500 m
Average speed 

time
60 s
Avg. speed= 8 m/s
Direction does not matter!
Now we find the average velocity, which is the net
displacement divided by time. In this case, the direction
matters.
t = 60 s
x = -100 m
x1= +200 m
xo = 0
xo = 0 m; x = -100 m
Direction of final
displacement is to the left as
shown.
Average velocity:
Note: Average velocity is directed to the west.
Graphing velocity: uniform motion (constant velocity)
is represented by a horizontal line on a v-t graph.
B
C
A
D
Increasing speed?
V (m/s)
Constant speed?
Increasing speed?
Backwards and increasing
speed?
+
0
Backwards at a constant
speed?
Decreasing speed?
? What does the intersection of 2 lines on a v-t graph tell us?
Note: v-t graphs give NO information about position.
Checking Understanding
Here is a motion diagram of a car moving along a straight stretch
of road:
Which of the following velocity-versus-time graphs matches this
motion diagram?
A.
B.
C.
D.
A graph of position versus time for a
basketball player moving down the
court appears like so:
Which of the following velocity graphs matches the above
position graph?
A.
B.
C.
D.
Slide 2-15
A graph of velocity versus time for a
hockey puck shot into a goal appears
like so:
Which of the following position graphs matches the above
velocity graph?
A.
B.
C.
D.
slope
displacement vs. time
slope
velocity vs. time
Area underneath = Δd
acceleration vs. time
Area underneath = Δv
slope
displacement vs. time
slope
velocity vs. time
Area underneath = Δd
area:
slope:
acceleration vs. time
Area underneath = Δv
p-t
v-t
a-t
time (s)
time (s)
time (s)
none
m/s x s = m = Δd
m/s = v
m/s/s = a
Rate at which
displacement
changes
Rate at which
velocity changes
m/s2 x s = m/s = Δv
Acceleration
The rate of
change of
velocity
Think about this....
What are three ways to change the velocity of a car?
Accelerate
Decelerate
Change direction
average acceleration is the rate of change in velocity
between t0 and t1
a = ∆v
∆t
acceleration = change in velocity
change in time
acceleration (m/s2) = (vf) - (vi)
time
If a car moves at a
constant velocity, then its
acceleration is zero
Checking Understanding
These four motion diagrams show the motion of a particle along
the x-axis. Which motion diagrams correspond to a positive
acceleration? Which motion diagrams correspond to a negative
acceleration?
Slide 2-22
Checking Understanding
These four motion diagrams show the motion of a particle along
the x-axis. Rank these motion diagrams by the magnitude of the
acceleration. There may be ties.
Slide 2-21
Checking Understanding
These six motion diagrams show the motion of a particle along
the x-axis. Rank the accelerations corresponding to these motion
diagrams, from most positive to most negative. There may be
ties.
Slide 2-23
Acceleration
There is a difference between negative acceleration and
deceleration:
Negative acceleration is acceleration in the negative direction as
defined by the coordinate system.
Deceleration occurs when the acceleration is opposite in direction to
the velocity.
©2008 by W.H. Freeman and
Company
Acceleration
acceleration is the rate of change of velocity.
v2  v1 v
a

t 2  t1
t
Acceleration
Acceleration is a vector, although in onedimensional motion we only need the sign.
here is an example of deceleration:
a = v2 – v1
∆t
©2008 by W.H. Freeman
and Company
Helpful rule:
If the sign of both the acceleration and the
velocity are the same = speeding up
If the signs of the acceleration and
velocity are different = slowing down
THINK ABOUT THIS
Why do
highways
have speed
limits and not
velocity
limits?
or deriving for dum dums
Quick calculation:
1. A plane travels in a single direction on a
runway. It reaches the end of the runway in 20
seconds and its velocity is 80 m/s. What is
its acceleration?
+ 4 m/s2
2. A skateboarder is traveling at 8
m/s. He slows and comes to a stop in 4
sec. What was the acceleration?
- 2 m/s2
3. A sailboat is moving at 12 m/s when a gust of
wind changes its velocity to 18 m/s. The wind
lasts 10 seconds. For that 10 sec, what was its
acceleration?
0.6 m/s2
4. A speed skater accelerates at 5 m/s2. What
will his speed be 4 seconds later?
20 m/s
At highway speeds, a particular
automobile is capable of an
acceleration of about 1.6 m/s2. At
this rate, how long does it take to
accelerate from 80 km/h to 110
km/h?
.
The time can be found from the average acceleration,
a
v
t
 1m s 
t 
v
a

110 km h  80 km h
1.6 m s 2

 30 km h  

3.6 km h

1.6 m s 2
  5.208s  5 s
One more….
A chicken runs across the road at a
speed of 12 m/s. If the road is 36
meters across, how long does it take
for the chicken to cross the road?
3s
A. Displacement with constant acceleration
B. Velocity with constant acceleration
C. Displacement with constant acceleration
D. Final velocity after any displacement
Displacement with constant acceleration
Displacement depends on acceleration, velocity and time.
Vavg = Δd
Δt
For an object moving with constant
acceleration, the average velocity is equal to
the average of the initial and final velocities.
v-t
vf
vavg
V (cm/s)
Δd =
Δt
Vavg = vi + vf
2
Multiply both sides by ∆t
vi
0
0
time (s)
Δd = ½ (vi + vf) Δt
Try one:
A race car reaches a speed of 42 m/s. It then begins a
uniform negative acceleration, using its parachute and
breaking system, and comes to rest 5.5 s later. Find how
far the car moves while stopping.
Displacement with Constant Acceleration
A car accelerates uniformly from rest to a speed of 23.7
km/h in 6.5 s. Find the distance the car travels during
this time.
20 m
One more:
When Maggie applies the brakes of her car, the car
slows uniformly from 15.00 m/s to 0.00 m/s in 2.50 s.
How many meters before a stop sign must she apply
the brakes in order to stop at the sign?
18.8 m
Velocity with Constant Acceleration
a = vf -vi = vf -vi
tf - ti
∆t
 Rearrange the equation for acceleration to find vf
 Multiply by ∆t and add the initial velocity to both sides
vf = vi + a ∆t
 Final velocity = initial velocity + (acceleration x time interval)
Displacement with Constant Acceleration
vf = vi + a ∆t
 Substitute this expression in for vf into the displacement
with constant acceleration equation:
∆d = 1 (vi + a∆t +vi) ∆t
2
∆d = 1 (vf - vi) ∆t
2
 Distribute ∆t
∆d = 1 [2vi ∆t + a(∆t)2]
2
∆d = vi ∆t + 1 a (∆t)2
2
Displacement with Constant Acceleration
∆d = vi ∆t + 1 a (∆t)2
2
This equation is useful not only for finding how far an
object travels under constant acceleration but also for
finding the distance required for an object to reach a
certain speed or to come to a stop
Practice Problems:
 A boat with an initial speed of 23.7 km/h accelerates at a
uniform rate of 0.92 m/s2 for 3.6 s. Find the final speed and
the displacement of the boat during this time.
Hint: you need to solve using two equations
35.6 km/hr
28.9 m
displacement/velocity when time is unknown
∆d = 1 (vf + vi) ∆t
2
multiply by 2
2∆d = (vf + vi) ∆t
You can find the final
velocity of an accelerating
object without knowing
how long it has been
accelerating.
2∆d = ∆t
(vf + vi)
You can use this equation to derive another equation
for finding final velocity after any displacement.
Final Velocity after any Displacement

substitute the expression for ∆t into the final v equation
2∆d = ∆t
(vf + vi)
vf = vi + a ∆t
Subtract vi from
vf = vi + a ( 2∆d)
(vf + vi) both sides vf -vi = a ( 2∆x)
(vf + vi)
Multiply both
sides by (vf +
vi) to get
velocities on the
same side
vf2 - vi2 = 2a∆x
Solve
for vf2
vf2 = vi2 + 2a∆d
Final Velocity after any Displacement cont.
 When using this equation remember you MUST take
the square root of the right side to find the final
velocity!
practice
vf2 = vi2 + 2a∆d
A baby sitter pushing a stroller starts from rest and
accelerates at a rate of 0.500 m/s2. What is the velocity of
the stroller after it has traveled 4.75 m?
? vf ?
vi = 0 m/s
a = 0.500 m/s2 ∆d = 4.75m
vf2 = (0 m/s)2 + 2(0.500 m/s2)(4.75 m)
x+
vf2 = 4.75m2/s2
vf = 4.75m2/s2
= + 2.18 m/s
vf2 = vi2 + 2a∆d
2. An electron is accelerated from rest in an
accelerator at 4.5 X 107 m/s2 over a distance of
95 km. Assuming constant acceleration, what is
the electron’s final velocity?
watch your units!
9.2 x 105 m/s
2 part
acceleration
problems
Dinner at a Distance, Part I
Chameleons catch insects with their tongues, which they can
extend to great lengths at great speeds. A chameleon is aiming for
an insect at a distance of 18 cm. The insect will sense the attack
and move away 50 ms after it begins. In the first 50 ms, the
chameleon’s tongue accelerates at 250 m/s2 for 20 ms, then
travels at constant speed for the remaining 30 ms. Does its tongue
reach the 18 cm extension needed to catch the insect during this
time?
Slide 2-24
Dinner at a Distance, Part II
Cheetahs can run at incredible speeds, but they can’t keep up
these speeds for long. Suppose a cheetah has spotted a gazelle.
In five long strides, the cheetah has reached its top speed of 27
m/s. At this instant, the gazelle, at a distance of 140 m from the
running cheetah, notices the danger and heads directly away.
The gazelle accelerates at 7.0 m/s² for 3.0 s, then continues
running at a constant speed that is much less than the cheetah’s
speed. But the cheetah can only keep running for 15 s before it
must break off the chase. Does the cheetah catch the gazelle, or
does the gazelle escape?
Slide 2-25
F
A
L
L
I
N
G
O
B
J
E
C
T
S
©2008 by W.H. Freeman and
Company
What is my acceleration right now?
According to Galileo
With no air
resistance, all freely
falling objects have
the same constant
acceleration.
The Mechanical Universe
The magnitude of acceleration of a falling object:
g= -9.81 m/s/s
Galileo and Apollo
Aug. 2, 1971, astronaut David Scott
Falling Objects
Near the surface of the Earth, all objects experience
approximately the same acceleration due to gravity.
This is one of the most
common examples of
motion with constant
acceleration.
©2008 by W.H. Freeman and Company
 Consider the position of the
free-falling object at regular
time intervals, every 1
second.
 The fact that the distance
which the ball travels every
interval of time is increasing
is a sure sign that the ball is
speeding up as it falls
downward.
It is accelerating!
Falling Objects
The acceleration due to
gravity at the Earth’s
surface is approximately
9.80 m/s2.
Misconceptions about falling objects.
Because acceleration due to gravity is directed
downward to the center of the earth,
acceleration of a falling object is always negative
- even when the object is thrown up into the air!
+
-9.81 m/s2
+
Based on the usual axis, downward
direction is -.
 A curved line on a position vs. time graph signifies an accelerated motion.
Since a free-falling object is undergoing an acceleration of g = -9.81 m/s/s,
you would expect that its position-time graph would be curved.
 A closer look at the position-time graph - the object starts with a small
velocity (slow) and finishes with a large velocity (fast). Since the slope of
any position vs. time graph is the velocity of the object, the initial small
slope indicates a small initial velocity and the final large slope indicates a
large final velocity.
 Last, but not least, the negative slope of the line indicates a negative (i.e.,
downward) velocity.
Free Fall
 the velocity-time graph shows the object starts with a zero velocity
(from rest) and finishes with a large, negative velocity (the object is
moving in the negative direction and speeding up). An object moving
in the negative direction and speeding up has a negative acceleration.
 Analysis of the slope of the graph is consistent with this – moving
with a constant acceleration of -9.81 m/s2.
Objects thrown in the air have a
+ velocity and – acceleration.
This means the object is slowing down .
The object moves upward but with
smaller and smaller velocity.
At the top of its path, v has decreased and is momentarily 0.
Now it moves downward with – v and still – a (this means the object
is speeding up!)
If everything accelerates at the same rate, does
that mean everything falls at the same rate?
 Even if they have a different weight?
 yes
 Even if they are different sizes?
 yes
 Even if they are different shapes?
 not if you include air resistance
 If you put it in a vacuum, then yes
Free Fall Acceleration
Objects moving toward earth move faster and faster as they fall.
Acceleration Due to Gravity
 Every object on the earth
experiences a common force: the
force due to gravity.
 This force is always directed
toward the center of the earth
(downward).
 The acceleration due to gravity is
relatively constant near the Earth’s
surface.
g
W
Earth
Gravitational Acceleration
 In a vacuum, all objects fall
with same acceleration.
 Equations for constant
acceleration apply as usual.
 Near the Earth’s surface:
a = g = -9.80 m/s2 (-32 ft/s2 )
Directed downward (usually negative).
Acceleration due to Gravity
This acceleration
vector is the
same on the way
up, at the top,
and on the way
down!
Near the surface of the
Earth, all objects accelerate
at the same rate (ignoring air
resistance).
a = -g = -9.8 m/s2
9.8 m/s2
Interpretation: Velocity decreases by 9.8 m/s each second,
meaning velocity is becoming less positive or more
negative. Less positive means slowing down while going
up. More negative means speeding up while going down.
Example: A ball is thrown vertically upward with an initial
velocity of 30.0 m/s. What are its position and velocity after
4.00 s, and 7.00 s?
Find also the maximum height attained
Given: a = -Δ9.8 m/s2
vo = 30.0 m/s
+
t = 4.00 s; 7.00 s
a=g
Find:
Δy = ? – displacement
v=?
- final velocity
After those three “times”
Δy = ? – maximum height
vo = +30.0 m/s
Given: a = -9.8 m/s2; vo = 30.0 m/s
t = 2.00 s; 4.00 s; 7.00 s
Solutions:
For t = 4.00 s:
For t = 7.00 s:
Given: a = -9.8 m/s2; vo = 30.0 m/s
t = 2.00 s; 4.00 s; 7.00 s
Solutions:
For t = 4.00 s:
For t = 7.00 s:
Given: a = -9.8 m/s2; vo = 30.0 m/s
t = 2.00 s; 4.00 s; 7.00 s
Solutions:
For maximum height, v = 0 (the ball stops at maximum height):
Can use equations for constant acceleration to solve problems.
Example 1: A flower pot falls from a window 25.0 m above the
sidewalk
a. How fast is it moving when it hits the ground?
b. How much time does someone on the ground have to get
out of the way?
1. diagram the problem
2. Define what you know, and what you’re
looking for
Δd = 25 m vi = 0 m/s a = -9.8 m/s2
?vf ? ?∆t?
25 m
3. Choose the equation(s)
vf2 = vi2 + 2a∆y
vf = vi + a∆t
Example 1: A flower pot falls from a window 25.0 m above the sidewalk
the way?
a. How fast is it moving when it hits the ground?
b. How much time does someone on the ground have to get out of
4. Solve !
a. vf2 = vi2 + 2a∆y
vf2 = 0 + 2(-9.8)(25)
vf = √490 = 22 m/s
b. vf = vi + a∆t
∆t = vf /a
25 m
= 22 m/s
-9.8 m/s2
= 2.25 s
Example 2: Jan sets a volleyball straight up, leaving her hands at
12.0 m/s.
a. How high does it go?
b. If, when the ball is on the way down, it is spiked at the
same height at which it was set, how long was it in flight?
c. How fast is it traveling when it is spiked?
1. diagram the problem
0
Place starting point of the ball at the
origin (yi = 0 at t = 0)
floor
2. Define what you know, and what you’re looking for
di = 0 m vi = 12 m/s a = -9.8 m/s2
?vf ?
?∆t?
3. Choose the equation(s)
vf2 = vi2 + 2a∆y
4. solve!
vi = 12 m/s
a = - 9.81 m/s2
0
floor
a. At the top of its flight, the ball has instantaneous
velocity of 0. vf = 0
so: 0 = 122 + 2(-9.8 m/s) ∆y
= 144 + (-19.6)Δy
Δy = 7.35 m
b. To analyze the rest of the problem:
remember that the down half of the trip is a mirror image of the up half.
So, if, while going up, the ball passes through a particular height at a
particular velocity, on its way down it will pass through that height at the
same speed (with its velocity directed down). This means that the up half of
the trip takes the same time as the down half of the trip, so we could just
figure out how long it takes to reach its max height, and then double that to
get the total time.
OR, you can do the math: df = di + vit + ½ at2
0 = 0 + 12t + ½(-9.8)t2
0 = 12 – 4.9t
t = 12/4.9 = 2.45 s
0
floor
c. The answer has to be 12 m/s down, because of the mirror-image
relationship between the up half of the flight and the down half.
OR, you can do the math:
vf = vi + a∆t
vf = 12 + -9.8(2.45s)
= -12m/s
Practice!
 Reviewing Concepts
 Pg. 107-108, 1, 2, 4, 6-9, 12, 17-24
 website the Physics Classroom
 Practice Problems
 pg. 109 –
 Problems 27, 28, 34, 35, 40, 43, 45-47
 Questions like #49 – 59, 64- 76
Lab Time!!
Determining
acceleration