# CONSTANT ACCELERATION

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```					       Purpose: Find the relationship between the position (x) of an object as it rolls
down an incline and the time taken (t) to reach each position.
t2,x2
Picture represents equal
position intervals. If so, what
t1,x1                                             would happen to the time?

t3,x3
Procedure: 1) Set up the ramp.
2) Control the ramp angle and the
What’s required:
mass of the ball.
3) Perform 3 trials for each of the 5          1) Data Table 2) Original Graph
positions along the ramp.                      3) Modified graph. 4) Derivation of
4) Plot Position vs. time and then
modify. (Plot time on x axis)
5) Derive an equation for the
relationship.
One could speak of the average velocity of the object in
the graph, but since the object started very slowly and
steadily increased its speed, the term average velocity has
little meaning.

x

x

t         t
What happens as you shrink the time
interval t over which you calculate the
average velocity?

x               x               x

x              x            x
t            t
t
t               t
t
As one shrinks the interval, t to zero, the secant
becomes a tangent; the slope of the tangent is the
average velocity at this instant, or simply the
instantaneous velocity at that clock reading.

x

x

t     t
Instantaneous velocity (v1, v2, vf, vi, etc.) the velocity
at a given point (instant) in time. It is equal to the
slope of the tangent to the x vs. t graph.
Acceleration

x                                        Δv
v
Δt

t                           t

Δv
Slope =        = acceleration
Δt

The slope of a v vs. t graph is the acceleration.
Acceleration
Acceleration is a vector quantity that is defined as the
rate at which an object changes its velocity. An object is
accelerating if it is changing its velocity (magnitude or
direction).

Acceleration values are expressed in units of
velocity/time. Typical acceleration units include the
following:
m/s/s
mi/hr/s
km/hr/s
Changing speed, constant direction
Acceleration exists

Constant speed, changing direction
Acceleration exists

Changing speed, constant direction
Acceleration exists
Direction of Motion   Action/Sign of      Sign of
Velocity       Acceleration
New Graphs                  Written Description    Motion Map
(x vs t) (v vs t) (a vs t)   Starting Dir  (v)   (a)
point
x
x
t                              t2
• Position α time2
• y = mx2 + b
• Final Equation: x (m) = 2.5 m/s2 x time2 (s2)
• How does the position vs time graph differ from the
previous unit’s lab? There is no longer a direct
relationship between position and time.
• Explain how the position vs. time graph shows the
motion exhibited by the ball? The slope is no longer
constant, therefore the velocity is changing.
CONSTANT ACCELERATION
Galileo (1564 – 1642)

1
displacement 3
5
7
1 second = 1 displacement
9
2 second = 1 + 3 displacement = 2  2

3 second = 1 + 3 + 5 displacement = 32
4 second = 1 + 3 + 5 + 7 displacement = 42
5 second = 1 + 3 + 5 + 7 + 9 displacement = 52
Therefore, displacement a time2
x                                                        Time
(s)
Position
(m)
Instantaneous
velocity
Group Ave.
Velocity
(m/s)           (m/s)

0        0

1        2.5

1.5      5.63

2        10

2.5      13.6

3.0      22.5

0                                               3.6   t (s)
1. Draw tangent lines to the curve below at all but the last point.
2. Find the slope of each tangent line. Then slope of the tangent line is the instantaneous
velocity at that point in time Record the value of the velocity in the table above.
3. Average the slope values within your lab group for each point. Use Graphical analysis
to obtain a velocity vs. time graph using your groups average values. Draw a picture of
the graph on another sheet of paper. Label axis w/ units and record stats.
Compare Graphs                                  Same
original
x
graph                      t
• What is the difference between a X vs t2 graph & the V vs t graph.
• Derive the equation for both.

v

x
t2                                              t
4. Write the mathematical model for this
5. The equation for the curve is x=2.5m/s^2(t^2). graph underneath your V vs t graph.
How does the slope of your velocity vs. time graph
compare with the slope of the curve above.            Y = mX + b
Y = mX2 + b                                           vf = 5 m/s2 (time) + vi
Final Equation: X = 2.5 m/s2 (time2)                  Since slope = ave. accel.

So, for X vs t2 graph   slope = ½ a                   vf = a Δ t + vi or vf = vi + a Δ t
vf - vi
X = ½         at2                              a=
Δt
Motion Graphs

x                  v            a

t          t            t

x

0
Motion Graphs

x                         v                       a

t                       t                          t

x

0
Starting Point   Direction        Velocity           Acceleration

0         + forward       + V (speeding up)       +constant
Acceleration Example #1
V (m/s)   a (m/s/s)
0         +5
+5        +5
+10       +5
15       +5
Δv from +5 to +10 m/s requires
a +5 m/s/s acceleration!
Motion Graphs

x                      v                    a

t                     t                        t

x
0

Starting Point   Direction       Velocity             Acceleration

0          + forward       + V (slowing down)     -constant
Acceleration Example #2
V (m/s)   a (m/s/s)
+ 10       -5
+ 5       -5
0       -5

Δv from +10 to + 5 m/s requires
a - 5 m/s/s acceleration!
Motion Graphs

x                           v                      a

t                       t                         t

x
0

Starting Point       Direction       Velocity           Acceleration

above             backward        -V (speeding up)       -constant
Acceleration Example #3
V (m/s)   a (m/s/s)
0         -5
-5        -5
-10       -5
-15       -5
Δv from -5 to -10 m/s requires
a -5 m/s/s acceleration!
a negative acceleration does
not mean the velocity is
slowing down, it has to do
with direction!
Motion Graphs
a
v
x

t
t
t

x

0
Starting Point   Direction         Velocity               Acceleration

above         backward       - V (slowing down)         +constant
Acceleration Example #4
V (m/s)   a (m/s/s)
- 10       +5
- 5        +5
0        +5

Δv from -10 to -5 m/s requires
a +5 m/s/s acceleration!
More Conventions…
Speeding Up
+ v, + a
Same sign
- v, - a

Slowing Down
+ v, - a
Opposite sign
- v, + a
Motion Graphs

x                   v           a

t                           t
t

x

0
Speed is not acceleration
As the ball rolls down this hill
a) Its speed increases and its acceleration decreases.
b) Its speed decreases and acceleration increases.
c) Both increase
d) Both remain constant
e) Both decrease
Graphing Acceleration
B

v        A                               C

t                       D
A:       speeding up        + velocity ↑   + acceleration
B:       const motion   const. velocity    0 acceleration
C:       slowing down       + velocity ↓   - acceleration
D:       speeding up        - velocity ↑   - acceleration
Rearranging Equation #1
vf - vi
a=               Slope of an V vs t graph
Δt

a Δt = vf - vi

vf = vi + a Δt
Deriving the Acceleration Equations
2

• modify     x vs t graph
x

t                          x

Equation is: x = 2.5 m/s/s x   time2     t2

So, for x vs t2 graph m = ½ a
Therefore,    Δx = ½ at2 + vit
Deriving the Acceleration Equations
3

Through a complex derivation – when you
combine equations #1 and #2 you get:

vf2 = vi2 + 2aΔx
Constant Acceleration Equations
Missing variable

ΔX           vf = vi + a Δt

Vf           ΔX = ½ at2 + vit

Vf2   =   vi 2   + 2aΔx
t

a           ΔX = ½ (v f+vi)t
All the equations in one place
constant velocity        uniform acceleration

x                 v
v                 a
t                 t
x  vt           v f  at  v 0
x  v0t  at
1
2
2


v  v  2ax
2
f
2
0

1. A bicyclist has an acceleration of 2 m/s2.
If he starts from rest, determine his velocity
and position at t = 5s.

2. A car is traveling at a speed of 30 m/s
when the brakes are suddenly applied,
causing a constant deceleration of 4 m/s2.
Determine the time required to stop the car
and the distance traveled before stopping.

1.10 m/s, 25 m
2. 7.5 s, 112.5 m
Early one morning Farmer Joiner tries his hand at milking a new
breed of cow. He chooses one named Hackensack. (Yes he’s
milking Hackensack, the new Jersey). However, Hackensack fails to
realize how much pull Farmer Joiner has and reacts oddly. Every
time Dr. J milks her, she has trouble with her moo. It sounds like M-
m-m-ooo (a sort of udder stutter, you might say). Finally she can’t
take it any longer.
a) Starting from rest, she accelerates at 2 m/s2 for 4 seconds out the
barn door.
b) She then runs at a constant velocity for 5 minutes,
c) until Dr. J succeeds in throwing a rope around her neck, after
which she slows down to a stop at a rate of 1 m/s2.
When the ordeal is over, Dr. J admits that it was a mooving
experience indeed.
Qualitatively sketch the following graphs for Hackensack’s trip.
A. position vs. time graph      B. velocity vs. time graph
C. acceleration vs. time graph D. motion map.
An automobile accelerates from rest at 2 m/s2 for 20 s. The
speed is then held constant for 20 s. Then there is an
acceleration of - 3 m/s2 until the automobile stops.
Find:
a. the distance traveled between the 4th and 5th seconds.
b. the total distance traveled.
c. the total time traveled

a.     9.00 m
b. 1466.53 m
c.   53.33 s
Ella Vader is driving her flashy Porsche
down the highway at 30 m/s. She sees a
deer ahead and 2.0 s after she spots the
deer she hits the brakes. It takes her 7.5
more seconds to stop.
a) How far did Ella travel before stopping?

b) What was Ella’s acceleration while she
slowed down?
Terms for Constant Acceleration Test
Instantaneous Velocity:
 definition
 find from x vs t (tangent line, slope)
Acceleration:
 definition (a = Δv / Δt)
 find from a v vs t (slope)
For accelerated Motion:
 x vs t, v vs t, a vs t graphs
 motion map
Graphs & Tracks Exercise
Conceptual Questions – Acceleration & Velocity
Constant Acceleration Equations
vf = vi + at
Δx = ½at2 + vit
vf2 = vi2 + 2aΔx

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