U
Unit 3: Kinematics
Uniform Rectilinear Motion
Uniform Accelerated Rectilinear Motion
The Motion of Projectiles
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p228­229
We can use arrows to indicate direction and change of velocity
along a straight line
The length of the arrow is proportional to the speed.
A positive acceleration and negative direction
Negative acceleration and positive direction
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graphs: slope of a line
slope = rise/run
if y vs x then slope = Δy/Δx
If the graph is displacement versus time:
slope = Δd/Δt = d2 - d1/ t2 - t1 = velocity
If the graph is velocity versus time:
slope = Δv/Δt = v2 - v1/ t2 - t1 = acceleration
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Uniform Motion:
Constant speed: an object moves equal distances in equal time intervals
Uniform Motion(: an object moves with constant velocity (constant speed and direction)
Graphs:
Velocity vs Time: uniform motion
v
t
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remember constant velocity on a displacement vs time graph?
d
t
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UARM
UNIFORMLY
ACCELERATED
RECTILINEAR
MOTION
motion in a straight line accelerating at a constant rate
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constant acceleration
The car accelerates at 2 m/s2
means every second the car will go 2m/s faster
Time (s)
Velocity (m/s)
0
0
1
2
2
4
3
6
4
8
5
10
6
12
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If a car is slowing down the accelertation will be ­ 3m/s2
time (s)
Velocity
0
30
1
27
2
24
3
21
4
18
5
15
6
12
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Aristotle
4th century BC
2 kinds of motion • violent (throw, push)
• natural (fire rised, stones fall down)
Oresme 14 the century
predicted that if the initial velocity was zero, distance was proportional to time squared
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Galileo
1564-1642
Did experiments:
marble on the inclined plane
he noticed a marble on an inclined plane took more time
than a marble dropped (free fall)
"the gravitational force has been diluted"
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450 years later
Apollo 15 lands on the moon
A feather falls at the same time as a hammer
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Acceleration
the change in velocity over time
a = Δv/Δt = m/s2
a truck's speed changes from 5 m/s to 50 m/s, in 60 seconds, what is its acceleration?
50m/s ­ 5m/s / 60s = 45/60 m/s2
= .75m/s2
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Uniformly Accelerated Rectilinear Motion
velocity is no longer constant (more real)
velocity varies moment to moment
ti = initial time (s)
tf = final time (s)
xi = initial position (m)
xf = final position (m)
vi = initial velocity (m/s)
vf = final velcity (m/s)
a = acceleration (m/s2)
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Instantaneous velocity
velocity at a precise moment in time
On a graph:
position versus time,
UARM
d
t
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take the tangent of the curve
d
t
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a=
y2 - y1
x2 - x1
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velocity vs time graphs
v
t
There is a constant change in velocity, this object is speeding up with uniform acceleration
a = Δv/Δt = ?
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velocity vs time graphs
Slope
Δv/Δt gives acceleration
Area
v
Δv*Δt = displacement
t
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Equations of Uniform Acceleration
Equation 1: a = Δv/Δt = v2 - v1/Δt
a * Δt = v2 - v1
a * Δt + v1 = v2
v2 = v1 + at
Equation 1:
v2 = v1 + at
Equation 2: xf = xi + ½(vi + vf)Δt
Equation 3: xf =xi + v1t + ½at2
Equation 4: vf2 = vi2 + 2aΔx
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How to solve kinematic problems:
textbox p232
• draw a diagram
• origin of x axis, at starting point, xi = 0
• id ti and tf
• id known parameters (be sure to indicate
signs (+ or -)
• Find one of the 4 equations where the
quantity sought is the only unknown
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Example:
The driver of a car which is moving
east at 25m/s applies the brakes
and begins to decelerat at 2.0m/s2
How far does the car travel in
8.0s?
a = -2.0 m/s2
vi = 25 m/s
t = 8.0 s
d=?
which formula?
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d = vit + 1/2 a t2
d = 25(8.0) + 1/2(-2.0)(8.02)
d = 200 - 64
d = 136 m (+)
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Examples in text book p 232
A:
B:
Practice: Section 10.2
p. 234
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p. 234
1. What do we know?
Δd = 402m vi= 0m/s
Δt = 6.0s xi = 0m xf = 402m
?a
? vf (km/h)
Look for formulas with only one
unknown....
xf = xi + (viΔt + 1/2(aΔt2))
find a
402 = 0 + (0*6 + 1/2(a*6^2)
402 = 1/2(a*36)
804 = 36a
a = 22.3m/s2 = 22m/s2
vf
vf2 = vi2 +2aΔx
vf2 = 0 + 2* 22.3 * 402
vf = √17929.2
vf = 133.9 m/s = 130 m/s
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Sample Problems
1. A ball rolling down a hill at 4.0 m/s accelerates
at 2.0 m/s2 What is its velocity 5.0s later
Given:
v1 = 4.0 m/s
a = 2.0 m/s2
t = 5.0s
We can use equation 1
Equation 1:
v2 = v1 + at
v2 = 4.0m/s + 2.0 m/s2*5.0s
v2 = 4.0 m/s + 10.0 m/s = 14.0 m/s
The ball reaches a velocity of 14
m/s in 5.0 s.
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2. A car travelling at 10 m/s (2 Sig figs) accelerates at 4.0 m/s2 for 8.0s. What is its displacement during this interval?
Given v1 = 10 m/s
a = 4.0 m/s2
t = 8.0s
find Δd
We can use equation # 3
Equation 3: xf =xi + v1t + ½at2
2
Δx =vit + 1/2 at
=10 m/s* 8.0s + 1/2 * 4.0m/s2 * 8.0s2
= 80m + 128m
= 208 m
= 2.1 x 102m
The cars displacement for 8.0s is 2.0 x 102 m
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3.
A car accelerating at 5.0 m/s2 has a displacement of 114m in 6.0s. What was its velocity at the beginning of the interval?
given:
a = 5.0 m/s2
t = 6.0s
Δd = 114
find v1
We can use equation 3
Equation 3: xf =xi + v1t + ½at2
Δd = x f - x i
2
Δd = vit + 1/2at
114m = vi(6.0s) + 1/2 (5.0 m/s2)(6.0s)2
114m = 6.0s(vi) + 90 m
combine like terms
114m - 90m = 6.0s (vi)
24m = 6.0s (vi)
vi = 4.0 m/s
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4. A ball rolls at an initial velocity of 4.0 m/s up a hill. five seconds later it is rolling down the hill at 6.0 m/s2. Equation 1:
Find the following:
v2 = v1 + at
Equation 2: xf = xi + ½(vi + vf)Δt
a) acceleration
b) displacement at 5.0 s.
Equation 3: xf =xi + v1t + ½at2
a) assuming up the hill is positive and down the hill is negative (in terms of displacement and therefor speed) given: v1 = 4.0 m/s
v2 = ­6.0 m/s
t = 5.0s
Equation 4: vf2 = vi2 + 2aΔx
find a
a = (v2 ­ v1)/Δt
= (­6.0 m/s ­ 4.0 m/s)/5.0s
= ­10m/s/5s = ­2m/s2
b) Use a to find displacement
Equation 2: xf = xi + ½(vi + vf)Δt
Equation #2 or # 4
Equation 4: vf2 = vi2 + 2aΔx
Using equation 2
Δx =( 1/2)(v1 +v2)*t
0.5* (-6.0m/s +4.0m/s)(5.0s)
= 0.5(-2.0m/s)(5.0s)
=-5m
The ball is 5.0 m down the hill from
its starting pint after 5.0s
Using equation 4
(-6.0m/s)2 = (4.0m/s)2 + 2(-2m/s2)(Δx)
36 = 16 + -4x
20 = -4x
x = -5m
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