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UUnit 3: Kinematics
Uniform Rectilinear MotionUniform Accelerated Rectilinear MotionThe 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 timethan 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 - y1x2 - 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
v
t
Slope
Δv/Δt gives acceleration
AreaΔv*Δt = displacement
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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 3: xf =xi + v1t + ½at2
Equation 2: xf = xi + ½(vi + vf)Δt
Equation 4: vf2 = vi
2 + 2aΔx
Equation 1: v2 = v1 + at
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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/st = 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 - 64d = 136 m (+)
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Examples in text book p 232
A:
B:
Practice: Section 10.2p. 234
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p. 2341. 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 a402 = 0 + (0*6 + 1/2(a*6^2)402 = 1/2(a*36)804 = 36aa = 22.3m/s2 = 22m/s2
vfvf2 = vi2 +2aΔxvf2 = 0 + 2* 22.3 * 402
vf = √17929.2vf = 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/sa = 2.0 m/s2t = 5.0s
We can use equation 1Equation 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/sa = 4.0 m/s2t = 8.0s find Δd
We can use equation # 3
Equation 3: xf =xi + v1t + ½at2
Δx =vit + 1/2 at2
=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/s2t = 6.0sΔd = 114
find v1
We can use equation 3
Equation 3: xf =xi + v1t + ½at2
Δd = xf - xi
Δd = vit + 1/2at2
114m = vi(6.0s) + 1/2 (5.0 m/s2)(6.0s)2
114m = 6.0s(vi) + 90 mcombine 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.
Find the following:
a) accelerationb) displacement at 5.0 s.
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/sv2 = 6.0 m/st = 5.0s
find aa = (v2 v1)/Δt = (6.0 m/s 4.0 m/s)/5.0s = 10m/s/5s = 2m/s2
b) Use a to find displacementEquation #2 or # 4
Equation 4: vf2 = vi
2 + 2aΔx
Equation 3: xf =xi + v1t + ½at2
Equation 2: xf = xi + ½(vi + vf)Δt
Equation 4: vf2 = vi
2 + 2aΔx
Equation 1: v2 = v1 + at
Equation 2: xf = xi + ½(vi + vf)Δt
Using equation 2Δx =( 1/2)(v1 +v2)*t0.5* (-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 + -4x20 = -4xx = -5m
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