Download - CBA #1 Review 2013-2014
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CBA #1 Review 2013-2014
Graphing Motion 1-D Kinematics Projectile Motion Circular Motion Gravity
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Graphing Motion
Distance vs. Time
Velocity vs. time
Acceleration vs. time
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Displacement Dx
Dx = xf - xi
In this case, xi = -1m, xf = 3m,So Dx = 3 – (-1) = 4m
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Average Velocity vavg
Vavg = Dx / Dt
In this case, xi = -1m, xf = 3m,so Dx = 3 – (-1) = 4m , andDt = 4 – 0 = 4s.
So, vavg = 4m/4s = 1m/s
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Average velocity is the slope of the x vs. t graph.
Compare the velocities for the three graphs.
The graph tells you
1. The direction of motion.
2. The relative speed.
Vavg = Dx / Dt
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Acceleration a = Dv / Dt
The acceleration of an object tells you how much the velocity changes every second.
The units of acceleration are m/s2.
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The acceleration is the slope of a velocity vs. time graph.
a = Dv / Dt = rise / run = 3/5 m/s2.
+ slope = speeding up-slope = slowing downzero slope = constant speed
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Summary
Displacement Dx = xf - xi
Average Velocity Vavg = Dx / Dt
Acceleration a = Dv / Dt
Average velocity is the slope of the x vs. t graph.
Acceleration is the slope of the v vs. t graph.
The graph tells you1. The direction of motion.2. The relative speed
The acceleration of an object tells you how much the velocity changes every second.
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Constant Acceleration GraphsA cart released from rest on an angled ramp.
An object dropped from rest.
9
Posi
tion
Velo
city
Acce
lera
tion
Time Time Time
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1-D Kinematics
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ExampleA car starts from rest and accelerates at 4 m/s2 for 3 seconds.
1. How fast is it moving after 3 seconds?
2. How far does it travel in 3 seconds?
1. 4 = (vf – 0) / 3 , vf = 12 m/s
2. Dd = 0(3) + .5(4)(32) = 18m
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ExampleA car starts from rest and obtains a velocity of 10m/s after traveling 15m. What is its acceleration?
a = (102 – 0) / ( 30) = 3.33 m/s2
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Projectile Motion
Vertical Motion: vA = -vC vB = 0
From B to C : d = ½gt2 v = gt
t = Time from A to B = Time from B to C
Total time in air = 2t
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Example
A projectile is fired upwards and hits the ground 10 seconds later.
1. How high did it go?
2. What speed was it fired at?
1. Note : t = 5 seconds d = ½gt2 = .5(9.8)(52) = 122.5m
2. v = gt = 9.8(5) = 49 m/s
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Example
A ball moving at 5 m/s rolls off of a table 1m tall and hits the ground.
1. How long was it in the air?
2. What horizontal distance did it travel?
1. d = ½gt2 1 = .5(9.8)t2 t = = .45s
2. x = vt = 5(.45) = 2.26m
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Circular Motion
Example: A car rounds the circular curve (r = 50m) in 10 seconds.
1. What is the velocity?2. What is the centripetal acceleration while in the curve?
1. V = d/t = (pr/t ) = (3.14)(50)/10 = 15.7 m/s
2. a = v2/r = (15.7)2/50 = 4.93 m/s2
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Step 1: Identify all of the forces acting on the object Step 2 : Draw a free body Diagram Step 3: Break every force into x and y components. Step 4: Apply the second law: SFx = max SFy = may
This usually gives 2 equations and 2 unknowns. Step 5: If needed, apply the kinematic equations. xf = xi +vit +1/2at2 vf = vi + at
Applying Newton’s Laws of Motion
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Problems With Acceleration
A box (m = 20kg) is pushed to the right with a force of 50N. A frictional force of 20N acts to the left. What is the acceleration of the box?
P = ( 50, 0 )
W = ( 0, - 196 )
N= ( 0, N )
f = ( -20,0)
SFx = max , 50 – 20 = 20 ax , ax = 1.50 m/s2
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Acceleration Down an Inclined Plane with Friction
Find an expression for the acceleration down the plane and for the normal force.
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Sketch the forces………..Make the free-body diagram.
N = ( 0, N)
T = (-f, 0)
W = ( mgsin q , -mgcosq )
0 – f + mgsin q = 0 ; max = mgsin q – fN + 0 - mgcosq = 0 ; N = mgcosq
Apply the 2nd Law :
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EXAMPLE
Find the net force down the plane.
max = mgsinq – f = 40sin(30) – 10 = 20 – 10 = 10N
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Universal Gravity
F = m1m2G/r2
Newton’s Law of Gravity : Every two objects attract each otherwith a gravitational force given by:
m1 = mass of the first object in kg m2 = mass of the second object in kg r = distance between the two masses in meters G = 6.67 x 10-11
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Example
Find the force between these two masses.
F = m1m2G/r2 = (10)(10)(6.67 x10-11)/22 =
1.67 x 10-9 Newtons