CBA #1 Review 2014-2015
Graphing Motion
1-D Kinematics
Projectile Motion
Circular Motion
Gravity
Graphing Motion
Distance vs. Time
Velocity vs. time
Acceleration vs. time
Average velocity is the slope of
the x vs. t graph.
The graph tells you
1. The direction of motion.
2. The relative speed.
Vavg = Dx / Dt
Compare the velocities for the three graphs.
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.
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 down
zero slope = constant
speed
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.
The graph tells you
1. The direction of motion.
2. The relative speed
Acceleration is the slope of the v vs. t graph.
The acceleration of an object tells
you how much the velocity changes
every second.
Constant Acceleration Graphs
A cart released from
rest on an angled ramp.
Acceleration
Velocity
Position
An object dropped from
rest.
Time
Time
Time
7
1-D Kinematics
Example
A 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.
2.
4 = (vf – 0) / 3 ,
vf = 12 m/s
Dd = 0(3) + .5(4)(32) = 18m
Example
A 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
Fired Horizontally
d = ½ gt2
x = vt
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
2.
1 = .5(9.8)t2 t = 1/4.9 = .45s
x = vt = 5(.45) = 2.26m
Example
A ball rolls off of a table and hits the ground 1.5m away after falling for .5 seconds.
What was its initial velocity?
x = vt , so v =x/t = 1.5m/.5s = 3. 00 m/s
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
Applying Newton’s Laws of Motion
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
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?
N= ( 0, N )
P = ( 50, 0 )
f = ( -20,0)
SFx = max , 50 – 20 = 20 ax , ax = 1.50 m/s2
W = ( 0, - 196 )
EXAMPLE
Find the net force down the plane.
max = mgsinq – f = 40sin(30) – 10 = 20 – 10 = 10N
Universal Gravity
Newton’s Law of Gravity : Every two objects attract each other
with a gravitational force given by:
F = m1m2G/r2
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
Example
Find the force between these two masses.
F = m1m2G/r2 = (10)(10)(6.67 x10-11)/22 =
1.67 x 10-9 Newtons