1. CBA #1 Review 2013-2014 Graphing Motion 1-D Kinematics
1-D Kinematics
Projectile Motion
Circular Motion
Gravity

2. Graphing Motion Distance vs. Time Velocity vs. time
Acceleration vs. time

3. Displacement Dx Dx = xf - xi In this case, xi = -1m, xf = 3m,
So Dx = 3 – (-1) = 4m

4. Average Velocity vavg Vavg = Dx / Dt In this case, xi = -1m, xf = 3m,
so Dx = 3 – (-1) = 4m , and
Dt = 4 – 0 = 4s.
So, vavg = 4m/4s = 1m/s

5. Average velocity is the slope of the x vs. t graph.
The graph tells you
The direction of motion.
The relative speed.
Vavg = Dx / Dt
Compare the velocities for the three graphs.

6. The acceleration of an object tells you how much the velocity changes
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.

7. 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

8. 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
The direction of motion.
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.

9. Constant Acceleration Graphs
A cart released from rest on an angled ramp. An object dropped from rest.
Position
Velocity
Acceleration
Time

10. 1-D Kinematics

11. Example A car starts from rest and accelerates at 4 m/s2 for
3 seconds.
How fast is it moving after 3 seconds?
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

12. Example A car starts from rest and obtains a velocity of 10m/s
after traveling 15m. What is its acceleration?
a = (102 – 0) / ( 30) = m/s2

13. Projectile Motion t = Time from A to B = Time from B to C
Vertical Motion: vA = -vC vB = 0
From B to C : d = ½gt v = gt
t = Time from A to B = Time from B to C
Total time in air = 2t

14. Example
A projectile is fired upwards and hits the ground 10 seconds later.
How high did it go?
What speed was it fired at?
Note : t = 5 seconds d = ½gt2 = .5(9.8)(52) = 122.5m
2. v = gt = 9.8(5) = 49 m/s

15. Example
A ball moving at 5 m/s rolls off of a table 1m tall and hits the ground.
How long was it in the air?
What horizontal distance did it travel?
d = ½gt = .5(9.8)t2 t = 1/4.9 = .45s
2. x = vt = 5(.45) = 2.26m

16. Circular Motion
Example: A car rounds the circular curve (r = 50m) in 10 seconds.
What is the velocity?
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

17. 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 +1/2at vf = vi + at

18. 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, )

19. Acceleration Down an Inclined Plane with Friction
Find an expression for the acceleration down the plane and for the normal force.

20. 0 – f + mgsinq = 0 ; max = mgsinq – f N + 0 - mgcosq = 0 ; N = mgcosq
Sketch the forces………..
Make the free-body diagram.
N = ( 0, N)
T = (-f, 0)
W = ( mgsinq , -mgcosq )
Apply the 2nd Law :
0 – f + mgsinq = ; max = mgsinq – f
N mgcosq = ; N = mgcosq

21. EXAMPLE Find the net force down the plane.
max = mgsinq – f = 40sin(30) – 10 = 20 – 10 = 10N

22. Universal Gravity m2 = mass of the second object in kg
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

23. Example 1.67 x 10-9 Newtons Find the force between these two masses.
F = m1m2G/r2 = (10)(10)(6.67 x10-11)/22 =
1.67 x 10-9 Newtons