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

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Presentation on theme: "CBA #1 Review 2013-2014 Graphing Motion 1-D Kinematics Projectile Motion Circular Motion Gravity Graphing Motion 1-D Kinematics Projectile Motion Circular."— Presentation transcript:

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

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

3 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. V avg =  x /  t

4 Acceleration a =  v /  t The acceleration of an object tells you how much the velocity changes every second. The acceleration of an object tells you how much the velocity changes every second. The units of acceleration are m/s 2.

5 The acceleration is the slope of a velocity vs. time graph. a =  v /  t = rise / run = 3/5 m/s 2. a =  v /  t = rise / run = 3/5 m/s 2. + slope = speeding up -slope = slowing down zero slope = constant speed

6 Summary Displacement  x = x f - x i Average Velocity V avg =  x /  t Acceleration a =  v /  t Average velocity is the slope of the x vs. t graph. Acceleration is the slope of the v vs. t graph. The graph tells you 1.The direction of motion. 2.The relative speed The acceleration of an object tells you how much the velocity changes every second.

7 Constant Acceleration Graphs A cart released from rest on an angled ramp. An object dropped from rest. 7 Position Velocity Acceleration Time

8 1-D Kinematics

9 Example A car starts from rest and accelerates at 4 m/s 2 for 3 seconds. 1.How fast is it moving after 3 seconds? 2.How far does it travel in 3 seconds? 1. 4 = (v f – 0) / 3, v f = 12 m/s 2.  d = 0(3) +.5(4)(3 2 ) = 18m

10 Example A car starts from rest and obtains a velocity of 10m/s after traveling 15m. What is its acceleration? a = (10 2 – 0) / ( 30) = 3.33 m/s 2

11 Fired Horizontally d = ½ gt 2 x = vt

12 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? 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? 2. x = vt = 5(.45) = 2.26m

13 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? 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 = m/s

14 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 = (  r/t ) = (3.14)(50)/10 = 15.7 m/s 2. a = v 2 /r = (15.7) 2 /50 = 4.93 m/s 2

15 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:  F x = ma x  F y = ma y This usually gives 2 equations and 2 unknowns. Step 5: If needed, apply the kinematic equations. x f = x i +v i t +1/2at 2 v f = v i + at Applying Newton’s Laws of Motion

16 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, ) N= ( 0, N ) f = ( -20,0)  F x = ma x, 50 – 20 = 20 a x, a x = 1.50 m/s 2

17 EXAMPLE Find the net force down the plane. ma x = mgsin  – f = 40sin(30) – 10 = 20 – 10 = 10N

18 Universal Gravity F = m 1 m 2 G/r 2 Newton’s Law of Gravity : Every two objects attract each other with a gravitational force given by: Newton’s Law of Gravity : Every two objects attract each other with a gravitational force given by: m 1 = mass of the first object in kg m 2 = mass of the second object in kg r = distance between the two masses in meters G = 6.67 x m 1 = mass of the first object in kg m 2 = mass of the second object in kg r = distance between the two masses in meters G = 6.67 x

19 Example Find the force between these two masses. F = m 1 m 2 G/r 2 = (10)(10)(6.67 x )/2 2 = 1.67 x Newtons

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