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

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

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

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

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.

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.

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 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.

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

1-D Kinematics

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

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

Projectile Motion t = Time from A to B = Time from B to C 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

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

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 = ½gt2 1 = .5(9.8)t2 t = 1/4.9 = .45s 2. x = vt = 5(.45) = 2.26m

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

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 )

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

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 = 0 ; max = mgsinq – f N + 0 - mgcosq = 0 ; N = mgcosq

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

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

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