CBA #1 Review 2013-2014

Graphing Motion 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 , andDt = 4 – 0 = 4s.

So, vavg = 4m/4s = 1m/s

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

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 downzero 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.

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.

Constant Acceleration GraphsA cart released from rest on an angled ramp.

An object dropped from rest.

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1-D Kinematics

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

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

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

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

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

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

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

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

Acceleration Down an Inclined Plane with Friction

Find an expression for the acceleration down the plane and for the normal force.

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 :

EXAMPLE

Find the net force down the plane.

max = mgsinq – f = 40sin(30) – 10 = 20 – 10 = 10N

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

Example

Find the force between these two masses.

F = m1m2G/r2 = (10)(10)(6.67 x10-11)/22 =

1.67 x 10-9 Newtons