Motion – many examples surround us an ice skater coasting along a car screeching to a halt a ball

Roberto Ramos

Description

Question of the DayThe magnitudes of two vectors A and B are 12 and 8 units, respectively. Which of the following pairs of numbers represent the largest and smallest possible magnitude of the resultant vector R = A + B ?a. 14.4 units, 4 units b. 12 units, 8 units c. 20 units, 4 units d. None of the above 0% 0% 0% 0%10 14.4 units, 4 units 12 units, 8 units 20 units, 4 units None of the aboveIf at least one component of a vector is a positive number, the vectora. Cannot have any component that is negative.b. Cannot be zero c. Cannot have 3 dimensions0% 0% 0%10Cant hven .. noa ay. Cannot be zero Cannot have 3 dime...Mechanics – study of motion KinematicsDynamics (mathematical description,(cause of motion)ignores cause of motion)Approach: Use particle model (objects treated as point-like:has mass but infinitesimal size)Focus: Motion along a Straight Line (1-dimensional)Goal: Find position x(t), velocity v(t) z(t)x(t)y(t)v (t) v (t)y z v (t)x , vx o o Goal of Classical Mechanics:Predict the future motion of a particle, given its initial conditions(x , v ) and the effect of the environment (acceleration a)o o Defining Position x(t)origin• Defined in terms of a coordinate system – One dimensional: the x- or y-axis • object’s “position” - location with respect to the coor- dinate system A Position-Time Graph – ...

Informations

• Publié par: Nifong
• Nombre de lectures: 29
• Langue: English

Extrait

Question of the Day The magnitudes of twovectors AandBare 12 and 8 units, respectively. Which of the following pairs of numbers represent the largest and smallest possible magnitude of the resultant vectorR=A+B?

a. b. c. d.

14.4 units, 4 units 1   2 units, 8 units 2   0 units, 4 units N   one of the above  

0

0

0

0

If at least one component of a vector is a positive number, the vector

a.

b. c.

Cannot have any c   omponent that is negative. Cannot be zero C   annot have 3 d   imensions

0

0

0

Mechanics of motion– study  

Kinematics (mathematical description, ignores cause of motion)

Dynamics (cause of motion)

Approach:Use particle model (objects treated as point-like: has mass but infinitesimal size)

Focus:

Goal:

Motion along a Straight Line (1-dimensional)

Find positionx(t), velocityv(t)

x o , v o

 

x(t)

y(t)

( v y t)

z(t)

v z (t)  

( v x t)

Goal of Classical Mechanics: Predict the future motion of a particle, given its initial conditions (x o , v o and the effect of the environment (acceleration) a)  

•

•

Defining Position x(t)

Defined in terms of a coordinate system  

–One dimensional: the x- or y-axis     object’s “position”    with- location respect to the coor-   dinate system  

origin

APosition-Time Graph

– shows   

The smooth curve is a guess as to what happened between data points.

x vs t graph  

Defining Displacement∆x

•Defined as the change in position during s   ome time interval xx f x i

•This occurs in a time∆t so that we can d   efine an average velocityv ave

Defining Average Velocity v ave

•Theaverage velocityis rate at which the is l d    occursp acement v average   t x  x f   x t i •dimensions are length/time [m/s] •th   e slope of the line in positi h   •Gives limited information  

Average Velocity is different from average Speed

Average Speed = Total Distance/Total Time

(a scalar)

=d/t

AverageVelocity= Displacement/Total Time

(a vector)

=x/t

PROBLEM

A person walks first at a constant speed of 5.00m/s along a straight line from Point A to Point B and then back along the line from B to A at a constant speed of 3.00 m/s.

a. What is her average speed over the entire trip ? b. What is her average velocity over the entire trip ?

A

d

B

Soln:

PROBLEM A person walks first at a constant speed of 5.00m/s along a straight line from Point A to Point B and then back along the line from B to A at a constant speed of 3.00 m/s.

a. What is her average speed over the entire trip ? b. What is her average velocity over the entire trip ?

A

d

a.Let the time to go from A to B = t 1  m/s= d/5 Let the time to go from B to A = t 2 = d/3 m/s    

B

  average speed= Total Distance/Total Time = 2d/{d/5m/s) + (d/3m/s)} =3.75m/s

b.average velocity v ave = 0 since the total displacement = 0.