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INTRODUCTION & RECTILINEAR KINEMATICS: CONTINUOUS MOTION Today’s Objectives : Students will be able to: 1. Find the kinematic quantities (position, displacement, velocity, and acceleration) of a particle traveling along a straight path. In-Class Activities : Check Homework Reading Quiz Applications Relations between s(t), v(t), and a(t) for general rectilinear motion. Relations between s(t), v(t), and a(t) when acceleration is constant. Concept Quiz Group Problem

INTRODUCTION & RECTILINEAR KINEMATICS: CONTINUOUS MOTION

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INTRODUCTION & RECTILINEAR KINEMATICS: CONTINUOUS MOTION. Today’s Objectives : Students will be able to: Find the kinematic quantities (position, displacement, velocity, and acceleration) of a particle traveling along a straight path. In-Class Activities : • Check Homework • Reading Quiz - PowerPoint PPT Presentation

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Page 1: INTRODUCTION &  RECTILINEAR KINEMATICS: CONTINUOUS MOTION

INTRODUCTION & RECTILINEAR KINEMATICS: CONTINUOUS MOTION

Today’s Objectives:Students will be able to:1. Find the kinematic quantities

(position, displacement, velocity, and acceleration) of a particle traveling along a straight path.

In-Class Activities:• Check Homework• Reading Quiz• Applications• Relations between s(t), v(t), and a(t) for general rectilinear motion.• Relations between s(t), v(t), and a(t) when acceleration is constant.• Concept Quiz• Group Problem Solving• Attention Quiz

Page 2: INTRODUCTION &  RECTILINEAR KINEMATICS: CONTINUOUS MOTION

READING QUIZ

1. In dynamics, a particle is assumed to have _________.

A) both translation and rotational motions

B) only a mass

C) a mass but the size and shape cannot be neglected

D) no mass or size or shape, it is just a point

2. The average speed is defined as __________.

A) r/t B) s/t

C) sT/t D) None of the above.

Page 3: INTRODUCTION &  RECTILINEAR KINEMATICS: CONTINUOUS MOTION

APPLICATIONS

The motion of large objects, such as rockets, airplanes, or cars, can often be analyzed as if they were particles.

Why?

If we measure the altitude of this rocket as a function of time, how can we determine its velocity and acceleration?

Page 4: INTRODUCTION &  RECTILINEAR KINEMATICS: CONTINUOUS MOTION

APPLICATIONS (continued)

A sports car travels along a straight road.

Can we treat the car as a particle?

If the car accelerates at a constant rate, how can we

determine its position and velocity at some instant?

Page 5: INTRODUCTION &  RECTILINEAR KINEMATICS: CONTINUOUS MOTION

An Overview of Mechanics

Statics: The study of bodies in equilibrium.

Dynamics:1. Kinematics – concerned with the geometric aspects of motion2. Kinetics - concerned with the forces causing the motion

Mechanics: The study of how bodies react to forces acting on them.

Page 6: INTRODUCTION &  RECTILINEAR KINEMATICS: CONTINUOUS MOTION

RECTILINEAR KINEMATICS: CONTINIOUS MOTION(Section 12.2)

A particle travels along a straight-line path defined by the coordinate axis s.

The total distance traveled by the particle, sT, is a positive scalar that represents the total length of the path over which the particle travels.

The position of the particle at any instant, relative to the origin, O, is defined by the position vector r, or the scalar s. Scalar s can be positive or negative. Typical units for r and s are meters (m) or feet (ft).

The displacement of the particle is defined as its change in position.

Vector form: r = r’ - r Scalar form: s = s’ - s

Page 7: INTRODUCTION &  RECTILINEAR KINEMATICS: CONTINUOUS MOTION

VELOCITY

Velocity is a measure of the rate of change in the position of a particle. It is a vector quantity (it has both magnitude and direction). The magnitude of the velocity is called speed, with units of m/s or ft/s.

The average velocity of a particle during a time interval t is

vavg = r / t

The instantaneous velocity is the time-derivative of position.

v = dr / dt

Speed is the magnitude of velocity: v = ds / dt

Average speed is the total distance traveled divided by elapsed time: (vsp)avg = sT / t

Page 8: INTRODUCTION &  RECTILINEAR KINEMATICS: CONTINUOUS MOTION

ACCELERATION

Acceleration is the rate of change in the velocity of a particle. It is a vector quantity. Typical units are m/s2 or ft/s2.

As the book indicates, the derivative equations for velocity and acceleration can be manipulated to get a ds = v dv

The instantaneous acceleration is the time derivative of velocity.

Vector form: a = dv / dt

Scalar form: a = dv / dt = d2s / dt2

Acceleration can be positive (speed increasing) or negative (speed decreasing).

Page 9: INTRODUCTION &  RECTILINEAR KINEMATICS: CONTINUOUS MOTION

SUMMARY OF KINEMATIC RELATIONS:RECTILINEAR MOTION

• Differentiate position to get velocity and acceleration.

v = ds/dt ; a = dv/dt or a = v dv/ds

• Integrate acceleration for velocity and position.

• Note that so and vo represent the initial position and velocity of the particle at t = 0.

Velocity:

t

o

v

vo

dtadv s

s

v

v oo

dsadvvor t

o

s

so

dtvds

Position:

Page 10: INTRODUCTION &  RECTILINEAR KINEMATICS: CONTINUOUS MOTION

CONSTANT ACCELERATION

The three kinematic equations can be integrated for the special case

when acceleration is constant (a = ac) to obtain very useful equations.

A common example of constant acceleration is gravity; i.e., a body

freely falling toward earth. In this case, ac = g = 9.81 m/s2 = 32.2 ft/s2

downward. These equations are:

ta v v co

yieldst

oc

v

v

dtadvo

2coo

s

t(1/2) a t v s s yieldst

os

dtvdso

)s - (s2a )(v v oc

2o

2 yieldss

sc

v

v oo

dsadvv

Page 11: INTRODUCTION &  RECTILINEAR KINEMATICS: CONTINUOUS MOTION

EXAMPLE

Plan: Establish the positive coordinate, s, in the direction the

particle is traveling. Since the velocity is given as a

function of time, take a derivative of it to calculate the

acceleration. Conversely, integrate the velocity

function to calculate the position.

Given: A particle travels along a straight line to the rightwith a velocity of v = ( 4 t – 3 t2 ) m/s where t is in seconds. Also, s = 0 when t = 0.

Find: The position and acceleration of the particle when t = 4 s.

Page 12: INTRODUCTION &  RECTILINEAR KINEMATICS: CONTINUOUS MOTION

EXAMPLE (continued)Solution:

1) Take a derivative of the velocity to determine the acceleration.

a = dv / dt = d(4 t – 3 t2) / dt =4 – 6 t => a = – 20 m/s2 (or in the direction) when t = 4 s

2) Calculate the distance traveled in 4s by integrating the velocity using so = 0:

v = ds / dt => ds = v dt => => s – so = 2 t2 – t3 => s – 0 = 2(4)2 – (4)3 => s = – 32 m ( or )

t

o

s

s

(4 t – 3 t2) dtdso

Page 13: INTRODUCTION &  RECTILINEAR KINEMATICS: CONTINUOUS MOTION

CONCEPT QUIZ

1. A particle moves along a horizontal path with its velocity varying with time as shown. The average acceleration of the particle is _________.A) 0.4 m/s2 B) 0.4 m/s2

C) 1.6 m/s2 D) 1.6 m/s2

2. A particle has an initial velocity of 30 ft/s to the left. If it then passes through the same location 5 seconds later with a velocity of 50 ft/s to the right, the average velocity of the particle during the 5 s time interval is _______.A) 10 ft/s B) 40 ft/sC) 16 m/s D) 0 ft/s

t = 2 s t = 7 s

3 m/s 5 m/s

Page 14: INTRODUCTION &  RECTILINEAR KINEMATICS: CONTINUOUS MOTION

GROUP PROBLEM SOLVING

Given:Ball A is released from rest

at a height of 40 ft at the

same time that ball B is

thrown upward, 5 ft from the

ground. The balls pass one

another at a height of 20 ft.

Find:The speed at which ball B was

thrown upward.

Plan: Both balls experience a constant downward acceleration

of 32.2 ft/s2 due to gravity. Apply the formulas for

constant acceleration, with ac = -32.2 ft/s2.

Page 15: INTRODUCTION &  RECTILINEAR KINEMATICS: CONTINUOUS MOTION

GROUP PROBLEM SOLVING (continued)Solution:

1) First consider ball A. With the origin defined at the ground,

ball A is released from rest ((vA)o = 0) at a height of 40 ft

((sA )o = 40 ft). Calculate the time required for ball A to drop to

20 ft (sA = 20 ft) using a position equation.

sA = (sA )o + (vA)o t + (1/2) ac t2

So,

20 ft = 40 ft + (0)(t) + (1/2)(-32.2)(t2) => t = 1.115 s

Page 16: INTRODUCTION &  RECTILINEAR KINEMATICS: CONTINUOUS MOTION

GROUP PROBLEM SOLVING (continued)Solution:

2) Now consider ball B. It is throw upward from a height of 5 ft

((sB)o = 5 ft). It must reach a height of 20 ft (sB = 20 ft) at the

same time ball A reaches this height (t = 1.115 s). Apply the

position equation again to ball B using t = 1.115s.

sB = (sB)o + (vB)ot + (1/2) ac t2

So,

20 ft = 5 + (vB)o(1.115) + (1/2)(-32.2)(1.115)2

=> (vB)o = 31.4 ft/s

Page 17: INTRODUCTION &  RECTILINEAR KINEMATICS: CONTINUOUS MOTION

ATTENTION QUIZ

2. A particle is moving with an initial velocity of v = 12 ft/s and constant acceleration of 3.78 ft/s2 in the same direction as the velocity. Determine the distance the particle has traveled when the velocity reaches 30 ft/s.

A) 50 ft B) 100 ftC) 150 ft D) 200 ft

1. A particle has an initial velocity of 3 ft/s to the left at s0 = 0 ft. Determine its position when t = 3 s if the acceleration is 2 ft/s2 to the right.

A) 0.0 ft B) 6.0 ftC) 18.0 ft D) 9.0 ft

Page 18: INTRODUCTION &  RECTILINEAR KINEMATICS: CONTINUOUS MOTION