Motion in One DimensionChapter 2

2.1 Displacement & VelocityLearning Objectives

• Describe motion in terms of displacement, time, and velocity

• Calculate the displacement of an object traveling at a known velocity for a specific time interval

• Construct and interpret graphs of position versus time

Scalar Quantities & Vector Quantities

• Scalar quantities have magnitude

• Example: speed 15 m/s

• Vector quantities have magnitude and direction

• Example: velocity 15 m/s North

Displacement

• Net change in position of an object• Straight line distance from beginning to end• Is not the same as distance

Displacement

• Displacement is a vector quantity • Indicates change in location of a body

∆x = xf - xi

• It is specified by a magnitude and a direction.

• Displacement does not take into account the path followed between xi and xf .

Displacement is change in position

www.cnx.org

Describing MotionDescribing motion requires a frame of reference

http://www.sfu.ca/phys/100/lectures/lecture5/lecture5.html

Determining DisplacementIn these examples, position is determined with respect to the origin, displacement wrt x1

http://www.sfu.ca/phys/100/lectures/lecture5/lecture5.html

Indicating Direction of Displacement

When sign is used, it follows the conventions of a standard graph

Positive Right Up

Negative Left Down

Direction can be indicated by sign, degrees, or geographical directions.

Displacement• In this example, the body moved 10 units to get from A to B.• Distance = length traveled = 10 units• Displacement = net change in position ≈ 5.1 units, NE

Velocity

t

xx

t

xv if

in time change

ntdisplaceme velocity average

Velocity• Example

• A squirrel runs in a westerly direction from one tree to another, covering 55 meters in 32 seconds. Calculate the squirrel’s average velocity

• vavg = ∆x / ∆t

• vavg = 55 m / 32 s

• vavg = 1.7 m/s west

Velocity is not Speed

Velocity is a vector quantity

Speed is a scalar quantity

Magnitude Magnitude

Direction No direction

= ∆ displacement /∆ time

vavg = ∆x /∆t

= distance traveled/ time

= d /∆t

Position-time graphs(x-t graphs)

Velocity can be interpreted graphically: Position Time Graphs

Velocity can be interpreted graphically: Position Time Graphs

Calculate the average velocity for the entire trip

Position-Time Graphs

dev.physicslab.org

Object at rest?

Traveling slowly in a positive direction?

Traveling in a negative direction?

Traveling quickly in a positive direction?

Instantaneous Velocity

• Is not average velocity

• Is velocity of an object at any given moment in time or at a specific point in the object’s path

Position-time when velocity is not constant

Instantaneous Velocity

• Velocity at any point on an x-t graph• Velocity at a given point in time• Is the slope of a line tangent to the x-t curve

Average velocity compared to instantaneous velocity

Instantaneous velocity is the slope of the tangent line at any particular point in time.

2.2 Acceleration

2.2 AccelerationLearning Objectives

• Describe motion in terms of changing velocity

• Compare graphical representations of accelerated and non-accelerated motions

• Apply kinematic equations to calculate distance, time, or velocity under conditions of constant acceleration

X-t graph when velocity is changing

Changes in Velocity

• Acceleration is the rate of change of velocity

• a = ∆v/∆t

• a = (vf – vi) / (tf – ti)

• Since velocity is a vector quantity, velocity can change in magnitude or direction

• Acceleration occurs whenever there is a change in speed or direction of movement.

Dimensions of acceleration

• a = ∆v/∆t

• = meters per second per second

• = m/s/s

• = m/s2

• Sample problems 2B

Acceleration is a vector quantity

• Has magnitude and direction

• Both velocity & acceleration can have (+) and (-) values

• Whether –a indicates slowing down or speeding up depends upon the sign of velocity

www.gcsescience.com

Velocity-Time Graphs

What would a position-time graph look like?

Acceleration-time graph?

Free-fall motion graphs

x-t graph v-t graph

www.gcsescience.com

dev.physicslab.org

Velocity & Acceleration

Velocity acceleration_ __ Motion

+ + speeding up

(-) (-) speeding up

+ (-) slowing down

(-) + slowing down

Displacement with Constant Acceleration

tvvx

tvv

x

vv

t

x

vv

vvv

t

xv

fi

fi

fi

avgavg

fiavgavg

2

1 Or

2

Thus

2 Then

Since2

and

tavvThent

vvaSince

if

if

Final velocity of an accelerating object

2Δ2

1ΔΔ tatvx i

Displacement during constant acceleration

Final velocity after any displacement

xavv if 222

A baby sitter pushes a stroller from rest, accelerating at 0.500 m/s2. Find the velocity after the stroller travels 4.75m. (p. 57)

Identify the variables.Solve for the unknown.Substitute and solve.

Practice 2E, p. 58

Kinematic Equations

xavvtatvx

tavvtvvx

t

va

t

xvxxx

ifi

iffi

if

2 2

1

)(2

1

222

2.3 Falling Objects

Objectives

1. Relate the motion of a freely falling body to motion with constant acceleration.

2. Calculate displacement, velocity, and time at various points in the motion of a freely falling object.

3. Compare the motions of different objects in free fall.

Free Fall

• In the absence of air resistance, all objects fall to earth with a constant acceleration

• The rate of fall is independent of mass

• In a vacuum, heavy objects and light objects fall at the same rate.

• The acceleration of a free-falling object is the acceleration of gravity, g

• g = 9.81m/s2 memorize this value!

• Free fall is the motion of a body when only the force due to gravity is acting on the body.

• The acceleration on an object in free fall is called the acceleration due to gravity, or free-fall acceleration.

• Free-fall acceleration is denoted with the symbols ag (generally) or g (on Earth’s surface).

Free Fall

• Free-fall acceleration is the same for all objects, regardless of mass.

• This book will use the value g = 9.81 m/s2.• Free-fall acceleration on Earth’s surface is –

9.81 m/s2 at all points in the object’s motion. • Consider a ball thrown up into the air.

– Moving upward: velocity is decreasing, acceleration is –9.81 m/s2

– Top of path: velocity is zero, acceleration is –9.81 m/s2

– Moving downward: velocity is increasing, acceleration is –9.81 m/s2

Free Fall Acceleration

Free-Fall Problem• A tennis ball is tossed vertically with an

initial velocity 0f 6.0 m/s. How high will the ball go? How long will the ball be in the air until it returns to its starting point?

• Known: vi = 6.0 m/s; a = -g = -9.81 m/s2

• How high? vf2 = vi

2 + 2aΔy solve for Δy• vf = vi + aΔt solve for Δt• Since the ball goes up Δt and comes down

Δt, time in the air is 2Δt

Sample Problem

• Falling Object• Jason hits a volleyball so that it

moves with an initial velocity of 6.0 m/s straight upward.

• If the volleyball starts from 2.0 m above the floor,

• how long will it be in the air before it strikes the floor?

Sample Problem, continued

1. DefineGiven: Unknown:

vi = +6.0 m/s Δt = ?

a = –g = –9.81 m/s2 Δ y = –2.0 m

Diagram: Place the origin at the Starting point of the ball

(yi = 0 at ti = 0).

2. Plan Choose an equation or situation:

Both ∆t and vf are unknown.

We can determine ∆t if we know vf

Solve for vf then substitute & solve for ∆t 3. Calculate Rearrange the equation to isolate the unknowns:

yavv if 222 tavv if

yavv if 22a

vvt if

vf = - 8.7 m/s Δt = 1.50 s