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Chapter 2

Motion in One Dimension

Motion in One DimensionSections2-1 Displacement and Velocity2-2 Acceleration2-3 Falling objects

Displacement and VelocityMotion One-dimensional motion is the simplest form Motion takes place over time and depends upon

the frame of referenceFrame of reference – a coordinate system for

specifying the precise location of objects in spaceChoose a reference frame and use it consistentlySome reference frames simplify the problem

Displacement and VelocityDisplacement Displacement is a change in position

It is the straight-line distance (with direction) from the starting point to the ending point

Greek letter delta (∆) denotes a change in something

Displacement

displacement = change in position = final position – initial positionSI Unit of Displacement: meter (m)

�x = xf � xi

Displacement and VelocityDisplacement Not always equal to the distance traveled Can be positive or negative (direction)

positive directions: to the right; upwardnegative directions: to the left; downward

Displacement and VelocityVelocity Average velocity is displacement divided by the

time interval

Positive or negative, depending on displacement

Average Velocity

SI Unit of Velocity: meter per second (m/s)

vavg =�x

�t

=xf � xi

tf � ti

average velocity =

change in position

change in time

=

displacement

time interval

Displacement and VelocityWorkbook Problem 2AThe fastest fish, the sailfish, can swim 1.2 × 102 km/h.

Suppose you have a friend who lives on an island 16 km away from the shore. If you send a message using a sailfish as a messenger, how long will it take for the message to reach your friend?

Displacement and VelocityVelocity Velocity is not the same as speed

Average speed tells how fast

Average velocity tells how fast and what direction Velocity can be interpreted graphically

If position vs. time is plotted on a graph, the average velocity can be found by finding the slope of the line

Displacement and VelocityVelocitySuppose a bicyclist is riding with

constant velocity of v = +4 m/s. After riding for one second, his displacement is +4 m. Two seconds later, it is +12 m. The straight line in the graph shows his displacement x for any time t.

Recall that vavg = ∆x/∆t. Notice that ∆x/∆t is the slope of the line in the position vs. time graph. So, the slope of the position vs. time graph is equal to the velocity.

Image credit: Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.

Displacement and Velocity

The graphs of the three segments of the bicyclist's trip each have different slopes. Where the slope is positive, the velocity is also positive. Zero slope means zero velocity. Negative slope indicates negative velocity.

Image credit: Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.

Displacement and VelocityVelocityThe graph of an accelerating object

would look something like this one. The slope of the line tangent to a point is the instantaneous velocity (the velocity at that point in time).

Image credit: Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.

AccelerationChanges in Velocity Acceleration measures the rate of change in

velocity, or how quickly velocity changes

Acceleration can be positive or negative and will agree with the direction of the change in velocity

Average Acceleration

SI Unit of Acceleration: meter per second per second (m/s2)

aavg =�v

�t=

vf � vitf � ti

average acceleration =

change in velocity

time required for change

AccelerationWorkbook Problem 2BIn 1977 off the coast of Australia, the fastest speed by a

vessel on the water was achieved. If this vessel were to undergo an average acceleration of 1.80 m/s2, it would go from rest to its top speed in 85.6 s. What was the speed of the vessel?

AccelerationChanges in Velocity Acceleration has direction and magnitude

When ∆v is negative, the acceleration is negativeWhen ∆v is positive, the acceleration is positive

The slope and shape of the graph describe the object's motionThe slope of the velocity

vs. time graph yields average acceleration.

Image credit: Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.

AccelerationMotion with Constant Acceleration Displacement depends on acceleration,

initial velocity, and timeball moves top to bottom with constant

acceleration (what is the cause?)velocity increases and so does distancevelocity increases the same amount in each

intervaldisplacement increases the same amount in

each interval

and

so

solving for ∆x:

Displacement with Constant Uniform Acceleration

�x = 12(vi + vf)�t

displacement =

12(initial velocity + final velocity)(time interval)

vavg =�x

�t

vavg =vi + vf

2

�x

�t

= vavg =vi + vf

2

AccelerationWorkbook Problem 2CIn England, two men built a tiny motorcycle with a wheel

base (the distance between the centers of the two wheels) of just 108 mm and wheel's measuring 19 mm in diameter. The motorcycle was ridden over a distance of 1.00 m. Suppose the motorcycle has constant acceleration as it travels this distance, so that its final speed is 0.800 m/s. How long does it take the motorcycle to travel the distance of 1.00 m? Assume the motorcycle is initially at rest.

AccelerationVelocity with Constant Uniform Acceleration The average acceleration equation can be

rearranged to yield a useful equation...

useful when displacement is not known

a = �v�t = vf�vi

�t

Velocity with Constant Uniform Acceleration

vf = vi + a�t

final velocity = initial velocity + (acceleration ⇥ time interval)

AccelerationVelocity with Constant Uniform Acceleration Substitute this expression for vf into the previous

displacement equation...

useful when final velocity is not known

�x = 12 (vi + vf)�t

�x = 12 (vi + (vi + a�t))�t

Velocity with Constant Uniform Acceleration

�x = vi�t+ 12a�t

2

displacement = (initial velocity⇥ time interval) +

12acceleration⇥ (time interval)

2

AccelerationWorkbook Problem 2D

AccelerationWorkbook Problem 2D

AccelerationWorkbook Problem 2D

vi = �15.0 m/s

AccelerationWorkbook Problem 2D

vi = �15.0 m/s

vf = +15.0 m/s

AccelerationWorkbook Problem 2D

vi = �15.0 m/s

vf = +15.0 m/s

a = +2.5 m/s2

AccelerationWorkbook Problem 2D

vi = �15.0 m/s

vf = +15.0 m/s

a = +2.5 m/s2

�t = ?

AccelerationWorkbook Problem 2Dvi = �15.0 m/s

vf = +15.0 m/s

a = +2.5 m/s2

�t = ?

AccelerationWorkbook Problem 2Dvi = �15.0 m/s

vf = +15.0 m/s

a = +2.5 m/s2

�t = ?

a = vf�vi�t • appropriate equation

AccelerationWorkbook Problem 2Dvi = �15.0 m/s

vf = +15.0 m/s

a = +2.5 m/s2

�t = ?

a = vf�vi�t

�t = vf�via = 15.0�(�15.0)

2.5 = 12 s

• appropriate equation

• isolate unknown, sub & calc

Falling ObjectsFree Fall Freely falling bodies undergo constant

acceleration All objects dropped near the surface of a

planet fall with the same constant acceleration – this is free fall (neglecting air resistance)

With air resistance, objects do not fall at the same rate.

Free-fall acceleration is denoted as g Near the surface of Earth, g = 9.81 m/s2

Down is negative: a = –g = –9.81 m/s2

Falling ObjectsFree Fall What goes up must come down

Gravity begins working on a thrown object as soon as it is released.

Freely falling objects always have the same downward accelerationThe force of gravity causes a

downward acceleration on a freely falling object of 9.81 m/s2

Falling ObjectsWorkbook Problem 2FThe famous Gateway to the West Arch in St. Louis,Missouri,

is about 192 m tall at its highest point. Suppose Sally, a stuntwoman, jumps off the top of the arch. If it takes Sally 6.4 s to land on the safety pad at the base of the arch,what is her average acceleration? What is her final velocity?