Section 1 – Displacement and Velocity

Pages

To simplify the concept of motion, we will first consider motion that takes place in one direction.

One example is the motion of a commuter train on a straight track.

To measure motion, you must choose a frame of reference. A frame of reference is a system for specifying the precise location of objects in space and time.

For motion equations we will use the following symbols:

x = horizontal displacement (distance – d)

y = vertical displacementvi = initial velocityvf = final velocityt = time intervalaacceleration

x = xf – xi displacement = final position – initial position

• Displacement is a change in position.• Displacement is not always equal to the

distance traveled.• The SI unit of displacement is the meter,

m.

Average velocity is the total displacement divided by the time interval during which the displacement occurred.

f iavg

f i

x xxv

t t t

average velocity = change in position

change in time =

displacement

time interval

• In SI, the unit of velocity is meters per second, abbreviated as m/s.

Velocity describes motion with both a direction and a numerical value (a magnitude).

Speed has no direction, only magnitude.

Average speed is equal to the total distance traveled divided by the time interval.

distance traveledaverage speed =

time of travel

Object 1: positive slope = positive velocity

Object 2: zero slope= zero velocity Object 3: negative slope = negative

velocity

• For any position-time graph, we can determine the average velocity by drawing a straight line between any two points on the graph.

• If the velocity is constant, the graph of position versus time is a straight line. The slope indicates the velocity.

The instantaneous

velocity at a given time can be determined by measuring the slope of the line that is tangent to that point on the position-versus-time graph.

The instantaneous velocity is the velocity of an object at some instant or at a specific point in the object’s path.

Guess Format!!!!

You and your friend are walking to Piston Petes at an average velocity of 0.98 m/s. If it takes you 34 minutes to arrive, what was your displacement?

t

xv

Two students walk in the same direction along a straight path, at a constant speed – one at 0.90 m/s and the other at 1.90 m/s. Assuming that they start at the same point and the same time, how much sooner does the faster student arrive at a destination 780 m away?

Section 2 – Acceleration

Acceleration is the rate at which velocity changes over time.

f iavg

f i

v vva

t t t

change in velocityaverage acceleration =

time required for change

• An object accelerates if its speed, direction, or both change.

• Acceleration has direction and magnitude. Thus, acceleration is a vector quantity.

With an average acceleration of -1.2 m/s2, how long will it take a cyclist to bring a bicycle with an initial speed of 6.5 m/s to a complete stop?

t

va

It takes 4.8 seconds for a car’s speed to increase by 10. m/s. What is its acceleration?

Given:

v = 10.m/s

Δt = 4.8 sec

a = ?

Formula

t

va

Substitution

ssm

a8.4

.10

Answer: a = 2.1 m/s2

When velocity changes by the same amount during each time interval, acceleration is constant.

Since velocity, displacement, and acceleration are all vector quantities, you must keep direction in mind when you substitute values into the equations. Up or to the right are considered to be positive directions. Down or to the left are considered to be negative directions.

A car is traveling at 50.0 m/s must slow down to 30.0 m/s in the next 10.0 m. What deceleration must the car have?

Given:

vi = 50.0 m/s

vf = 30.0 m/s

d = 10.0 m

a = ?

Formula

x

vva

xvv

if

if

2

222

22

Substitution

)0.10(2

)0.50()0.30( 22

msm

sm

a

Answer: 20.80s

ma

A rocket is capable of accelerating at 800. m/s2. How long after lift off will the rocket reach 500. m/s?

Given:

a = 800. m/s2

vf = 500. m/s

vi = 0 m/s

t = ?

Formula

a

vvt

tavv

if

if

Substitution

2

2

800

500

800

0500

smsm

t

smsm

sm

t

Answer: t = 0.625 sec

A person pushing a stroller starts from rest, uniformly

accelerating at a rate of 0.500 m/s2. What is thevelocity of the stroller after it has traveled 4.75 m?

Given: vi = 0 m/s

a = 0.500 m/s2

x = 4.75 m

Unknown: vf = ?

Equation:

Substitution:

Solution:

2 2 2f iv v a x

2 2(0 m/s) 2(0.500 m/s )(4.75 m)fv

2.18 m/sfv