KinematicsKinematics

Kinematics is the study of the motion ofobjects which includes distance,displacement, speed, velocity,

acceleration,and time.

Vocabulary Used In KinematicsVocabulary Used In KinematicsFrame of Reference

A set of axes or a point from which you make measurements.

These measurements can be position, distance, displacement, speed,

velocity, etc.

Position

Separation between a frame of reference and an object measured in

cm, m, km.

Distance

Length measurement of separation.

Displacement

A change in position of an object with respect to starting point.

Do not consider distance and displacement synonymous!

If you walk across a room, turn around

and return to your starting point, you have traveled the 2 × width of the

room.

However, your displacement is zerobecause you ended up where youstarted.

Vector or Scalar?Vector or Scalar?

In everyday life, when you measure thingslike mass, volume, length, temperature, etc.you often forget the unit and never providedirection.

Quantities that only have magnitude (size) are called scalar quantities.

In physics, we often make measurements thathave both magnitude and direction and neverforget the units!

Examples of vector quantities are displacement,

velocity, and acceleration.

For example, when you say that you aretraveling at 35 km/h N, you are stating yourdirection as well as your rate.

When you weigh something and find it to be25 N (newtons), you are giving the magnitudeand direction (understood to be straight

down).

Vector vs ScalarVector vs Scalar

Two very important measured quantities arespeed and velocity.

Average speed is a scalar quantity becauseonly the magnitude is given with no specifieddirection.

average speed

100 m/s

distancetime

s= =

Average velocity is a vector quantity because inaddition to magnitude, direction is always

given.

average velocity

100 m/s N

Velocity can either be positive or negative.

If the object is moving in the direction that has been defined as positive, then the velocity is positive, otherwise it is negative.

displacementtime=v =

This graph shows the displacement of anobject that started next to you at the origin.

The displacement is positive throughout the time

interval (t = 0-16 s) which means the object moved

in a positive direction with respect to you.

When the object comes to a stop at 16 s, itsmaximum displacement is 35 m N.

N

S

From 0-4 s, the object travels with a constantvelocity as indicated by the constant slope.

From 4-6 s, the object has stopped (the slope is 0),

from 6-10 s, the object travels at a higher constant

velocity than 0-4 s, and from 10-16 s, it moves the

slowest of its three velocities.

v =ΔsΔt

v =3.3 m/s N

=10. m3.0 s

=vΔsΔt

4-6 s:

vΔsΔt

= =0 m/s

6-10 s:

18. m4.0 s

=vΔsΔt

= = 4.5 m/s N

0-4 s:

Instantaneous vs Average Instantaneous vs Average VelocityVelocity

Instantaneous velocity is the velocity at oneparticular instant of time.

When driving a car, each time you look at the speedometer, you are reading the instantaneous velocity provided you are not lost.

To determine your average velocity, you wouldneed to use your odometer and a watch.

How fast are you goingat 2.0 s?Ans. 3.3 m/s N

How fast are you goingat 5.0 s?Ans. 0 m/s

How fast are you going at 13.0 s?Ans. Find the slope of the line between 10.0 and 16.0 s. The slope is constant so the

velocity is constant. If the velocity is constant,

then the object must still be moving north.

Constant Velocity ProblemsConstant Velocity Problems

A train leaves Providence traveling at aconstant velocity of 36.0 m/s due west.

(a) How long does it take the train to travel

1620.0 m?

vave = 36.0 m/s Δs = 1620.0 m

vave = Δs

Δt

.

36.0 m/s = 1620.0 mΔt

Δt = 45.0 s

(b)What is the velocity of the train in km/h?

vave = ms

36.0 1 km×103 m

× 3600 s1 h

vave = 130. km/h W

You plan on driving from the west coast to the

east coast of the United States (4600 km) at

55 mi/h and you plan to stop every 2.0 h to rest

and refuel which will take 30. min.

(a) How long will the trip take?vave = 55

mih × 1 mi

1760 yd× 36 in

1 yd ×2.54 cm1 in

×1 m

102 cm ×1 km103 m

=88 km/h

.

vave =ΔsΔt

88kmh = 4600 km

Δt

Δt =52.27 h

ΔtT=52.27 h+52.27 h×1 stop

2 h×

30 min

1 stop

×1 h

60 min= 65.3 h

(b) What is the vave for the entire trip?

vave =ΔsΔt

= 4600 km65.3 h

=70.4 km/h

AccelerationAcceleration

Acceleration is defined as the rate at which the

velocity changes.

A change in velocity can be either in magnitude, direction, or both.

A rate always implies a time interval - /s,

/h, /fortnight.

This means acceleration always has two

units of time and they do not have to be

the same units although many times they

are.

An example of acceleration would be 9.80 m/s/s = 9.80 m/s2 or -9.80 m/s2.

The average acceleration is given byaave = Δv

Δt=

vf - vi

tf - ti

=m/ss

=ms2

Acceleration is a vector quantity meaning it has

both magnitude and direction.

If the acceleration is positive and in the same direction of the velocity, the

object speeds up.

If the acceleration is negative and in the opposite direction of the velocity, slows down.

As with velocity, average acceleration must

always have two points to be calculated and

instantaneous acceleration has only one point.

Interpreting Acceleration Interpreting Acceleration ProblemsProblems

If the problem says:

“… starts from rest …”, vi = 0 m/s

“… moves at a constant velocity …“, aave = 0.

“… comes to rest …”, vf = 0.

This graph shows the velocity of an from0-16.0 s.

From 0-4.0 s, the objectis accelerating positively.

From 4.0-6.0 s, the object is accelerating at a

greater rate.

From 6.0-9.0 s, the object is moving at a constant

velocity.

From 9.0-11.0 s, the object is decelerating.

From 11.0-13.0 s, theobject is again speedingup.

From 13.0-16.0 s, the object is accelerating at alower rate than from 11.0-13.0 s.

All of the above descriptions were determined by

looking qualitatively at the slope of the graph ineach time interval.

Where is the object in relation to you?

You don’t know!

Think about it. The object could have started next to you or started from

rest 100 m in front of you.

Would the same graph apply in either scenario?

You do know that the object continues to

get further from you.

a =ΔvΔt

a =1.25 m/s2 N

=5.0 m/s4.0 s

=ΔΔt

4-6 s:

aΔvΔt

= =

6-8.5 s:

=aΔvΔt

= 0

0-4 s:

12.5 m/s2.0 s

=6.25 m/s2 N

Acceleration Due To GravityAcceleration Due To Gravity

All objects accelerate at the same rate when air

resistance is ignored.

This is true no matter how large or small an object is, no matter the height from which it is dropped.

a = g = 9.80 m/s2 = 980 cm/s2 = 32 ft/s2

g changes from location to location.

g is a vector quantity meaning it has both

magnitude and direction.

g changes from location to location.

The sign convention can be very confusing and usually there is not a

lot of uniformity from resource to

resource.

Symmetry ConsiderationsSymmetry Considerations

A

B

D

C Point A:

vi > 0g = -9.80 m/s2

Point B:

vi > 0g = -9.80 m/s2

Point C:

vi = 0g = 9.80 m/s2

vi = vf

g = 9.80 m/s2

Point D:

Symmetry ConsiderationsSymmetry Considerations

A

B

D

C Point A to Point B:

Velocity is decreasing but is positive.

g = -9.80 m/s2

Point C to Point D:

Velocity is increasing but is negative.

g = 9.80 m/s2

Kinematics EquationsKinematics Equations

The following equations are for constantacceleration.

vf = vi + aΔt

x = xi + vit + ½aΔt2

vf2 = vi

2 + 2a(x-xi)

v=vf + vi

2

Some thoughts about the previous equations:

The symbols i and f stand for initial and final respectively.

Sometimes, a subscript 0 is used for initial and no subscript is used for

final.

It really does not matter, it is all what you

get used to.

x and xi represent the position and sometimes s is used in place of x.

Many times xi = 0 which simplifies the equation.

x – xi represents the displacement.

Many times x is used to represent horizontal displacement and y is substituted for x when the

displacement is vertical.

Some g ProblemsSome g Problems

You throw a ball straight up to the top of aflagpole and it takes 4.0 s for the ball to

return.

(a) What is the height of the flagpole in meters?ΔtT = Δtup + Δtdown = 4.0 s

Δtup = Δtdown = 2.0 s

g = ± 9.80 m/s2

x = xi + vit + ½aΔt2

Δx = 0 + ½ × 9.80 m/s2 × (2.0 s)2 = 20. m

Some final thoughts:

From symmetry considerations, you may

use the upward or downward path.

The downward path was chosen which

meant both xi and vi = 0.

Many times there may be an alternate formula(s) which could be used to solve the problem which is fine.

(b) What is the velocity of the ball if you catch it from the same height you threw it?

vf2 = vi

2 + 2a(x-xi) = vi2 + 2gΔx

vf2 = 0 + 2 × 9.80 m/s2 × 20. m

vf = 20. m/s

Another g ProblemAnother g Problem

A skydiver jumps from an airplane and hisparachute opens after 1.8 s.

(a) What is his speed when his parachute opens?

vi = 0 g = 9.80 m/s2

vf = vi + gΔt = 0 + 9.80 m/s2 × 1.8 s

vf = 18 m/s

(b) What was the average speed during the first

1.8 s?

(c) How far does the skydiver fall in 1.8 s?

vave =vi + vf

2=

0 + 18 m/s2 = 9.0 m/s

vave =Δs

Δt

Δs =9.0 m/s × 1.8 s=16 m

(d) Immediately after the parachute is opened,

what happens to the skydiver’s acceleration? Justify your answer.

As soon as the parachute is opened, the skydiver is no longer in free fall. The parachute creates air resistance which means that the weight of the skydiver is

not the only force acting. The acceleration decreases.

(e) Immediately after the parachute is opened,

what happens to the skydiver’s velocity? Justify your answer.

At the instant the parachute is opened, the

acceleration decreases but it is still positive

meaning the velocity still is increasing. The

velocity will be increasing by less than 9.80 m/s2.

Graphing Uniform AccelerationGraphing Uniform Acceleration

When the acceleration of an object is uniform, its displacement varies withthe square of the time.

This results in the graph of Displacement vs Time to be a parabola. The displacement increases much faster forequal sized intervals than the time does.

You can perform the same type of graphicalanalysis as done before, but somewhatdifferently.

As shown in the next graph, the instantaneousvelocity does not equal the average velocity.

This only occurs when you have uniform motion.

Uniform motion is either moving in a straight line at constant speed or not moving at all (a = 0).

.

Displacement vs Time

0.0

400.0

800.0

1200.0

1600.0

2000.0

0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 20.0

Time (s)

Dis

pla

cem

ent

(m)

To find theinstantaneousvelocity at 17.0 s,one must draw atangent to the curveat 17.0 s.

Taking the slope of the tangent, gives:

v =Δs

Δt=

1416 m – 920 m3.0 s

=165 m/s

Don’t get confused!

We did use two points on the tangent line

to determine the slope but the slope was

computed for only one point on the parabola.

To determine the average velocity during a time

interval, you use the same calculation asbefore

vave =Δs

Δt

And The Analysis Goes On …And The Analysis Goes On …

Another graph used extensively withaccelerated motion is Velocity vs Time.

.

One thing tonote is that theslope is constantwhich is indicativeof uniform orconstant velocity.

The slope of the straight line is given by:

a =Δv

Δt=

98.0 m/s – 0 m/s10.0 s

=9.80 m/s2

.

One thing tonote is that theslope is constantwhich is indicativeof uniform orconstant velocity.

The slope of the straight line is given by:

a =Δv

Δt=

98.0 m/s – 0 m/s10.0 s

=9.80 m/s2

The Speed TrapThe Speed Trap

A motorist is speeding on Rt. 95N at a constant

velocity of 30.0 m/s. He zips by a speed trapand it takes the state police officer 5.00 s tofinish his coffee. He then pursues the

motoristaccelerating at a constant 5.20 m/s2.

(a) How long does it take for the police officer to catch the speeder?

Δtm = t + 5.00 s Δtp = Δtp

vave = 30.0 m/s vi = 0Δxm = Δxp ap = 5.20 m/s2

Δxp = Δxm, xi = 0

Δxm = vave Δtm = 30.0(t + 5.00 s)

x = xi + vit + ½aΔt2

Δxp = 0 + ½ × 5.20 m/s2 × Δt2

30.0(t + 5.00) = ½ × 5.20 × Δt2

It is not uncommon that in problems such asthese, you omit the units as you must use thequadratic formula.

There is a quadratic solver program availablethat you can download into your TI-83 or TI-84found at Pick-Up Window: Freeware or learn touse the solver feature already installed on yourTI-83 or TI-84 and the instructions can be foundat Chemistry Reference Sheets.

2.60Δt2 – 30.0t – 150. = 0

Δt = 15.3 s

(b) How far did the police have to travel in his pursuit?

x = xi + vit + ½aΔt2

Δxp = 0 + ½ × 5.20 m/s2 × (15.3 s)2 = 609 m

A Graphical Look At The Speed A Graphical Look At The Speed TrapTrap

.

The point of intersection represents the point at whichthe police officer catches up with the speeder.

.

A = b × h = 30.0 m/s × 20.3 s = 609 m

A = ½ × b × h = ½ × 79.6 m/s × 15.3 s = 609 m

Interpreting Graphs SummaryInterpreting Graphs Summary

Type of Graph Slope Area Under Graph

Disp vs Time vave N/A

Velocity vs Time aave Δs

Accel vs Time N/A Δv

A = bh(rectangle)

A = ½bh (triangle)

Wrap Up QuestionsWrap Up Questions

If the velocity of an object is zero, can theacceleration ever be nonzero?

Yes, if an object is moving in a straight line,stops, and reverses direction. A good exampleof this is when a ball is thrown straight up in

theair and reaches its highest point. Theinstantaneous velocity is 0 but the

accelerationis 9.80 m/s2 because the ball is changingdirection.

If the velocity of an object is nonzero, can theacceleration be zero?

Yes, if an object is moving with a constantvelocity, then the acceleration is zero.

Can the average velocity and instantaneousvelocity of an object ever be equal?

Yes, if the object moves at a constant velocityfor a specified amount of time, then theinstantaneous and average velocity are thesame during the time interval.