Kinematics Kinematics Describing Motion Reference Frames Measurements of position, distance or speed must be with respect to a frame of reference. Measurements.

KinematicsKinematics

Describing Motion Describing Motion

Reference FramesReference Frames

Measurements of position, distance or Measurements of position, distance or speed must be with respect to a speed must be with respect to a frame of frame of reference.reference.

What is the speed of a person with respect to the ground if she walks toward the back of the train at 5km/h while the train moves forward at 40 km/h?

Coordinate axes are used to represent the frame of reference

DisplacementDisplacement

Defined as Defined as change in positionchange in positionWhat is the displacement of a person who walks 100 m East then 60 m West?

Answer: 40 m East

Displacement has both magnitude and Displacement has both magnitude and direction – it is a direction – it is a vectorvector

Representing DisplacementRepresenting Displacement

Let xLet x11 be position of object at time t be position of object at time t11

Let xLet x22 be position at time t be position at time t22

Then displacement Then displacement x = xx = x22 – x – x11

Greek letter delta) means changeGreek letter delta) means change

If a person starts at x1 = 40m and walks to the left until reaching x2 = 10m what is the displacement?

Answer: x = x2 – x1 = 10m – 40m = -30m

Average Speed and VelocityAverage Speed and Velocity

Velocity is speed Velocity is speed andand direction direction Average speed = distance traveled Average speed = distance traveled ÷ time ÷ time

elapsedelapsed Average velocity = displacement Average velocity = displacement ÷ time ÷ time

elapsedelapsed Not always equalNot always equal

Find average speed and velocity for a trip 60m North followed by 40m South in 10 seconds

Answers: 10 m/s; 2 m/s N

ExampleExample

During a four second interval a runner’s During a four second interval a runner’s position changes from xposition changes from x11 = 50m to x = 50m to x22 = =

10m. What was the average velocity?10m. What was the average velocity? vvav av = = x/x/t = (xt = (x22 – x – x11 )/4s = -40m/4s = -10 m/s )/4s = -40m/4s = -10 m/s

What is the average speed? (always positive!)What is the average speed? (always positive!)Challenge: convert this velocity to kilometers per hour

-10 m/s x 1 km/1000m x 3600 sec/ 1h = -36 km/h

Pro tip: use conversion tables on inside front cover of your text

How Far?How Far?

How far could a How far could a runner traveling at an runner traveling at an average speed of 36 average speed of 36 km/h go in 20. km/h go in 20. minutes?minutes?

20 min = 1/3 hour20 min = 1/3 hour DDxx = v = vavav tt = 36 km/h = 36 km/h

x 1/3h = 12 km = 1.2 x x 1/3h = 12 km = 1.2 x 101044 m m

Three WaysThree Ways

D = s x TD = s x T S = D/TS = D/T T = D/sT = D/s

Instantaneous VelocityInstantaneous Velocity

Velocity at a particular instant of timeVelocity at a particular instant of time Defined as average velocity over an Defined as average velocity over an

infinitesimally short time intervalinfinitesimally short time interval v = lim(as v = lim(as t t --> 0) --> 0) x/x/tt Finite because both numerator and Finite because both numerator and

denominator approach zero; limit denominator approach zero; limit approaches a definite valueapproaches a definite value

AccelerationAcceleration

Acceleration is how fast velocity changesAcceleration is how fast velocity changes Average acceleration = change of velocity Average acceleration = change of velocity ÷ ÷

time elapsedtime elapsed a a av av = (v= (v22 – v – v11)/(t)/(t22 –t –t11) = ) = v/v/tt

ExampleExample

Find average acceleration of a car that Find average acceleration of a car that accelerates along straight road from rest to accelerates along straight road from rest to 80 km/h in 5 seconds80 km/h in 5 seconds

aaav av = (80 km/h – 0 km/h)/5s = 16 km/h/s= (80 km/h – 0 km/h)/5s = 16 km/h/s

Convert to m/s/sConvert to m/s/s

80 km/h(1000m/1km)(1h/3600s) = 22.2 m/s80 km/h(1000m/1km)(1h/3600s) = 22.2 m/s a a av av = (22.2 m/s – 0.0 m/s)/5.0 s = 4.4 m/s/s = (22.2 m/s – 0.0 m/s)/5.0 s = 4.4 m/s/s

or 4. 4 m/sor 4. 4 m/s22

PronouncedPronounced “meters per second squared”“meters per second squared”

Challenge: Can an object with zero velocity Challenge: Can an object with zero velocity have non zero acceleration?have non zero acceleration?

Object Slowing DownObject Slowing Down

Called decelerationCalled deceleration Find average acceleration of a car moving to the Find average acceleration of a car moving to the

right(+x direction) 15.0 m/s when driver brakes to right(+x direction) 15.0 m/s when driver brakes to 5.0 m/s in 5.0 s?5.0 m/s in 5.0 s?

aaavav = =v/v/t = (5.0 m/s – 15.0 m/s) t = (5.0 m/s – 15.0 m/s) ÷ 5.0 s =÷ 5.0 s = -2.0 m/s-2.0 m/s2 2 (acceleration negative)(acceleration negative) Would acceleration still be negative if car was Would acceleration still be negative if car was

moving to the left? moving to the left? NO! Its acceleration vector would then point to the NO! Its acceleration vector would then point to the

right and be positive.right and be positive.

Constant(Uniform) AccelerationConstant(Uniform) Acceleration

Let tLet t11 = 0, t = 0, t22 = t = elapsed time = t = elapsed time

Re-name xRe-name x11 = x = x0 0 ; v; v11 = v = v00

Average velocity vAverage velocity vav av = (x – x= (x – x00)/t)/t

aaavav = (v – v = (v – v00)/t )/t

From these it is possible to derive (next From these it is possible to derive (next slide)slide)

For Uniform Acceleration, aFor Uniform Acceleration, a

v = vv = v00 +at +at (a)(a)

x = xx = x00 +v +v00t + 1/2att + 1/2at22 (b)(b)

vv22 = v = v00 22+ 2a(x-x+ 2a(x-x00)) (c)(c)

vvavav = (v = (v00 + v)/2+ v)/2 (d)(d)

Each equation can be solved for any of the Each equation can be solved for any of the variablesvariables

Problems can be solved more than one wayProblems can be solved more than one way

v = vv = v00 +at (a) +at (a)

Describes change in velocity under uniform acceleration.

tells how fast a particle will be going at time t if at time zero its velocity was v0

x = xx = x00 +v +v00t + 1/2att + 1/2at2 2 (b) (b)

Describes change in position under uniform acceleration

Sometimes called "equation of motion." tells where a particle will be at time t, if at

time zero it was at x0 moving with velocity

v0

vv22 = v = v00

22+ 2a(x-x+ 2a(x-x00) (c)) (c)

Velocity equationVelocity equation Tells what velocity will be after a particle Tells what velocity will be after a particle

with initial velocity vwith initial velocity v0 0 accelerates a distance accelerates a distance

x - xx - x00 with uniform acceleration a with uniform acceleration a

Take square root to find vTake square root to find v Simplifies to vSimplifies to v2 2 == 2ad when v 2ad when v0 0 = 0 and = 0 and

d = displacementd = displacement

vvavav = (v = (v00 + v)/2 (d)+ v)/2 (d)

Average (mean) velocity of particle for a Average (mean) velocity of particle for a trip under uniform accelerationtrip under uniform acceleration

Provides shortcut way to solve certain Provides shortcut way to solve certain problemsproblems

Special Case of Free Fall*Special Case of Free Fall*

v = vv = v00 –gt a = -g –gt a = -g

y = yy = y00 +v +v00t - 1/2gtt - 1/2gt22

vv22 = v = v00 2 2 - 2g(y-y- 2g(y-y00) ) With y up positive, g = 9.80 m/s/s With y up positive, g = 9.80 m/s/s *Free Fall is motion under the influence of gravity alone

This form of equations assumes down is positive

Concept Check (1)Concept Check (1)

The velocity and acceleration of an objectThe velocity and acceleration of an object– (a) must be in the same direction(a) must be in the same direction– (b) must be in opposite directions(b) must be in opposite directions– (c) can be in the same or opposite directions(c) can be in the same or opposite directions– (d) must be in the same direction or zero(d) must be in the same direction or zero

(c) Can be in the same or opposite directions. Example: a rock thrown upward

Concept Check(2)Concept Check(2)

At the top of its pathAt the top of its path the velocity and the velocity and acceleration of a brick thrown upward areacceleration of a brick thrown upward are

(a) both non zero(a) both non zero

(b) both zero(b) both zero

(c) velocity is zero acceleration is non zero(c) velocity is zero acceleration is non zero

(d) acceleration is zero, velocity is non zero(d) acceleration is zero, velocity is non zero

(c) is correct; v = 0, a = 9.80 m/s/s downward

Problem Solving TipsProblem Solving Tips

Read and re-read the problemRead and re-read the problem Make a diagram with all given infoMake a diagram with all given info Ask yourself “what is problem asking?”Ask yourself “what is problem asking?” Ask which physics principles applyAsk which physics principles apply Look for most applicable equationsLook for most applicable equations Be sure problem lies within their range of Be sure problem lies within their range of

validityvalidity

Special TricksSpecial Tricks

Break problem up into parts,Break problem up into parts,

like up and down part of path of an object like up and down part of path of an object thrown upthrown up

Use symmetryUse symmetry Take situation to an extreme and look for a Take situation to an extreme and look for a

constraintconstraint Choose reference frame that makes problem Choose reference frame that makes problem

easiesteasiest

Problem solving…Problem solving…

Do algebraic calculationsDo algebraic calculations Be aware you may have to solve equations Be aware you may have to solve equations

simultaneouslysimultaneously Do arithmetic at endDo arithmetic at end Check units and significant figuresCheck units and significant figures Ask, “is answer reasonable?”Ask, “is answer reasonable?”

Your ChoiceYour Choice

You may choose up positive or negative; You may choose up positive or negative; same with left or rightsame with left or right

You may put the zero of coordinates You may put the zero of coordinates anywhere you chooseanywhere you choose

Generally make choices that minimize Generally make choices that minimize number of negative quantitiesnumber of negative quantities

ExamplesExamples

What time is required for a car to travel What time is required for a car to travel 30.0 m while accelerating from rest at a 30.0 m while accelerating from rest at a uniform 2.00 m/suniform 2.00 m/s22

x = ½ a tx = ½ a t22

tt22 = 2x/a = 2x/a t = (2x/a)t = (2x/a)1/21/2

t = (2(30m)/2.00 m/st = (2(30m)/2.00 m/s22) = 5.48s) = 5.48s

a = 2.00m/s2

x0 = 0

v0 = 0

x = 30m

Distance to BrakeDistance to Brake

A car traveling 28 m/s brakes at -6.0 m/sA car traveling 28 m/s brakes at -6.0 m/s22. What . What distance is required to stop.distance is required to stop.

Method 1. Use Method 1. Use vv22 = v = v00 22+ 2a(x-x+ 2a(x-x00))

Solve for x = xSolve for x = x00 + (v + (v22 – v – v0022) ÷ 2a) ÷ 2a

x = (0 – 28 m/sx = (0 – 28 m/s22)) ÷ 2(-6.0 m/s÷ 2(-6.0 m/s22) = 65m) = 65m Method 2. Use vMethod 2. Use vavav = (v = (v00 + v)/2 = 14 m/s+ v)/2 = 14 m/s

v = vv = v00 +at ; t = (v – v +at ; t = (v – v00)/a = -28m/s ÷ -6.0 m/s)/a = -28m/s ÷ -6.0 m/s2 2 = 4.67 = 4.67

ss

x = vx = vav av t = 14 m/s (4.67s) = 65 mt = 14 m/s (4.67s) = 65 m

Moral of StoryMoral of Story

There is always more than one way There is always more than one way to solve a (kinematics) problem!to solve a (kinematics) problem!

You do not have to use the teacher’s way or You do not have to use the teacher’s way or the textbook way so long as the method and the textbook way so long as the method and answer are correct.answer are correct.

Warning: Sometimes it is possible to get a Warning: Sometimes it is possible to get a correct answer using an incorrect method – correct answer using an incorrect method – no creditno credit

Free FallFree Fall

Which falls faster, an elephant or a mouse?Which falls faster, an elephant or a mouse?

Galileo’s ExperimentGalileo’s Experiment

He asked, which He asked, which would reach the would reach the ground first, a marble ground first, a marble or a cannonball?or a cannonball?

Courtesy Dan Heller Photography

Galileo’s DiscoveryGalileo’s Discovery

All objects accelerate to earth equally All objects accelerate to earth equally (regardless of mass)(regardless of mass)

Air resistance must be neglected for this to Air resistance must be neglected for this to be truebe true

Acceleration due to gravity a =Acceleration due to gravity a =

g = 9.80 m/sg = 9.80 m/s22

Velocity ReachedVelocity Reached

In free fall an object’s In free fall an object’s speed increases by speed increases by about 10 m/s in each about 10 m/s in each second. second.

v = gt v = gt ~~ 10t 10t

g g ~~ 10 m/s 10 m/s22

Time to Time to fall(s)fall(s)

Speed (m/s)Speed (m/s)

00 00

11 1010

22 2020

33 3030

44 4040

55 5050

Distance Fallen From RestDistance Fallen From Rest

Distance fallen in t Distance fallen in t secondsseconds

d = 1/2gtd = 1/2gt22

timetime distancedistance

11 55

22 2020

33 4545

44 8080

55 125125

Example: Ball Thrown UpwardExample: Ball Thrown Upward

Person throws ball upward with v = 15.0 Person throws ball upward with v = 15.0 m/s. (a) How high does it go? (b) How long m/s. (a) How high does it go? (b) How long is it in the air?is it in the air?

Choose y positive up, negative downChoose y positive up, negative down Choose y = 0 at throw heightChoose y = 0 at throw height Use vUse v22 = v = v00

22 +2ay +2ay

y = (vy = (v22 – v – v0022)/2a = (0 – (15.0m/s))/2a = (0 – (15.0m/s)22)/2(-)/2(-

9.80m/s9.80m/s2) 2) = 11.5 m= 11.5 m

Time In The AirTime In The Air

Use y = vUse y = v00t + ½ a tt + ½ a t22

0 = (15.0 m/s)t + ½(-9.80 m/s0 = (15.0 m/s)t + ½(-9.80 m/s22)t)t22

Factor: (15.0 m/s – 4.90 m/sFactor: (15.0 m/s – 4.90 m/s22t)t = 0t)t = 0 Two solutions: t =0 corresponds to Two solutions: t =0 corresponds to

instant ball throwninstant ball thrown T = 15.0 m/s T = 15.0 m/s ÷ 4.90 m/s÷ 4.90 m/s2 2 = 3.06 s = 3.06 s

corresponds to instant ball returns to groundcorresponds to instant ball returns to ground

Another Way?Another Way?

Can you think of another way?Can you think of another way? Find time to rise from Find time to rise from v = vv = v00 +at +at

t = (v – vt = (v – v00)/a = (15m/s – 0)/9.80 m/s = 1.53s)/a = (15m/s – 0)/9.80 m/s = 1.53s

Then use average velocity 7.5 m/s x 1.53s = 11.5 m Then use average velocity 7.5 m/s x 1.53s = 11.5 m to find height.to find height.

Total time in air is double 1.53s by symmetry. Total time in air is double 1.53s by symmetry. Gravity gives back speed on the way down at the Gravity gives back speed on the way down at the same rate it took it away on the way up.same rate it took it away on the way up.

PuzzlersPuzzlers

What is the velocity of a ball thrown What is the velocity of a ball thrown upward at its maximum height?upward at its maximum height?

What is its acceleration at that height?What is its acceleration at that height? Are velocity and acceleration always in the Are velocity and acceleration always in the

same directionsame direction

Useful Form of Kinematics Useful Form of Kinematics Equations for Free FallEquations for Free Fall

v = vv = v00 - gt - gt (a)(a)

y = yy = y00 +v +v00t - 1/2gtt - 1/2gt22 (b)(b)

vv22 = v = v00 2 2 - 2g(y-y- 2g(y-y00)) (c)(c)

vvavav = (v = (v00 + v)/2+ v)/2 (d)(d)

Here assuming up is positiveHere assuming up is positive g = 9.80 m/sg = 9.80 m/s22

Graphing MotionGraphing Motion

Position Graph (x or y vs. t)Position Graph (x or y vs. t)– Slope is velocitySlope is velocity– If curved slope defined as slope of tangent to If curved slope defined as slope of tangent to

the curve at that pointthe curve at that point Velocity GraphVelocity Graph

– Slope is accelerationSlope is acceleration– Area under graph is displacementArea under graph is displacement

Acceleration GraphAcceleration Graph– Constant and non zero for uniform accelerationConstant and non zero for uniform acceleration

Concept Check (3)Concept Check (3)

Which of these could be the Which of these could be the velocityvelocity graph graph of a rock thrown upward, then falling of a rock thrown upward, then falling downward? (Assume up is positive)downward? (Assume up is positive)

(a)(a) (b) (b) (c)(c)

(d)(d) (e)(e)