Clickers Setup• Turn on your clicker (press the power button)

• Set the frequency:– Press and hold the power button– Two letters will be flashing– If it’s not “AC”, press “A” and then “C”

• If everything works, you should see “Welcome” and “Ready”

Kinematics of 1-dimensional motionDisplacement: 0xxx

0x

)(txx

x

0

t

x

tt

xxv x

12

12Average velocity:

Instantaneous velocity:

dt

dx

t

xv x

lim0t

Average acceleration:

t

v

tt

vva xxxx

12

12

Instantaneous acceleration:

dt

dv

t

va xxx

lim0t

x

t txv

t

t

x

2

2

dt

xda x

Curvature

Equations of Kinematics for Constant Acceleration

tavtvt

vva xxx

xxx

0

0 )( xv

t

)1(

)2(

tavvv

v

tvxxt

xxv

xxxx

x

xx

2

1

2 00

00

)3( 200 2

1)( tatvxtx xx

)4( )(2 020

2 xxavv xxx

Proof:x

xx

x

xxxxxx

a

vv

a

vvvvt

vvxx

222

20

2000

0

Result: Position x, velocity vx, acceleration ax, and time t are related by the kinematic equations. There are two initial conditions x0 and v0x .

xv

Parabola

Freely Falling BodiesAcceleration due to gravity g=9.8 m/s2

Example 2.6: Find velocity and position ofa coin freely falling from rest after 1, 2, 3 s.

Solution: 1. Choose axes and plot trajectory.2. Indicate known and unknown data

y0 v0y y vy ay t

0 0 ? ? -g=-9.8m/s2 1, 2, 3 s

By choiceof axis

Implied data

3. Write and solve equations in symbolsgtvtavtv yyyy 0)()1(

)3( 2200 5.05.0)( gtytatvyty yy

4. Do numerical calculations.5. Check the answer: dimensions, signs, functional dependencies, scales…

Problem Solving Steps1. Geometry & drawing: trajectory, vectors ,coordinate axes free-body diagram, …

2. Data: a table of known and unknown quantities, including “implied data”.

3. Equations ( with reasoning comments ! ), their solution in algebraic form, and the final answers in algebraic form !!!

4. Numerical calculations and answers.

5. Check: dimensional, functional, scale, sign, … analysisof the answers and solution.

,...,,, jFavr

,,, zyx

Exam Example 1 : Coin Toss Vy=0y

00v

0vv

ay0 v0y y vy ay t

0 +6m/s ? ? -g=-9.8m/s2 ?Questions:(a) How high does the coin go?

)4( )(2 020

2 yyavv yyy msm

sm

a

vvy

y

y 8.1/8.92

)/6(

2 2

220

2

(b) What is the total time the coin is in the air?

)1(t

vva yyy

0 s

sm

sm

a

vvt

y

y 6.0/8.9

/62

0

Total time T= 2 t = 1.2 s(c) What is its velocity when it comes back at y=0 ?

)4( )(2 020

2 yyavv yyy for y=0 and vy<0 yields vy2 = v0

2 → vy = -v0 = - 6m/s

(problem 2.85)

Position and Velocity by Integration

f f

f

i

x

x

t

t

i

t

ttiif

dttvxdxx

tvxxxx

0 0

0

)(00

00

)(tv

0t ftit

iv

f fv

v

t

t

dttavdvvtv0 0

)()( 00

tav Constant-acceleration formulas via integration

atvdtavdttavtvtt

0

00

00 )()(

2/)()( 200

0

0

0

00 attvxdtatvxdttvxxtt

)1(

)3(

Example of kinematic problem with two possible solutions

x0 v0 x vx ax t

0 +3250m/s 215 km ? -10m/s2 ?

Accelerating spacecraft data

Question: Find velocity at a displacement x=215 km ?

X=215 km = 215 000 m

a

0v

smv /2500

smv /2500

x0v

)4( )(2 020

2 xxavv xxx smxavv x /2500220

Two segments of motion with different accelerations

x01 v01 x1 v1 a1 t1

0 0 +120 m ? +2.6 m/s2 ?

Segment 1

Segment 2

x02 v02 x2 v2 a2 t2

x1 v1 ? +12m/s -1.5 m/s2 ?

Question: Find total displacement D ?

)4(sm

xavv

/25

2 112011

msm

smsm

a

vvx

160)/5.1(2

)/25()/12(

2)4(

2

22

2

202

22

2

x

2v

2a

1a

mxxx 1200111 mxxx 1600222

mmmxxD 28016012021

0x1=x02

Segment 1 Segment 2

Exam Example 2: Accelerated Car (problems 2.7 and 2.17)

Data: x(t)=αt+βt2+γt3, α=6m/s, β=1m/s2, γ = -2 m/s3, t=1sFind: (a) average and instantaneous velocities;(b) average and instantaneous accelerations;(c) a moment of time ts when the car stops.Solution: (a) v(t)=dx/dt= α+2βt +3γt2 ; v0=α;

(b) a(t)=dv/dt= 2β +6γt; a0=2β;

(c) v(ts)=0 → α+2βts +3γts2=0

2/)( ttttxv

ttaatvtva 322/))((/))(( 00

ssm

smtt ss 17.1

/6

/)61(

3

33

22

x0

v

a

V(t)

t0ts

α

a(t)

2β

Exam Example 3: Truck vs. Car (problem 2.34)

x0

constv

cv

ca

Data: Truck v=+20 m/sCar v0=0, ac=+3.2 m/s2

Questions: (a) x where car overtakes the truck;(b) velocity of the car Vc at that x;(c) x(t) graphs for both vehicles;(d) v(t) graphs for both vehicles.

Solution: truck’s position x=vt, car’s position xc=act2/2 (a) x=xc when vt=act2/2 → t=2v/ac → x=2v2/ac (b) vc=v0+act → vc=2v

x

t0

truckcar

truckcar

t

V(t)

v

0

t=2v/ac

v/ac

vc=2v

t=2v/ac

Exam Example 4: Free fall past window (problem 2.84)Data: Δt=0.42 s ↔ h=y1-y2=1.9 m, v0y=0, ay= - g

Find: (a) y1 ; (b) v1y ; (c) v2y

y0

y1

y2

V0y=0

V1y

V2y

ayh

1st solution: (b) Eq.(3) y2=y1+v1yΔt – gΔt2/2 → v1y= -h/Δt + gΔt/2(a) Eq.(4) → v1y

2 = -2gy1 →

y1 = - v1y2 /2g = -h2/[2g(Δt)]2 +h/2 – g(Δt)2 /8

(c) Eq.(4) v2y2 = v1y

2 +2gh = (h/Δt + gΔt/2)2

2nd solution:(a)Free fall time from Eq.(3): t1=(2|y1|/g)1/2 , t2=(2|y2|/g)1/2 → Δt+t1=t2

2

11

2

11 22

1||||2

2

)(||||

2

tg

t

h

gyhygt

tghyy

gt

(b) Eq.(4) → ||2 11 ygv y (c) Eq.(4) → )|(|2 12 hygv y

Exam Example 5: Relative motion of free falling balls (problem 2.94) y

0

H

0v aData: v0=1 m/s, H= 10 m, ay= - g

Find: (a) Time of collision t;(b) Position of collision y;(c) What should be H in order v1(t)=0.

1

2

Solution: (a) Relative velocity of the balls is v0 for they have the same acceleration ay= –g → t = H/v0

(b) Eq.(3) for 2nd ball yields y = H – (1/2)gt2 = H – gH2/(2v02)

(c) Eq.(1) for 1st ball yields v1 = v0 – gt = v0 – gH/v0 , hence, for v1=0 we find H = v0

2/g

Pulsars – Rotating Neutron StarsDiscovery (1967) : “Little green men”

t Radio, optical, and X-ray pulses

tPeriod T~ 1ms – 4s is astonishingly stable !!!

T

Pulse duration ssTt 1.01004.0

10 km

n, p+,e-

kmss

kmtcL 30010000300 3

Compare: Rsun ~ 700 000 km

PhotonsNeutronstar

Earth

Pulsar maps have been included on the two Pioneer Plaques as well as the Voyager Golden Record. They show the position of the Sun, relative to 14 pulsars, so that our position both in space and in time can be calculated by potential extraterrestrial intelligences. Pulsar positioning could create a spacecraft navigation system independently, or be an auxiliary device to GPS instruments.