Today:’Review’forExam’1’...

Today: Review for Exam 1 Wednesday: Chapter 5

Exam Informa+on 10/1/14 – Wednesday 7:30 PM All Sec+ons 505-‐509 à MPHY 205 (this room, 30 min a.er class ends) Dura+on à 1 hour 15 min Ø  Please return to the classroom 20 min aLer class is dismissed and wait to be let into the room. Ø  Know your instructor’s name (S+egler) and your sec+on number (505 -‐ 509). Ø  Have your TAMU ID ready to show when turning in your exam. Ø  Calculators are allowed, no pre-‐saved data on them, if you are caught using previously saved data

it will be considered chea=ng and dealt with accordingly. Exam Structure Ø  5 mul+ple choice ques+on

•  You do not need to show your work on these. •  There is no par+al credit.

Ø  4 long problems •  The problems will be similar to homework problems of medium difficulty. •  There will be a mix of ‘numeric’ and ‘symbolic’ problems. •  Par+al credit is given, you must show your work clearly.

T. S+egler 09/29/2014 Texas A&M University

Problem 1

Find the magnitude and direc+on of the sum R of the three vectors shown in the figure. R = A + B + C The vectors have the following magnitudes: A = 5.0m, B = 9.5m, and C = 6.0m. Express the direc+on of the vector sum by specifying the angle it makes with the posi+ve x-‐axis, with the counterclockwise angles taken to be posi+ve.

Problem 2

x(t) = t4 – 2t2 + 3 (in SI units) describes the posi+on of a par+cle moving along a line. (a) What is the average velocity between 0 and 2 s? (b) What is v(t) (b) What is the accelera+on at t = 3 s?

General math:

log (x/y) = log (x)! log (y)

log (xy) = log (x) + log (y)

log (xn) = n log (x)

lnx " loge x

x = 10(log10 x)

x = e(ln x)

ha = h cos ! = h sin"

ho = h sin ! = h cos"

h2 = h2a + h2

o tan ! =ho

#A = Axi+Ay j +Az k

#A · #B = AxBx +AyBy +AzBz = AB cos ! = A!B = AB!

#A# #B = (AyBz!AzBy )i+ (AzBx!AxBz)j + (AxBy!AyBx)k

= AB sin ! = A"B = AB"

dt= natn#1

If f(t) = atn, then

% t2t1

f(t)dt = an+1

tn+12 ! tn+1

ax2 + bx+ c = 0 $ x =!b±

%b2 ! 4ac

" + ! = 90!" + ! = 180!

Equations of motion:translational rotational

constant (linear/angular) acceleration only:

#r(t) = #r$ + #v$t+12#at

#v(t) = #v$ + #at

v2f = v2$ + 2a(r ! r$)

#r(t) = #r$ +12 (#vi + #vf )t

!(t) = !$ + $$t+12%t

$(t) = $$ + %t

$2f = $2

$ + 2%(! ! !$)

!(t) = !$ +12 ($i + $f )t

always true:

&#v' = !r2#!r1t2#t1

#v = d!rdt

&#a' = !v2#!v1

t2#t1#a= d!v

dt =d2!rdt2

#r(t) = #r$ +% t0 #v(t) dt

#v(t) = #v$ +% t0 #a(t) dt

&$' = "2#"1t2#t1

$ = d"dt

&%' = #2##1

t2#t1%= d#

dt =d2"dt2

!(t) = !$ +% t0 #$(t) dt

$(t) = $$ +% t0 #%(t) dt

Forces, Energy and Momenta:translational rotational

W = #F ·!#r =%

#F · d#r

P = dWdt = #F · #v

#pcm = m1#v1 +m2#v2 + . . .

= M#vcm#J =

#Fdt = !#p(

#Fext = M#acm = d!pcm

#Fint = 0

Ktrans =12Mv2cm

#& = #r # #F and |#& | = F"r

W = & !! =%

P = dWdt = #& · #$

#L = I1#$1 + I2#$2 + . . .

= Itot#$

= #r # #p(

#&ext = Itot#% = d!Ldt

#&int = 0

Krot =12Itot$

— Both translational and rotational —

W = !K = Ktrans,f +Krot,f !Ktrans,i !Krot,i

Etot,f = Etot,i +Wother ( Kf + Uf = Ki + Ui +Wother

U = !)

#F · d#r ; Ugrav = Mgycm ; Uelas =12k(r ! requil)

Fx(x) = !dU(x)/dx #F = !#)U = !*

$U$x i+

$U$y j +

$U$z k

Circular motion: arad =v2

s = R! vtan = R$ atan = R%

Relative velocity:#vA/C = #vA/B + #vB/C

#vA/B = !#vB/A

Constants/Conversions:

g = 9.80 m/s2 = 32.15 ft/s2 (on Earth’s surface)

G = 6.674# 10#11 N ·m2/kg2

R% = 6.38# 106 m M% = 5.98# 1024 kgR& = 6.96# 108 m M& = 1.99# 1030 kg

1 km = 0.6214 mi 1 mi = 1.609 km1 ft = 0.3048 m 1 m = 3.281 ft1 hr = 3600 s 1 s = 0.0002778 hr

1 kgms2 = 1 N = 0.2248 lb 1 lb = 4.448 N

1 J = 1 N·m 1 W = 1 J/s1 rev = 360$ = 2' radians 1 hp = 745.7 W

10#9 nano- n10#6 micro- µ10#3 milli- m10#2 centi- c

103 kilo- k106 mega- M109 giga- G

Forces: Newton’s:, #F = m#a, #FB on A = !#FA on B

Hooke’s: #Felas = !k(#r ! #requil)

friction: |#fs| * µs|#n|, |#fk| = µk|#n|

Centre-of-mass:

#rcm =m1#r1 +m2#r2 + . . .+mn#rn

m1 +m2 + . . .+mn

(and similarly for #v and #a)

Gravity:

Fgrav = GM1M2

Ugrav = !GM1M2

R12T =

2'a3/2%GM

Phys 218 — Final Exam Formulae

Problem 3

A rocket is launched from rest on the ground with an constant upward accelera+on of 5 m/s2. 6 s aLer the launch the rocket’s engine shuts down. What is the maximum height reached by the rocket? (Neglect air resistance.)

General math:

lnx " loge x

x = 10(log10 x)

x = e(ln x)

h2 = h2a + h2

o tan ! =ho

#A = Axi+Ay j +Az k

df/dt = natn#1

If f(t) = atn, then

% t2t1

f(t)dt = an+1 (t

n+12 ! tn+1

1 ) (for n $= !1)%

f(t)dt = an+1 t

n+1 + C (for n $= !1)

ax2 + bx+ c = 0 % x =!b±

&b2 ! 4ac

" + ! = 90!" + ! = 180!

Equations of motion:

constant acceleration only:

#r(t) = #r$ + #v$t+12#at

#v(t) = #v$ + #at

v2x = v2x,0 + 2ax(x! x0)(and similarly for y and z)

#r(t) = #r$ +12 (#vi + #vf )t

always true:

'#v( = !r2#!r1t2#t1

#v = d!rdt

'#a( = !v2#!v1

t2#t1#a= d!v

dt =d2!rdt2

#r(t) = #r$ +% t0 #v(t

%) dt%

#v(t) = #v$ +% t0 #a(t

%) dt%

Circular motion: |#arad| =v2

Relative velocity: #vA/C = #vA/B + #vB/C

#vA/B = !#vB/A

1 kgms2 = 1 N = 0.2248 lb 1 lb = 4.448 N

Forces: Newton’s:& #F = m#a, #FB on A = !#FA on B

Phys 218 — Exam I Formulae

Problem 4

A ball is dropped (from rest) from a window at height h and is seen to reach the ground in a certain +me. The ball-‐dropper then climbs to a height 2h but wants the ball to reach the ground in the original +me. Find the velocity v0 that must be given to this ball to achieve the goal.

General math:

lnx " loge x

x = 10(log10 x)

x = e(ln x)

h2 = h2a + h2

o tan ! =ho

#A = Axi+Ay j +Az k

df/dt = natn#1

If f(t) = atn, then

% t2t1

f(t)dt = an+1 (t

n+12 ! tn+1

1 ) (for n $= !1)%

f(t)dt = an+1 t

n+1 + C (for n $= !1)

ax2 + bx+ c = 0 % x =!b±

&b2 ! 4ac

" + ! = 90!" + ! = 180!

#r(t) = #r$ + #v$t+12#at

#v(t) = #v$ + #at

#r(t) = #r$ +12 (#vi + #vf )t

always true:

'#v( = !r2#!r1t2#t1

#v = d!rdt

'#a( = !v2#!v1

t2#t1#a= d!v

dt =d2!rdt2

#r(t) = #r$ +% t0 #v(t

%) dt%

#v(t) = #v$ +% t0 #a(t

%) dt%

#vA/B = !#vB/A

1 kgms2 = 1 N = 0.2248 lb 1 lb = 4.448 N

Problem 5

A basketball player releases the ball from a height h1 at an angle θ and ini+al velocity v0 in an aeempt to put the ball into the basket which is at height h2 and a horizontal distance d. Calculate the distance d if the ball is to make it into the basket. (find in terms of h1, h2, θ, and v0) General math:

lnx " loge x

x = 10(log10 x)

x = e(ln x)

h2 = h2a + h2

o tan ! =ho

#A = Axi+Ay j +Az k

df/dt = natn#1

If f(t) = atn, then

% t2t1

f(t)dt = an+1 (t

n+12 ! tn+1

1 ) (for n $= !1)%

f(t)dt = an+1 t

n+1 + C (for n $= !1)

ax2 + bx+ c = 0 % x =!b±

&b2 ! 4ac

" + ! = 90!" + ! = 180!

#r(t) = #r$ + #v$t+12#at

#v(t) = #v$ + #at

#r(t) = #r$ +12 (#vi + #vf )t

always true:

'#v( = !r2#!r1t2#t1

#v = d!rdt

'#a( = !v2#!v1

t2#t1#a= d!v

dt =d2!rdt2

#r(t) = #r$ +% t0 #v(t

%) dt%

#v(t) = #v$ +% t0 #a(t

%) dt%

#vA/B = !#vB/A

1 kgms2 = 1 N = 0.2248 lb 1 lb = 4.448 N

General math:

lnx " loge x

x = 10(log10 x)

x = e(ln x)

h2 = h2a + h2

o tan ! =ho

#A = Axi+Ay j +Az k

df/dt = natn#1

If f(t) = atn, then

% t2t1

f(t)dt = an+1 (t

n+12 ! tn+1

1 ) (for n $= !1)%

f(t)dt = an+1 t

n+1 + C (for n $= !1)

ax2 + bx+ c = 0 % x =!b±

&b2 ! 4ac

" + ! = 90!" + ! = 180!

#r(t) = #r$ + #v$t+12#at

#v(t) = #v$ + #at

#r(t) = #r$ +12 (#vi + #vf )t

always true:

'#v( = !r2#!r1t2#t1

#v = d!rdt

'#a( = !v2#!v1

t2#t1#a= d!v

dt =d2!rdt2

#r(t) = #r$ +% t0 #v(t

%) dt%

#v(t) = #v$ +% t0 #a(t

%) dt%

#vA/B = !#vB/A

1 kgms2 = 1 N = 0.2248 lb 1 lb = 4.448 N

Problem 6

A daring 510N swimmer dives off a cliff with a running horizontal leap, as shown. What must her minimum speed just as she leaves the top of the cliff, v0 , so that she will miss the ledge at the boeom, which is 1.75m wide and 9.00m below the top of the cliff.

General math:

lnx " loge x

x = 10(log10 x)

x = e(ln x)

h2 = h2a + h2

o tan ! =ho

#A = Axi+Ay j +Az k

df/dt = natn#1

If f(t) = atn, then

% t2t1

f(t)dt = an+1 (t

n+12 ! tn+1

1 ) (for n $= !1)%

f(t)dt = an+1 t

n+1 + C (for n $= !1)

ax2 + bx+ c = 0 % x =!b±

&b2 ! 4ac

" + ! = 90!" + ! = 180!

#r(t) = #r$ + #v$t+12#at

#v(t) = #v$ + #at

#r(t) = #r$ +12 (#vi + #vf )t

always true:

'#v( = !r2#!r1t2#t1

#v = d!rdt

'#a( = !v2#!v1

t2#t1#a= d!v

dt =d2!rdt2

#r(t) = #r$ +% t0 #v(t

%) dt%

#v(t) = #v$ +% t0 #a(t

%) dt%

#vA/B = !#vB/A

1 kgms2 = 1 N = 0.2248 lb 1 lb = 4.448 N

Problem 7

A jet plane comes in for a downward dive. The boeom part of the path is a quarter circle having a radius of curvature of 350 m. According to medical tests, pilots lose consciousness at an accelera+on of 5.50 g. (a) At what speed (in m/s) will the pilot black out for this dive? (b) At what speed (in mph) will the pilot black out for this dive?

General math:

lnx " loge x

x = 10(log10 x)

x = e(ln x)

h2 = h2a + h2

o tan ! =ho

#A = Axi+Ay j +Az k

df/dt = natn#1

If f(t) = atn, then

% t2t1

f(t)dt = an+1 (t

n+12 ! tn+1

1 ) (for n $= !1)%

f(t)dt = an+1 t

n+1 + C (for n $= !1)

ax2 + bx+ c = 0 % x =!b±

&b2 ! 4ac

" + ! = 90!" + ! = 180!

#r(t) = #r$ + #v$t+12#at

#v(t) = #v$ + #at

#r(t) = #r$ +12 (#vi + #vf )t

always true:

'#v( = !r2#!r1t2#t1

#v = d!rdt

'#a( = !v2#!v1

t2#t1#a= d!v

dt =d2!rdt2

#r(t) = #r$ +% t0 #v(t

%) dt%

#v(t) = #v$ +% t0 #a(t

%) dt%

#vA/B = !#vB/A

1 kgms2 = 1 N = 0.2248 lb 1 lb = 4.448 N

Problem 8

Passengers on a carnival ride move at constant speed in a horizontal circle of radius 14.0 m, making a complete circle in 10.0 s. (a) What is their accelera+on? (b) Draw the accelera+on vector at each point (A, B, C, D). (c) How would your answer in (b) change if the speed were not constant? (Explain in words and/or drawing)

General math:

lnx " loge x

x = 10(log10 x)

x = e(ln x)

h2 = h2a + h2

o tan ! =ho

#A = Axi+Ay j +Az k

df/dt = natn#1

If f(t) = atn, then

% t2t1

f(t)dt = an+1 (t

n+12 ! tn+1

1 ) (for n $= !1)%

f(t)dt = an+1 t

n+1 + C (for n $= !1)

ax2 + bx+ c = 0 % x =!b±

&b2 ! 4ac

" + ! = 90!" + ! = 180!

#r(t) = #r$ + #v$t+12#at

#v(t) = #v$ + #at

#r(t) = #r$ +12 (#vi + #vf )t

always true:

'#v( = !r2#!r1t2#t1

#v = d!rdt

'#a( = !v2#!v1

t2#t1#a= d!v

dt =d2!rdt2

#r(t) = #r$ +% t0 #v(t

%) dt%

#v(t) = #v$ +% t0 #a(t

%) dt%

#vA/B = !#vB/A

1 kgms2 = 1 N = 0.2248 lb 1 lb = 4.448 N

Problem 8 – cont.

(b) Draw the accelera+on vector at each point (A, B, C, D). (c) How would your answer in (b) change if the speed were not constant? (Explain in words and/or drawing)

General math:

lnx " loge x

x = 10(log10 x)

x = e(ln x)

h2 = h2a + h2

o tan ! =ho

#A = Axi+Ay j +Az k

df/dt = natn#1

If f(t) = atn, then

% t2t1

f(t)dt = an+1 (t

n+12 ! tn+1

1 ) (for n $= !1)%

f(t)dt = an+1 t

n+1 + C (for n $= !1)

ax2 + bx+ c = 0 % x =!b±

&b2 ! 4ac

" + ! = 90!" + ! = 180!

#r(t) = #r$ + #v$t+12#at

#v(t) = #v$ + #at

#r(t) = #r$ +12 (#vi + #vf )t

always true:

'#v( = !r2#!r1t2#t1

#v = d!rdt

'#a( = !v2#!v1

t2#t1#a= d!v

dt =d2!rdt2

#r(t) = #r$ +% t0 #v(t

%) dt%

#v(t) = #v$ +% t0 #a(t

%) dt%

#vA/B = !#vB/A

1 kgms2 = 1 N = 0.2248 lb 1 lb = 4.448 N

Problem 9

A plane is flying north at 200 m/s in gale-‐force winds of 35.0 m/s which are blowing in a direc+on 30.0° south of east. How far off course are they in the east-‐west direc+on aLer 2.00 hrs? (P=plane, A=air, E=earth) General math:

lnx " loge x

x = 10(log10 x)

x = e(ln x)

h2 = h2a + h2

o tan ! =ho

#A = Axi+Ay j +Az k

dt= natn#1

If f(t) = atn, then

% t2t1

f(t)dt = an+1

tn+12 ! tn+1

ax2 + bx+ c = 0 $ x =!b±

%b2 ! 4ac

" + ! = 90!" + ! = 180!

#r(t) = #r$ + #v$t+12#at

#v(t) = #v$ + #at

v2f = v2$ + 2a(r ! r$)

#r(t) = #r$ +12 (#vi + #vf )t

!(t) = !$ + $$t+12%t

$(t) = $$ + %t

$2f = $2

$ + 2%(! ! !$)

!(t) = !$ +12 ($i + $f )t

always true:

&#v' = !r2#!r1t2#t1

#v = d!rdt

&#a' = !v2#!v1

t2#t1#a= d!v

dt =d2!rdt2

#r(t) = #r$ +% t0 #v(t) dt

#v(t) = #v$ +% t0 #a(t) dt

&$' = "2#"1t2#t1

$ = d"dt

&%' = #2##1

t2#t1%= d#

dt =d2"dt2

!(t) = !$ +% t0 #$(t) dt

$(t) = $$ +% t0 #%(t) dt

W = #F ·!#r =%

#F · d#r

P = dWdt = #F · #v

#pcm = m1#v1 +m2#v2 + . . .

= M#vcm#J =

#Fdt = !#p(

#Fint = 0

Ktrans =12Mv2cm

#& = #r # #F and |#& | = F"r

W = & !! =%

P = dWdt = #& · #$

#L = I1#$1 + I2#$2 + . . .

= Itot#$

= #r # #p(

#&int = 0

Krot =12Itot$

U = !)

Fx(x) = !dU(x)/dx #F = !#)U = !*

$U$x i+

$U$y j +

$U$z k

#vA/B = !#vB/A

G = 6.674# 10#11 N ·m2/kg2

R% = 6.38# 106 m M% = 5.98# 1024 kgR& = 6.96# 108 m M& = 1.99# 1030 kg

1 kgms2 = 1 N = 0.2248 lb 1 lb = 4.448 N

Centre-of-mass:

#rcm =m1#r1 +m2#r2 + . . .+mn#rn

m1 +m2 + . . .+mn

Gravity:

Fgrav = GM1M2

Ugrav = !GM1M2

R12T =

2'a3/2%GM

Problem 10

In a triathlon, a contestant swims from start to finish in a +me T. In order to do so, the contestant has to swim against the flow of the river which has a constant speed V rela+ve to ground as shown. What is the x-‐component of the swimmer’s velocity, vx, with respect to the water?

&inish

swimmer’s path

General math:

lnx " loge x

x = 10(log10 x)

x = e(ln x)

h2 = h2a + h2

o tan ! =ho

#A = Axi+Ay j +Az k

dt= natn#1

If f(t) = atn, then

% t2t1

f(t)dt = an+1

tn+12 ! tn+1

ax2 + bx+ c = 0 $ x =!b±

%b2 ! 4ac

" + ! = 90!" + ! = 180!

#r(t) = #r$ + #v$t+12#at

#v(t) = #v$ + #at

v2f = v2$ + 2a(r ! r$)

#r(t) = #r$ +12 (#vi + #vf )t

!(t) = !$ + $$t+12%t

$(t) = $$ + %t

$2f = $2

$ + 2%(! ! !$)

!(t) = !$ +12 ($i + $f )t

always true:

&#v' = !r2#!r1t2#t1

#v = d!rdt

&#a' = !v2#!v1

t2#t1#a= d!v

dt =d2!rdt2

#r(t) = #r$ +% t0 #v(t) dt

#v(t) = #v$ +% t0 #a(t) dt

&$' = "2#"1t2#t1

$ = d"dt

&%' = #2##1

t2#t1%= d#

dt =d2"dt2

!(t) = !$ +% t0 #$(t) dt

$(t) = $$ +% t0 #%(t) dt

W = #F ·!#r =%

#F · d#r

P = dWdt = #F · #v

#pcm = m1#v1 +m2#v2 + . . .

= M#vcm#J =

#Fdt = !#p(

#Fint = 0

Ktrans =12Mv2cm

#& = #r # #F and |#& | = F"r

W = & !! =%

P = dWdt = #& · #$

#L = I1#$1 + I2#$2 + . . .

= Itot#$

= #r # #p(

#&int = 0

Krot =12Itot$

U = !)

Fx(x) = !dU(x)/dx #F = !#)U = !*

$U$x i+

$U$y j +

$U$z k

#vA/B = !#vB/A

G = 6.674# 10#11 N ·m2/kg2

R% = 6.38# 106 m M% = 5.98# 1024 kgR& = 6.96# 108 m M& = 1.99# 1030 kg

1 kgms2 = 1 N = 0.2248 lb 1 lb = 4.448 N

Centre-of-mass:

#rcm =m1#r1 +m2#r2 + . . .+mn#rn

m1 +m2 + . . .+mn

Gravity:

Fgrav = GM1M2

Ugrav = !GM1M2

R12T =

2'a3/2%GM

Problem 11

A rocket starts from rest and moves upward from the surface of the earth. For the first 10.0s of its mo+on, the ver+cal accelera+on of the rocket is given by ay(t) = (2.80 m/s3)t, where the +y-‐direc+on is upward. (a) What is the height of the rocket above the surface of the earth at t = 10.0s? (b) What is the speed of the rocket when it is 325 m above the surface of the earth?

General math:

lnx " loge x

x = 10(log10 x)

x = e(ln x)

h2 = h2a + h2

o tan ! =ho

#A = Axi+Ay j +Az k

dt= natn#1

If f(t) = atn, then

% t2t1

f(t)dt = an+1

tn+12 ! tn+1

ax2 + bx+ c = 0 $ x =!b±

%b2 ! 4ac

" + ! = 90!" + ! = 180!

#r(t) = #r$ + #v$t+12#at

#v(t) = #v$ + #at

v2f = v2$ + 2a(r ! r$)

#r(t) = #r$ +12 (#vi + #vf )t

!(t) = !$ + $$t+12%t

$(t) = $$ + %t

$2f = $2

$ + 2%(! ! !$)

!(t) = !$ +12 ($i + $f )t

always true:

&#v' = !r2#!r1t2#t1

#v = d!rdt

&#a' = !v2#!v1

t2#t1#a= d!v

dt =d2!rdt2

#r(t) = #r$ +% t0 #v(t) dt

#v(t) = #v$ +% t0 #a(t) dt

&$' = "2#"1t2#t1

$ = d"dt

&%' = #2##1

t2#t1%= d#

dt =d2"dt2

!(t) = !$ +% t0 #$(t) dt

$(t) = $$ +% t0 #%(t) dt

W = #F ·!#r =%

#F · d#r

P = dWdt = #F · #v

#pcm = m1#v1 +m2#v2 + . . .

= M#vcm#J =

#Fdt = !#p(

#Fint = 0

Ktrans =12Mv2cm

#& = #r # #F and |#& | = F"r

W = & !! =%

P = dWdt = #& · #$

#L = I1#$1 + I2#$2 + . . .

= Itot#$

= #r # #p(

#&int = 0

Krot =12Itot$

U = !)

Fx(x) = !dU(x)/dx #F = !#)U = !*

$U$x i+

$U$y j +

$U$z k

#vA/B = !#vB/A

G = 6.674# 10#11 N ·m2/kg2

R% = 6.38# 106 m M% = 5.98# 1024 kgR& = 6.96# 108 m M& = 1.99# 1030 kg

1 kgms2 = 1 N = 0.2248 lb 1 lb = 4.448 N

Centre-of-mass:

#rcm =m1#r1 +m2#r2 + . . .+mn#rn

m1 +m2 + . . .+mn

Gravity:

Fgrav = GM1M2

Ugrav = !GM1M2

R12T =

2'a3/2%GM

Problem 11 -‐ cont.

General math:

lnx " loge x

x = 10(log10 x)

x = e(ln x)

h2 = h2a + h2

o tan ! =ho

#A = Axi+Ay j +Az k

dt= natn#1

If f(t) = atn, then

% t2t1

f(t)dt = an+1

tn+12 ! tn+1

ax2 + bx+ c = 0 $ x =!b±

%b2 ! 4ac

" + ! = 90!" + ! = 180!

#r(t) = #r$ + #v$t+12#at

#v(t) = #v$ + #at

v2f = v2$ + 2a(r ! r$)

#r(t) = #r$ +12 (#vi + #vf )t

!(t) = !$ + $$t+12%t

$(t) = $$ + %t

$2f = $2

$ + 2%(! ! !$)

!(t) = !$ +12 ($i + $f )t

always true:

&#v' = !r2#!r1t2#t1

#v = d!rdt

&#a' = !v2#!v1

t2#t1#a= d!v

dt =d2!rdt2

#r(t) = #r$ +% t0 #v(t) dt

#v(t) = #v$ +% t0 #a(t) dt

&$' = "2#"1t2#t1

$ = d"dt

&%' = #2##1

t2#t1%= d#

dt =d2"dt2

!(t) = !$ +% t0 #$(t) dt

$(t) = $$ +% t0 #%(t) dt

W = #F ·!#r =%

#F · d#r

P = dWdt = #F · #v

#pcm = m1#v1 +m2#v2 + . . .

= M#vcm#J =

#Fdt = !#p(

#Fint = 0

Ktrans =12Mv2cm

#& = #r # #F and |#& | = F"r

W = & !! =%

P = dWdt = #& · #$

#L = I1#$1 + I2#$2 + . . .

= Itot#$

= #r # #p(

#&int = 0

Krot =12Itot$

U = !)

Fx(x) = !dU(x)/dx #F = !#)U = !*

$U$x i+

$U$y j +

$U$z k

#vA/B = !#vB/A

G = 6.674# 10#11 N ·m2/kg2

R% = 6.38# 106 m M% = 5.98# 1024 kgR& = 6.96# 108 m M& = 1.99# 1030 kg

1 kgms2 = 1 N = 0.2248 lb 1 lb = 4.448 N

Centre-of-mass:

#rcm =m1#r1 +m2#r2 + . . .+mn#rn

m1 +m2 + . . .+mn

Gravity:

Fgrav = GM1M2

Ugrav = !GM1M2

R12T =

2'a3/2%GM

(a) What is the height of the rocket above the surface of the earth at t = 10.0s? (b) What is the speed of the rocket when it is 325 m above the surface of the earth?

Problem 12

The accelera+on of an object is given by a(t) = 2.0 m/s2 + (.75 m/s3)t while moving in the x-‐direc+on. If its ini+al posi+on is x = 3.0 m and its ini+al velocity is zero, find the velocity and posi+on when t = 2.5 s.

General math:

lnx " loge x

x = 10(log10 x)

x = e(ln x)

h2 = h2a + h2

o tan ! =ho

#A = Axi+Ay j +Az k

dt= natn#1

If f(t) = atn, then

% t2t1

f(t)dt = an+1

tn+12 ! tn+1

ax2 + bx+ c = 0 $ x =!b±

%b2 ! 4ac

" + ! = 90!" + ! = 180!

#r(t) = #r$ + #v$t+12#at

#v(t) = #v$ + #at

v2f = v2$ + 2a(r ! r$)

#r(t) = #r$ +12 (#vi + #vf )t

!(t) = !$ + $$t+12%t

$(t) = $$ + %t

$2f = $2

$ + 2%(! ! !$)

!(t) = !$ +12 ($i + $f )t

always true:

&#v' = !r2#!r1t2#t1

#v = d!rdt

&#a' = !v2#!v1

t2#t1#a= d!v

dt =d2!rdt2

#r(t) = #r$ +% t0 #v(t) dt

#v(t) = #v$ +% t0 #a(t) dt

&$' = "2#"1t2#t1

$ = d"dt

&%' = #2##1

t2#t1%= d#

dt =d2"dt2

!(t) = !$ +% t0 #$(t) dt

$(t) = $$ +% t0 #%(t) dt

W = #F ·!#r =%

#F · d#r

P = dWdt = #F · #v

#pcm = m1#v1 +m2#v2 + . . .

= M#vcm#J =

#Fdt = !#p(

#Fint = 0

Ktrans =12Mv2cm

#& = #r # #F and |#& | = F"r

W = & !! =%

P = dWdt = #& · #$

#L = I1#$1 + I2#$2 + . . .

= Itot#$

= #r # #p(

#&int = 0

Krot =12Itot$

U = !)

Fx(x) = !dU(x)/dx #F = !#)U = !*

$U$x i+

$U$y j +

$U$z k

#vA/B = !#vB/A

G = 6.674# 10#11 N ·m2/kg2

R% = 6.38# 106 m M% = 5.98# 1024 kgR& = 6.96# 108 m M& = 1.99# 1030 kg

1 kgms2 = 1 N = 0.2248 lb 1 lb = 4.448 N

Centre-of-mass:

#rcm =m1#r1 +m2#r2 + . . .+mn#rn

m1 +m2 + . . .+mn

Gravity:

Fgrav = GM1M2

Ugrav = !GM1M2

R12T =

2'a3/2%GM

Problem 12 – cont.

The accelera+on of an object is given by a(t) = 2.0 m/s2 + (.75 m/s3)t while moving in the x-‐direc+on. If its ini+al posi+on is x = 3.0 m and its ini+al velocity is zero, find the velocity and posi+on when t = 2.5 s.

General math:

lnx " loge x

x = 10(log10 x)

x = e(ln x)

h2 = h2a + h2

o tan ! =ho

#A = Axi+Ay j +Az k

dt= natn#1

If f(t) = atn, then

% t2t1

f(t)dt = an+1

tn+12 ! tn+1

ax2 + bx+ c = 0 $ x =!b±

%b2 ! 4ac

" + ! = 90!" + ! = 180!

#r(t) = #r$ + #v$t+12#at

#v(t) = #v$ + #at

v2f = v2$ + 2a(r ! r$)

#r(t) = #r$ +12 (#vi + #vf )t

!(t) = !$ + $$t+12%t

$(t) = $$ + %t

$2f = $2

$ + 2%(! ! !$)

!(t) = !$ +12 ($i + $f )t

always true:

&#v' = !r2#!r1t2#t1

#v = d!rdt

&#a' = !v2#!v1

t2#t1#a= d!v

dt =d2!rdt2

#r(t) = #r$ +% t0 #v(t) dt

#v(t) = #v$ +% t0 #a(t) dt

&$' = "2#"1t2#t1

$ = d"dt

&%' = #2##1

t2#t1%= d#

dt =d2"dt2

!(t) = !$ +% t0 #$(t) dt

$(t) = $$ +% t0 #%(t) dt

W = #F ·!#r =%

#F · d#r

P = dWdt = #F · #v

#pcm = m1#v1 +m2#v2 + . . .

= M#vcm#J =

#Fdt = !#p(

#Fint = 0

Ktrans =12Mv2cm

#& = #r # #F and |#& | = F"r

W = & !! =%

P = dWdt = #& · #$

#L = I1#$1 + I2#$2 + . . .

= Itot#$

= #r # #p(

#&int = 0

Krot =12Itot$

U = !)

Fx(x) = !dU(x)/dx #F = !#)U = !*

$U$x i+

$U$y j +

$U$z k

#vA/B = !#vB/A

G = 6.674# 10#11 N ·m2/kg2

R% = 6.38# 106 m M% = 5.98# 1024 kgR& = 6.96# 108 m M& = 1.99# 1030 kg

1 kgms2 = 1 N = 0.2248 lb 1 lb = 4.448 N

Centre-of-mass:

#rcm =m1#r1 +m2#r2 + . . .+mn#rn

m1 +m2 + . . .+mn

Gravity:

Fgrav = GM1M2

Ugrav = !GM1M2

R12T =

2'a3/2%GM

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