Problem #l

The motion of a particle is defined by the relation x = 2t3 - l2t2 - 72t - g0,where@and[bre expressed in meters and seconds, r.ryin.(a)ry@)thgv'9-|.o.9i!'y,thg.aqce!eration.andthetotaldistance traveied when x = 0.

(€ rlc\oc,fyl * qI : 6b2_ ?q\, _12v- ()t

V=o : 6La*?Ktr*12.-'11- . II b= 6_5$ t=-1{s1 yr[-.,.sel

(A) \l€Loqtl .= c, ^+ t = 6 s er

0oo A{ x=o }t3 -\Lt2 -ltu -to =e.

t = los(c t= -?Str, yt[..,,l,ul-'-"'_

,-""'_ '

I X=o AL t = [<.rS€(

68 V g 6L? *?q t {f a{ L= \o{,ct-\ = (r t ro), _?.t x tr: _:l zV= Zgn ,n /s1q

A= A\V|L ._ l2L *"tq c\+ t, t' \cc

A -- \?* lo -tttfr= )f rnftccL

% t%&L td{a{ rJ;sfo-co }ro,.r, trd o.r J:-, F,vr4 V = o ar [=6l"-t".1.J;sto, o = I Av. I\ orcl* I \xr,,u

1

= I k -* [t \*,' -K6t = [ -6rr -(-8"r1 + \@*:€-6rp][

(. = -5tr rr, xlrr= o .

Xo:'-?@rn

Problem #2

Xbd ?^ y" \ '/uX+L,xr,r*z dt = \ VrJv.

JJO 'L'5

I u' - \h(2, )a = loo Y,' +- Suvt { q

I { vr : 5 i; ytt* + \.;,: x2 + 3, \?5

The acceleration of p5rinr A is defined by the relationlf:iOO, (L.tdtrwherla and 5 are expreiiid in roA' and meters, respcctively, and 6,ii aconstant. Knowing that the vSl-AEfty*o_f A*is 2J m/s when I = 0 a1d_J m/s_when {= 0,11 p, determine the value of /t.

% ft. ? ({): v*

b[ 0\: ']o.X tf+ ufy,l ) : V,lu 2or{ t \-t vrxt\ [* = Utvi!(

\i,h-|n V- 5r.h/(e1 _= \ (r,) r ,l- -"''r -t5)

I= bo1fo,\q.;q+1*(,1,15)2 1?,\?5( =u,lSrn )-6{ . vt : t?,5 _ Z,l5 - ?,\"t! = 5=+lr ' -A

\ L{: ?3\ - tt8 \ rn-t ,e arl

2

Problem #3

The position x of a particle in lectilinear motion is given as shown:

{r lrlfp,,r in6]

{}F

Over the time interval 0 < t < 8 sec. determine:

The displacement. the total distunce traveled and the average velocitv.

Sketch the morion.

%'+ha tisptaLrrm+ : ( - qLj-hqt" +1\Gtr \r

v<\uut3 = d[ = \?E? * LolL r?t6.0b

(f ue\,,,a t r,' =.,-.J \Z!1-\o.6t +t\6 _o .

,hI'E= 6 sec, it=3reL

% ovg{ 1-rrt tryr\[ iv,tVr-_r o<t < qt

drSlthCrrcru\ =,(($-!L(g'): qtr)3-5q.(Bl z+?rGtt)+ \1 *\+= 37Pi in\(h

oP a,vevogr v(l,u,i- _

X3 = 3:l rlr(,hL = lf t'nM"

.,e 1"t{ dic4-o.n-oq -\rwetcd

Yz =L8+ \,^-(^^

Y6 = 2t< wru

loK,,', \ + \ A*,,(\ + \r:(0, j I

- [Xi-v,1 t \Vc -xr \ + \Xq.r{(): lzrr -ret+ \ z.r -?qit + | trt _ 231\

131-\1 - Lt.inb^f rerab

&J' :8-o

'= t{tq iir GA

/t? SY\elC,hin3 trta

'-trttvrnbqvt Ut d:, (,

ri?Qrpr(n{ iq1^," U oCrvro\rr,.

,"^ J.^ tvvvv0 \(vY\ t

8)

ll,WA

L=oV = 'l [G iq/lct

t-- ----iU*0 = t[o il'r /ssc

,ti. r.Lr J lhL ̂

L--G 5cc

tl = |p-,r;>

281rr0.t" 3 l,:.Y =Zgva.

35+ rvr0a

u=%Jc(v -- thi" l\cC

Problem #4The velocity curveof a particle in rectilinear motion is shown in two cases.For each case,LT,--ltO-fr\ draw the acceleration (a-t) and position (x-t)curves of the particle over the interval 0 < t < 1-5 sec

,\,ce\ag"o^\ and then determine:v'-\ ?-il-Lr an{xr ol the particle during 0 S r ( ll.xr"I <'- b) ThE fwo values t1 and t2 of time for which the particle passes through the'\v\-. (^ Oflgln.

,\i\\oot'u'

9 1 ii/sj

$n G)

fl Hn+ r.levn..r5\ )ettrvrnai r

0.-t Vf \oufb =-g

r'5slopr"fl tin. e_Ce11l- ta

=19?i"G 11L\

Rr o5-f *c r

Frx tt4t (t

.iA"tfi5)

f{fl

Xz-x\--\6i K2 = -\Cl l.G = .r',.

(.a -Yr = -\6Xr-=-l({r" =-[6

V.t -Y r. I{2=3-\6=:1

W 'ttt ( t\

tu*6\ --t

Fl= 2u

i at u-.,'

-f ttt t-tl )I

-L$!=€scc

(.\-L l-u ffio r aLx1dg^- C-trvr,,e

nr\\: -?,xrty1t i " _,1,6 (+Ahr . - [{t-r xttr = -l( ?+

Ar = + 3+ 6{11, = )p+4.1 = \,\ Y6='tqt{

L4 crtrir)

L.l

t?IC

A

-t

*1

N* \\"(t(rs Kt - X1 ='1t,

Y^ - ?L^1 -1 - \1/- t -

\5,(t-t

',

\i/

lr?e t

e

Ib

s\oPg =

S\ote :S\o? P -t\"?e

:i*2\

tiv'ne-

XZ *Yt = L(

-3*r.{- o

i-1I L\

h;[ {-_\

-to.,

At'trS um der t-u-ru ue

Ar= -W*r,tXVz-*B?+Az= +(-{-t)xs =-t5ptA3 = ---+ ?vil wenr,"ci}e{

ftl-vctrutl = u,

V+2-,tt+Z t.r'_ , \2_5 )( t _g)

l\= zE -t^tt V=" t = \o Sec-

A:= *t(z+L{r= -r.g+A^-- (ltr.(t jrt =qP*Ar = t+rt = rtP+

%v

nYt,t

%,

o<U<L{ yz_yr=-q

U<t-< 3 Xr-Yr =-\53<LCtr.i \r2-x,=-L

KL=-B+l.G= 3e+

Xz= -\St3=-{[{{l= -a -\ ='8P+

Xt ' tt -t = -'tPL

{z- -q+\t'8€+

Y.

Itel

g

*qa

-1

7u, lo < t(\2--

lz<rt- rsfov Xr -Yr= (L

\tl( /rouj gr* tw S\of e r

t1€e -*

S\..r2^

-ii::,\*c>

= -?f('(

--L

: O.t{

b(?,

(v:

e,

?, ^t

G,s\c'& = _o__(t) -z\ i;- "r

5'\;f g = 2-

i\ "p ? ,: (,:.

l --1-- o \J4 t itw lil (a I disl ta u^inn t a{ r, < F ( \L_

6b A.e^ t{^d{" c^*re f.^.* t= \\ t \.=\?_ = 4* x.€__ e ?+ac [r1 ?nttih],_\ tr", ur,w \r

X)_Kr = (XZ_= C_1= _t{.1

r- \L| - ln\_,_\s =_(1tb

l( = difflooy\h^{ = lK-ti,_rK o' =

t--"t".t rJ, \towrG _A tK Lcutu( r( 8

,3 ( t e \L

Xr ?or o([rrL =

\ Y1= q;

tul $^* ?',*fl^ u q {_E X=. A+ 'ft,= q.sec \

t'TJ o\/\r,, t

fr.o* ?re?h Xr= \({{n(r = \(f+Xr : tL{+

l(r trr\Di++i

- __,_, _I

$re '\*1o r/'-)' E A

qLt=tttct

\\ S tE iJ

-+ (; t^,,/\ |' \[-= tL'l(l sgC\.--

-j

O

II

-(-+)=qA-1 i

1"6an ewn : a

-\ A= (+(t-tt;

i[^ ?i1*ufl(&t di't2ln0,u,rttrt oS L< LZ-

+-*r, (-[ o.rv{P Dt - {rr- Ko tt-l \-\ ( : J4_g\

{r7c} kl *-l d;ttu*c, X r 'Y1= \ 6."''\Xr= tt

s \-s\ + l-ttl

(t

'tsD

t, , Lr-

\ A_-8

dq\irv G U*tr =

r/4u\= 2 (t-u.\L\,c-

\rl- z-l eci"-. a E - t_.p-.rJ5

i

/*t+tt \,I : \t\ rrt\ 9-q)LE-*'

I

e| A - \h ( -q+ ?y-?s\ (U-kr'"\ -5- 1-),-

vYQr*t

Ib?(,L +vr* atu;t*C,t\n t'= r,

h.* V- t <.unv o.t qsb E I

Qri,iria{. X2-\t =rv -r fr-' , t

t'r nt pYO rut,l) . ^atL,t

" r^ 4g\ il,-t^.) {*\b= (tt =/ \o*tru-ttt -r/.8)(L-t^,

LVL -5(! + \?"1 =-3o\ 7V' -sbL rtal -* o

-f / ar r a

lbr = u.ur 161\ '\ tz= ?t'1Jcc

I-ro )(-r Lz:-=;-=*::,:-, _tl-t c,^r^"

\n U -\, cr,l/u(:-

A= Q* ( L-rzrt-t1 . 1

I t_ rt\cc

\t{ L < \(

T',=q.

'i',,: "'" oi'

"[=\t'

\ tr,- \1 se .\

Xr -X, ='\ko' A = o _ (-(l

i

I A=L{ r

Llg= wsvnxV -tfrgtr

Problem #5A projectile is fired with an uitial velocity of 250 m/s at a target B locatecl600 m above the gun A and at a horizontal distance of 4 km. Neglecting airresistance, Determine:-1) The value of the firing angle o,.

2) Time of flight from A to B3) Maximum height

Vo = ?5o rn / scc

Jg= 6tx*

h = (oou rn

?u. Gt ,[ c

V.r1 --- V \) \\\^ C(

t: 26oX-at -b

Sin z3,qos ._ \?,5 seC-5 t'b\

: 1$u* SynZ9,q.:f 4 tt,! - tlr4q,$\t\t,tr.= a(1,3b5f{\ ,,

f.;

z

F

G-\rtn

x=

,'t

4

E[[ m. rl,,

?w r(=(e,\\-vo c,rspt \2\

= V.cO o

%- ?ot

?,5 o Cot (9 '\\'

rAo- X\on^ct-qSz.,r - irrr, ,r,*

,i;" ,,-...-

T]{Km

P<r- D( = Z-A,t^ oiX.= \n C-ofa( t"

b* qplce

?S-uV= C"S Zg.\"

cuJ

:,{tArt1s (q*

= I S.355

/.0o@ -- t-\iruu{or,n v - \1-S s,6tt teLz ol

(u" - uLiaju tor' cn - \tqs,(%\?55,GEt>r{x

-.aqru {annn

tcr..U =.I__(Lt tom el

X = 6'9.\\" Cx;

- \?65' (%

't- \tss,6'3

= o"5G(29,\o(

torv'tt c:

- r'\

:l 3"rfYo"L

fravMMA hr//A\ 1 \J.r, =o i\J\

9[=tfd'(w'uL -ltLqvL

-- ?So* SU(g,\\".{ 9j'?:. A1( o , r{t ,y1

rf,t5in3 V1 =Vu(Y^d

'?'[2o l, (r^^ 6 s ,t-1.

3'tt

-ttt3,?l * ? l,?-r-

rAL

=t -- Zg-t:,Secbe

?5[ m/s

-1-

Problem #6A golfer aims his shot to crear the top of a tr.e,e by a distan ce h atthe peak ofa trajectory and to miss the pond on the opposire side. Knowine that thernqonifrr.lo ^f .' i. 1< < ^t^ )^t^,,,- - ' ,t ^ -l r

;fffi,i::illU is 25.5 m/s, determine the range of values

", O{{;l_ry;

12m

i

l't\tltlII

1

45m

W O --ZCrs. Vo5 -r ?*lrg-

:, "]t' t ,,,,',

' .. :";| ',',.r"tl'

?:

"-' " 'l*,-^.3m

o'd> G',t(r, ft&d I {*,

_:_ ftl ,end lorhf

fil tr = t4vn\ 1P= x

'8 6\,t-X= t{ltn..a:****-"-""-

vU,\np.t/nifi nunl" buor^/b r'd gl _ i

€ns <h< 6,i1 y\i\- )

1? = {t+t^'f

lrln cr -- lxe2vuz 6rh

2*25.( ? * {.-l'Z-h =-l?,31v,.,

\h = 5,3r 36 rn i

*e-:&tLhSL2* 2S.Sl *\Ls

Problem #7A golfer hits a golf ball with an initial velocity of 48 m/s at an angle of 25n

with the horizontal. Knowing that the fairway slopes downward at anaverage angle of 5n, determine the distance d between the golfer and point Bwhere the ball first lands.

,*=.:':4

w..!!!l=i

='n

x -?5oVo =

e'(3 /n lf.t

ot- V{= /o frsu. = Q4 f, {oJ ?s

'VJ = Vqfyraor = ttqrd fint5

og lX= dcasS.i ,-,TIILL

i-

-,-\ t$ =-d sin

lf - Ktcrrnor - S)..1**7 Vru L goiot r

- dr,n|u = dwtq- fo,^n,tS - 313\-^t d1 (c,ur51"

2* t,t$L t\ i (ia{ tsiu

d Cr'us -cotttamls) + ,J r- fz,s+zrd\,-r) =o.

d ="=JT!^

'*-'**--*--*--"1t'

I]

:%a..,,,,.. t;"'"""""' .. -..

4

\of

0

s -lt,I

rl I

sG.S4'

\

| ,o' 111

d = 2-tLt,( rn

-rJ_