PHYS 1300 - 2D Kinematics

Average Speed= -stane e 2I Sample problem Swimmer . cross a) Calculate the 0 . how average speed around the edge of the pool ? to formula Ot , know the only we diameter 10 min 225 82 m/s 3) at the same because a spes to Speedar-average = swimming pool of circular use what long would the information given to swimmer us , we take must to swim rearrange the have we d = Speedau (TT) 10 m = > - m because we are 0 82 m/s doing the whole pool circumfrance , . 64S = Checkpoint car of # goes arg Speed of . 72 kemh convert car = d for 20 same time . mis Which for half travelled an hour . Truck goes arg speed further? km/h : (t) = = to Cancel (Speedarg) (30 min) /20 36000 · Both Cars m m/s) = x totes outuni re Trucke 36 /em travelled the Same distance = d = = = (t) (Speedarg) (72 /em/k) 10. Sk) . 36 /em 2 D Kinematics &X & y ↑ Y2 : Y (vertical , displacement = Or Ux : Sample 22m displacement) displacements Chorizontal -X , , position : ↓ X = , X , = r 2 , 1) Total distance travelled 22 + 34 T +X 36 y 65 - ( .... when I separately. This X findd , x2 0X = = 17 . 8 + 14 4 . m 12 4-30 8 . . : ↳ The 32 Im = = 17 P - . = O = = is (32 2) . 37 m tan" 24 " + ( - 17 22 m = 17 . . 8 cos 36 m = 34cos(69) = 14 . Y 4 42 22 Sin = 12 4 = 34 Sin (68) . 30 8 . you always I treat to them find displacement 136) = = M allows must m m m negative in this (South) y-component facing down often 4) magnitude ( = . ye component because we made our DF , , end 0x20y & Oy using components you declare the xRy direction Displacement : 56m travelled = 34 m , 65'SE - Start-- OF X2 = "2 . + 2) Y 4 I problem : 36 NE lyz Oy & measured as , case it is measured negative sample problem : 4 2 . 1) surfer took 37S 1) find speed average . 2) Speedar find velocity average 2) Var instaneous velocity : magnitude the object if an to the object is moving of in a curve . 1 5 mis . 12 24's 0 m/s : motion -- of E path curry object at that moment the instanteneous , velocity is tangent 4 3 . A) ... the of find instanteneous velocity & ........ ... 1 -Ot- at that moment Sample problem A : : speed of : direction : the direction * : Speedar = & 2πr = ADB 2. 0x102 We dealing are 2π (1 5 x . : / circle 104) 2 8 X102 . = 3 4 m/S . PRAC Q : displacement * 4 (a b) of velocity . , person walks 38m 68m West - 300x 0 : tan" 04 .4 3 , find Or = = (*) = = = = tan" 61 : Or (t .. , use this /0x)" + (0y) 2 78 m waf is t constant , the the same in . magnitude And because both the same direction the both M/s through pythagreon (38)" + (68)" T068 78 M61' is points more velocity for . a F - 30 " O · North then because speed , points will remain 4 . /(b) travels glider horizontally 550 di -> - I point A-B I i 750 370 i Od dz : 370 : tan" (*) tan" / : m 1370)" + 175012 = - O 420-550 : = I kyB 750E , 420S N, 26 = 5644400 836 - The compass has m so A N d3 4 3 is still east our * axis y axis shifted the while South is our . a) clocl wall on Pocke,find = sweep , find w/ hand 2π(14) * to = 1 velocity when the amount that find one T Ov : 1S : 14sinG : 1 . tip C) 2 10 9 ⑳ · - - 8 s C L · from past centre of 6 : 00 pm &10 : 00pm the full hand covers : happens in 60s. 4 M/s 1S the on horizontally , to the will be the same as find tip average 112 h moves the instanteneous velocity pm for the Same speed in the Same moving& , 14cm is between ov = 1: 00 clock , 260 revolution · 10 that 5 m/s . know = tip see instanteneous OT a averag speed 60 b) second 45 : 00 direction , the arm however it will above horizontal , pm * come lef e back is be we 4 4 problem . Skater travels in straight a 8 Straight 21 line . Sx102m east of 25 north of east . , then north find resultant . x102m I a) 5 6 displacement , es S G . I -B I as - d , 0 Scos25' = 770 : 35 4 0 X d2 = . 6 . m 6 X10" cos(21) S22 =5 21 EofN(y) Sin (21) a) ..., resultant 56CN of -] m O (0x) = 1340 = + . 4 2 = . = S6 = . ↳ S X60 I 1410 Od - Ot 1340 262 = S 31 mr s . 6 Ot /4LU = (in Ed = 25 2 = (*) & Velocity find average speed 2S2 S = fan" = m b) 4 2 min , tan" : (04 2 (1124)" + (722)2 = DV displacement 722m = ·d - m X10" 200 *t (2 5 :. : A m 1124 = S = 181 ..... I . = by IS Noff(x) M 8 S sin 25 : = - . - 56 . 6 Not t = S S4m/S . is 1 . 3x10" m 4 53 practice S Os . to . slides down 4 5 guy , . m down a pole & runs 4 . S + 6 8 . 11 3 : 2. S M m . 6 8 M . average speed Os = 0x by = = 6 & . 26 s i displacement m . 4 Sm . = (0x32 = /6 812 . = 2 = resultant F (46 10y1 2 + + (4 . O = : = 3)2 24) + (20 25) tan" 33 ( : . . · d) Straight . firetrucle a) total distance moved b) 6 8m B 15 m : resultant below . displacement horizontal average velocity or : Od = O + = 0 . 15 5 OS . 8 . 15 = = s m . OS 1 . 63 m/S 33 below horizontal is 8 . 2 m 33 4 6 . avey velocity of west of South a) a cor during drive 72 km/h is find displacement OX Oy W 6 1 . 16 Sin = 72 = 30 15 . > - 132) S m . = :tan-/3061 . = Od 32 = = . 15)2 Hm/h 72 - 12 m (61 16) + 138 = E 3ih km 72cos(32) = = & min N ioy V 14 . 15 16 S : vt (72km/h) : ken32 /'min( wa 40 min 32 self Test 0 100 km . ? If under . Is so , , 73 m . run S Om/s, by . how safe to 0'Naf W much zone ? M : 04 z( T . B 85 7 . · X 05 4 cos = 76 . O =. = 8S 7 cos(73) = 25 / R x . 34. 0 (73) 02 0 m By 7 6 0 + 25 / b) we = : t = . = . have + H Ho 22 S : # R = (101) + (43) 10201 + : 1844 . to .., 104 7 - 110m . into 101 34 0 + 82 0 = s 4 . Ay m . = 2) . 8 S 7 sin Ry i & 05 4 m sin (27 - 73/ x BX . = 85 . . = 2) . = =. . A . Oy (27 . = I . direction Os 7 outside zone is he b) OS 4 m 27 2 South west Squirrel change a) (12) his home of . Chase I = . 1 110-100 m 43 solve m m for his of > 0 1000km X 1000 - = his . 10 m , ..., safe he was safe = out zone 100m within 18 zone . t he will not off m > convert - be anble making bracelet it back in time to turn 2) boat travells S2 Slem/h the same 30 2 . maintaining 27 S Not . a) d S2 = . 42 + speed W = . 5x = = . 3) 30 relative to . in * 2. in d 30 62 = IS 40 Sin(30 . <1 dy = - = - . . 2) Im . - x: ↓ . 46 /em = b) 86 kem 06 62 (09(27 . . s 76 83 rem 06 0y 11 : 62 sin /27 . - 36 . 2) 83 km (0x) = ~ 1ay)2 + (41 8612 . : . 20 Hm = + 1 - 36 . 83)2 SS 77 . tan" /Is it :, 42 from the v : I Vi . = direction to head . - . mins , starting point 06 62 km . 26 changes 35 for km/h) 0 d , = 30 62 CoS30 2 = N 62 k m 501min + 44 boat -of . average velocity S2 S x 4 4 = , 44 0 min for displacement b) boat . , east I 27 5 . 30 - 2 . 17 52 5 kem . Acceleration dar 20 in 0 : = t U2 + Gary Gary V - - +, OV 3x-component = OVY Eay-component = * For x-components & projectile only A So all 4. down considered /y-direction) formulas revolving ax ax acceleration , always is O is 0 = 4 straight The of stretches a horse of the semicircles end 14 of Determine m/s . C I D . arga from 160 160 m C ⑳ point 14 M/s pointd hower Both , the C approaches velocity . West , E - = = horse C a)c & a 160 apart speed . constant D4b) - m C - E = Gradis 00m I t : m - 250 the racing 250 long Horse gallops at r ⑳ as so will , motion As it of these D = 251 m Is 1 / 2 14 is appraches it remains the same changing - π(80) direction angles , are , south . perpendicular > = 0 distance . 46S- from distance C- D C-t from instanteneous acceleration or Tott = o finding = = -at vector addition finding Subtraction Vector 4 . 7(b) has car velocity of South the find Vf Vi Q - Up-V = 1 . 25m/s the = IS aarg 1-1 6) : 1Aargl - . = average mis . . = - = 1 1 6 . . tan" G + (1 M/s tan" . 6) : 45' = : ( (i) Waf S . for 7 . starts . 05 swimming east 6 4 mrs Determine the . , undergoes Gary = = = Vott because she (6 4) 12 0) (7 0) . m/s2 South is N moving south +Y - . . . 6 4 - 14 0 . . (6 = = > 2 0 final velocity - V is 25 acceleration 4 10 initially V velocity . ( 1 6) 2 26 Vf Vi (1 6) - : the 2) - I t 25 m/s = east , RISS later mis magnitude 2 direction of darg : = 25 4) + ( 14)2 - . E W - x +X 15. 4 m/S . Han ! ) Eilee s'Y 4 11 . ball thrown 0 8m/s" . from stationary initial unknown velocity. ball accelerates at 45 for 25 final velocity 14 M/S angle of a = . below the horizontal V = Votat Vo + ( 24 = V + ( . 24 + 14 6 43 6 Mrs 24 = - . . tan") = - 10 4 - 8)(2 0) . 19 6) . = Vo = Vo . 6) 24 34· . above the horizontal an