University:
University Of Guelph
Course:
PHYS 1300 | Fundamentals of Physics
Academic year:

2024
Views:

141

Pages:

6
Author:

Malz M.

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 Sample problem 2 2 . calculate average Velocity Tourist takes 90 Local driver same takes 4 0 min to travel is * spot as Displacement typically the Tourist = is Tourist = the tourist. Both ended in the same people smallest value when Vargen = given local = = = Speedarg 2 the alps Local in - Speedarg also min = s n = + : Islem - 0 30 lem/n , lem/h 100 = Ase = displacement values = = east to the distance finish Start to diff Vary going have the same Varg + mountain spot just different 10 lem/h Speedarg travelling , always the straight path from Varg cars of the tunnel at the bottom a 2 Rem . Started & they as the Speed for to travel 45 lem min * Note , the local tool exact & Speedarg = . /Sh 100 lem/h Acceleration Par : 0 - . cm . 10 0 X 10 0 110 ms 10 0 >m 11PX/0- 0 . 118 . - Speedarg 6s > - f . = 2 . 2 . 4/ x 10 S . ° 2 : . 41 : X100 kem Gixo e wrong 27 3 das & = 114ms M 157 Pm/s 27 3 0. S t : earth in muscle circumfrance 1 1 4 X 10 : a d : I moon · through days 24 x , x36002 x 1000 41x104 Speed aug d = 3 04 . = 2 I . x100m 36x106S : 2 . 3546 intestines 4 0 = 124 indigestible: speed arg m . 36h = 60 min 124x o 60 x = 43200 . s e = M 36 h 4 32X104 2160 4 0 = . X 60 min 2160 = -. = 2 00 . X 10-4 X 60 8 1 = . 24 min X10's e "om = 6 . 48 X 10-5 Q2 2(a) . Light 3 00 x = . 100mrs , time find insec d += S distance 1 += . 3 = from 50 x 10 " * 00 100 M/s . x d = 0 46 0 - . - = 1 50 x 10 . 0 . 84 "m x . = Sun 500 S Q 2 11 . to earth . & 12 0 + 0 12 . . = = Rest = 04 e Yo 0 0 . . + y(V , 46m + 33 m + v) t 1(0 + - opposite direct . : = takes - 0 104 M/s 0 46M - . . + V forward . final position 0 26 M Speed arg . 5 0 104)2 2 55 . : F i Var t ↳ -- ↳& ↳ acceleration is constant, velocity is Changing & uniform rate a t t, & 0 = x, = X = X Xo 3 Cotrl Viv x-xo : + 3 x xothcroft i : PRAC Q Q. / motorbike travels (sm/s of magnitude of Vo t = 3 6s a = 5 x : . x . 2 m/s2 O . far * We don't have ? I Speeds up . How 5 2 m/s" 3 15 m/s : , x = X = = xo + w) constant 3 6S . of acceleration ? final velocity Vott's at 2 (0) + (15m(s) (3 6) + !(5 . 08 during did she travel acceleration . m/s) 13 65)" 2 . M Q2 Rocket lanchpad & 0 = = Vo V a , up from Speed rest , reaching . What 630m/s = is O O = > 0 4 7 km . . =? * Note , time not given 0047m > - 4 . 7 X10' an altitude acceleration 630 m/s = x goes m 3 ↑ v of of C = I v O . kem above rocket ? Rearrange = 4 7 O > + 2(a)(X) v2 2x Gi 1630 miss I 2/4 7 . X (07) 42m/s2 Q3 hockey puck direction of as initial moves initial velocity velocity 26 m/s2 = x x 0 = vo (0) = 2 ( t =? . 28m 2: m 0 26 m/s2 It travels - V opposite of 26mrs" rest before coming to in a velocity initial 28 m = acceleration DbS travel time 0 m/s = a . 3 d Vo :? V constant 2 = g - Vo 0 - . . 26) (28) 26)/28) ( 2)( = = 2 ( 0 - I = - 0 . 3 if ↓ f 26)(28) 3 8 m/S . 3 8 m/s . Travel Time Vo V 3 8 m/s = * . 0 mis = = x = t I 28 m V = t Vo + at V = Rearrange a 0 m ? Vo - = 10 - - = /S S 3 0 . . 8 m/s) 26 M/sh u v is (t) then a isl is 1) , , then a is (t) .