# Application of kinematic equation:

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1 HELP: See me (office hours). There will be a HW help session on Monda nigh from 7-8 in Nicholson 109. Tuoring a #10 of Nicholson Hall. Applicaion of kinemaic equaion: a = cons. v= v0 + a = + v a v = v + a( ) 0 0

2 Eample: A red car and a green car, idenical ecep for he color, move oward each oher in adjacen lanes and parallel o he ais. A ime =0, he red car is a r =0 and he green car is a g = 0 m. If he red car has a consan veloci of 0 km/h, he cars pass each oher a =44.5 m, and if i has a consan veloci of 40 km/h, he pass each oher a =76.6m. Wha are (a) he iniial veloci and (b) he acceleraion of he green car? 0km Red car: a=0 so f (1) = v 1 1 v 1 = = 50 h 9 m / s 1 = 8.0s f () = v 1 40km 100 = 6.9s v = = m / s h 9 Now we mus simulaneousl solve wo equaions for he green car = v a 1 = = v 0 + a This Gives! = a=.0 m/ s v0 = 13.9 m/ s

3 Special Case: free-falling bod moion Close o he surface of he Earh all objecs move oward he cener of he Earh wih an acceleraion whose magniude is consan and equal o 9.8 m/s. We use he smbol g o indicae he acceleraion of an objec in free fall. B a a = -g v= v g 0 (eq. 1) g = o + v0 (eq. ) v v 0 = g ( o ) (eq.3) A

4 Quesion A person sanding a he edge of a cliff hrows one ball sraigh up and anoher ball sraigh down a he same iniial speed. Neglecing air resisance, which ball wih he greaer speed his he ground below he cliff? 1. upward.. downward. 3. neiher he boh hi a he same speed.

5 Kinemaics: Taking Advanage of Smmer v= v0 g = v g v = v g = v+ v 1 ( ) 0

6 Graphical Inegraion in Moion Analsis (nonconsan acceleraion) When he acceleraion of a moving objec is no consan we mus use inegraion o deermine he veloci v ( ) and he posiion ) ( of he objec. The inegraion can be done eiher using he analic or he graphical approach: dv a () = dv= ad () dv= ad () v v = ad () v = v + ad () d ad ( ) = [ Area under he a versus curve beween 0 and 1] 1 1 d v () = d = v () d d = v () d d 1 1 = vd = + vd vd = [ Area under he v versus curve beween 0 and 1 ] o

7 Eample: Acceleraion: (a) If he posiion of a paricle is given b = 0 5 3, where is in meers and is in seconds, when, if ever, is he paricle s veloci zero? (b) When is is acceleraion a zero? (c) For wha ime range (posiive or negaive) is a negaive? (d) For wha ime range (posiive or negaive) is a posiive? (e) Graph (), v() and a(). () = v() = d = 0 15 d dv d a() ( ) = = d d = 30

8 In phsics we have parameers ha can be compleel described b a number and are known as scalars. Temperaure and mass are such parameers. Oher phsical parameers require addiional informaion abou direcion and are known as vecors. Eamples of vecors are displacemen, veloci, and acceleraion. This chaper covers he basic mahemaical language o describe vecors. In paricular we need o know he following: Geomeric vecor addiion and subracion Geomeric vecor addiion and subracion i Resolving a vecor ino is componens The noion of a uni vecor Addiion and subracion vecors b componens Muliplicaion of a vecor b a scalar The scalar (do) produc of wo vecors The vecor (cross) produc of wo vecors

9 Vecor epressed b componens Vecor a can be wrien wih is componens and uni vecors a= a ˆi+ a ˆj (wo-dimensional case) a = a cos θ and a = asin θ. The quaniies a ˆi and a ˆj are called he vecor componens a a= a + a and an θ =. a Vecor a in hree-dimensional case a = a iˆ+ a ˆ j+ a kˆ a = acos θ ; a = asin θ ; a = asinθ z θ z z z θ = + + z ĵ θ a a az θ = θ = θ = a a a a cos ; cos ; cos a a a î ˆk a

10 O Adding vecors b componens ˆi ˆj a = a ˆi ˆ + a b = b + bj r = a+ b = rˆi+ r ˆj. r b i j ; i j. a The componens r and r are given b he equaions r = a + b and r = a + b. Subracing vecors b componens O a = a ˆi ˆj ˆi ˆ + a b = b + bj d = a b = d ˆ i+ d ˆj. d b i j ; i j. a The componens and are given b he equaions d = a b and d = a b. d d

11 Mulipling a Vecor b a Scalar Muliplicaion of a vecor a b a scalar s resuls in a new vecor b = sa. The magniude b of he new vecor is given b b = s a. If s > 0, vecor b has he same direcion as vecor a. If s< 0, vecor b has a direcion opposie o ha of vecor a. The Scalar Produc of Two Vecors The scalar produc ab of wo vecors aand bis given b ab = abcos φ. The scalar produc of wo vecors is also known as he "do" produc. The scalar produc in erms of vecor componens is given b he equaion ab ab+ ab+ ab. = z z Applicaion in phsics: Work: W = F d = Fdcosφ

12 Eample Wha is he angle φ beween a = 3.0iˆ 4.0 ˆjand b =.0iˆ + 3.0kˆ? Use ab = ab cos φ such ha cos φ = Since ab ab a = + = = + = 3.0 ( 4.0) 5.0 and b (.0) and a b= ab + ab + ab z z= ab = 6.0 ab 6.0 cos φ = = and φ = 109 ab a φ ˆk b ĵ îi

13 The Vecor Produc of Two Vecors The vecor produc c = a b is a vecor c. The magniude of c is given b he equaion c= ab sin φ. c is perpendicular o he plane P defined b a and b. The sense of he vecor c is given b he righ-hand rule: a. Place he vecors a and b ail o ail. b. Roae a in he plane P along he shores angle so ha i coincides wih b. c. Roae he fingers of he righ hand in he same direcion. d. The humb of he righ hand gives he sense of c. The vecor produc of wo vecors is also known as he "cross " prod uc.

14 The vecor produc c = a b in erms of vecor componens ˆi ˆj k, ˆ ˆi ˆj k, ˆ a= a ˆi ˆj kˆ + a + a z b = b + b + b z c = c + c + c z The vecor componens of vecor c are given b he equaions c = a b a b, c = ab ab, c = ab ab. z z z z z a a az a a a or c =, c, cz b b = b b = b b z z z Noe: Those familiar wih he use of deerminans can use he epression i j k a b= a a a z b b b z Noe: The order of he wo vecors in he cross produc is imporan: b a = ( a b) ). Applicaion in phsics: Torque: τ = r F Roaion ais r F

15 Eample Show a ( b c ) = b ( c a ) = c ( a b ) i j k a ( b c ) = ( a iˆ + a ˆ j + a kˆ ) b b b z z c c c z a a a z = a ( b c bc ) + a ( bc b c ) + a ( b c b c ) = b b b z z z z z b b bz = c c c = b ( c a ) z a a a z c c cz = a a a = c ( a b) z b b b z c c c z

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