Chaper 2. Moion in One-Dimension I Level : AP Physics Insrucor : Kim 1. Average Rae of Change and Insananeous Velociy To find he average velociy(v ) of a paricle, we need o find he paricle s displacemen x (=change in posiion) divided by he ime inerval v = x Q1) A paricle s moion can be described by he posiion funcion = ² +2. Find he average velociy beween =2 ~5s. Sol) v = x = x f x i = x(5) x(2) = 35 8 = 9m/s f i 5 2 3 = 2 +2 9m/s 35 The linear slope represens he average velociy beween he ime inerval = 2~5s 8 2 5 (s) However, finding he average velociy is no useful informaion. We are more ineresed in finding he velociy or speed a a cerain specific ime. In order o do ha, we can make he ime inerval shorer. Le s say we wan o find he velociy a a specific ime =2s. If we find he average velociy beween =2~2.1s, hen we can find he velociy coming close o he acual velociy a =2s. Example1)The average velociy during = 2~2.1s Posiion funcion is = ² +2, so = 2 +2 8.61 6.1m/s v = x = x(2.1) x(2) = 8.61 8 = 6.1m/s 2.1 2 0.1 The linear slope represens he average velociy beween he ime inerval = 2~2.1s 8 2 2.1 (s)
If we make he ime inerval much shorer and shorer, we will ge much closer o he velociy a =2s. Example2) The average velociy during = 2~2.0001s Posiion funcion : = ² +2 = 2 +2 6.0001m/s v = x = x(2.0001) x(2) = 8.0006 8 2.0001 2 0.0001 = 6.0001m/s The linear slope represens he average velociy beween he ime inerval = 2~2.0001s 8.61 8 2 2.0001 (s) From Example2), since = f i, = 0.0001s. Here, represens a very shor ime inerval. Then v = x = x(2+ ) x(2), where = 0.0001s The above equaion allows us o find velociy ha is fairly close o =2s. If we express an equaion o find he velociy a any specific ime = o, hen v = x = x( o+ ) x( o ), where is a very shor ime inerval If we evenually make he so close o zero( =0.00000 00001 ), hen we are acually find he acual velociy a = o. Making very very very close o zero is expressed as lim 0 So finding he insananeous velociy a any specific ime = o is x v =lim 0 = lim x( o + ) x( o ) : insananeous velociy a = 0 o Q2) A paricle s moion can be described by he posiion funcion = ² +2. Find he insananeous velociy a ime =2s. Mehod1) Using derivaives x v = lim 0 = x( o+ ) x( o ) x(2+ ) x(2) { (2+Δ) 2 +2(2+Δ)} {2 2 +2 2} = lim = lim 0 0 4+4Δ+Δ 2 +4+2Δ 8 Δ 2 +6Δ = lim 0 = lim 0 = lim(δ +6)=6m/s 0 Mehod2) Using x x o = v o + 1 2 ax² and v = vo + ax = ² +2 shows ha x o=0, v o=2m/s and a x=2m/s 2. So v = v o +a x = 2 + 2 2=6m/s
2. Rules for Differeniaing Funcions If we summarize he derivaive for all he posiion funcions in he pracice quesions, hen = 2 2 x => v = lim 0 = 4 = 2 + 4 + 3 2 x => v = lim 0 = 4 + 6 = 21 + 22 3 2 x => v = lim = 22 6 0 Following he paerns, we can devise a simple mehod of finding he derivaives of he posiion funcions. If x = n x, hen lim 0 = nn 1 Also, raher expressing he derivaives as lim x 0, a more simple expression is dx Δx vx = lim = dx : insananeous speed Δ 0 Δ d d Using he formulas : x x o =v o + 1 ax² v = vo + ax : can be used only when a=consan 2 Using derivaives : vx = dx d, where if x = n hen vx = dx d = n n 1 : can be used wheher a=consan or no Pracice Quesions 1. The posiion of a paricle moving along he x-axis = 2 2. i) Wha is he average velociy during he ime inerval =1s o =3s? Wha is he insananeous velociy a =3s? a) 6m/s, 12m/s b) 8m/s, 12m/s c) 6m/s, 14m/s d) 8m/s, 14m/s ii) Wha is he iniial posiion? Tha is, he posiion of he paricle a =0. a) 0m b) 2m c) 4m d) 6m iii) Wha is he iniial velociy? a) 0m/s b) 2m/s c) 4m/s d) 6m/s iii) Wha is he acceleraion? a) 10m/s 2 b) 8m/s 2 c) 6m/s 2 d) 4m/s 2
Using he formulas : x x o =v o + 1 ax² v = vo + ax : can be used only when a=consan 2 Using derivaives : vx = dx d, where if x = n hen vx = dx d = n n 1 : can be used wheher a=consan or no 2. The posiion of a paricle moving along he x-axis = 2 + 4 + 3 2. i) Wha is he average velociy during he ime inerval =1s o =3s? Wha is he velociy a =3s? a) 16m/s, 14m/s b) 12m/s, 14m/s c) 16m/s, 22m/s d) 12m/s, 22m/s ii) Wha is he iniial posiion? a) 0m b) 2m c) 4m d) 6m iii) Wha is he iniial velociy? a) 0m/s b) 2m/s c) 4m/s d) 6m/s iv) Wha is he acceleraion? a) 10m/s 2 b) 8m/s 2 c) 6m/s 2 d) 4m/s 2 3. A oy car sars a x = 0 and sops when i reaches x = 12m a = 3s. A x- graph is shown below. Calculae he average velociy during he ime inervals =0~3s. x(m) 12 3 (s) i) However, he graph above is no a realisic represenaion of he moion of he oy car. Why?
ii) Which of he x- graph below represens a gradual increase of speed? (a) (b) (c) (d) iii) Which of he x- graph below represens a gradual decrease of speed? (a) (b) (c) (d) iv) Draw a realisic x- graph below. (car sars a x = 0 and sops when i reaches x = 12m a = 3s) x(m) 12 (s) 3 v) Draw a v- graph below. v(m/s) 3 (s) 3. Acceleraion(1-Dim) The average acceleraion of a paricle is The insananeous acceleraion of a paricle is a x = Δv x Δ = v xf v xi f i Δv a x = lim = dv x Δ 0 Δ d
=> The insananeous acceleraion equals he derivaive of he velociy wih respec o ime => The insananeous acceleraion is he slope of he v- graph 4. A paricle moving along he x-axis according o he expression x=². Plo x-, v- and a- graph. (a) x- graph (b) v- graph (c) a- graph Using he formulas : x x o =v o + 1 2 ax² Using derivaives : vx = dx d and ax = dv d v = vo + ax : can be used only when a=consan : can be used wheher a=consan or no 5. Draw he graph of he = ( 1) 2 + 2 ( > 0). i) Derive he velociy funcion a any ime and he acceleraion funcion a any ime. Tha is, derive v() and a(). ii) Wha is he iniial posiion and he iniial velociy? iii) A wha ime is he velociy zero? iv) Draw, v() graph and a() graph x v a 2 2 1 2.414 1 2 3 1
The following quesion describes a moion where he acceleraion is no consan. So i can only be solved using calculus. 6. The posiion of a paricle moving along he x axis is given by = 6 2 3. i) Find he posiion of he paricle when =1s, 2s and 3s ii) Express he velociy funcion using vx = dx d iii) Find he velociy when =1s, 2s, 3s and 4s iv) Plo a 'velociy vs ime' graph below and find he posiion of he paricle when i achieves is maximum speed in he x direcion. A wha ime is he speed momenarily zero? Does he paricle reverse direcion? v() 12 9 6 3 2 4 6 v) Plo an 'acceleraion vs ime' graph below. A wha ime is he acceleraion is momenarily zero? a() 12 9 6 3 1 2 3
3. Rules of Inegraion The difference beween squaring and aking he square roo is ha if we square a posiive number and hen ake he square roo of he resul, he posiive square roo value will be he number ha you squared. ex) 4 2 =16 and 16 = 4 In he same way, if we can obain he velociy by differeniaing he posiion or obain he acceleraion by differeniaing he velociy, hen we can work opposie by obaining he posiion by ani-differeniaing he velociy or obain he velociy by ani-differeniaing he acceleraion. A synonym for ani-differeniaion is inegraion. Tha is, inegraion is he inverse of differeniaion. Basic Rule If x is any variable, and c is any consan, hen 7. Inegrae he x x n dx = 1 n+1 xn+1 + c, where n 1 8. Inegrae he x 3 9. Inegrae he 3x Deriving he posiion funcion x from he velociy funcion v from acceleraion funcion a. Since v = dx, so x = vd, d Deriving he velociy funcion v from he acceleraion funcion a. Since a = dv, so v = ad d
x x o =v o + 1 dx ax² v = vo + ax vx = 2 d x n dx = 1 n+1 xn+1 + c, where n 1 ax = dv d x = vd and v = ad Quesions 10-12 An objec moving in a sraigh line has a velociy v in meers per second ha varies wih ime in seconds according o he following funcion. v = 4 + 0.5 2 10. The insananeous acceleraion of he objec a = 2 seconds is (a) 2m/s 2 (b) 4m/s 2 (c) 5m/s 2 (d) 6m/s 2 (e) 8m/s 2 11. If he posiion of he objec was a x = 5m when = 0s, hen find he posiion of he objec a = 6s. (a) 60m (b) 65m (c) 70m (d) 75m (e) 80m 12. The displacemen of he objec beween = 0 and = 6 seconds is (a) 22m (b) 28m (c) 40m (d) 42m (e) 60m 13. A paricle moves along he x-axis wih a non-consan acceleraion described by a = 12(m/s 2 ). If he paricle sars from res so ha is speed v and posiion x are zero when = 0, where is i locaed when = 2 seconds? (a) x = 12m (b) x = 16m (c) x = 24m (d) x = 32m (e) x = 48m
Addiional Quesions 14. The velociy of a paricle moving along he x axis varies according o he expression v x= 40 5 2, where v x is in meers per second and is in seconds. See exbook p.34 for soluion and answer i) Find he average acceleraion in he ime inerval = 0 o = 2.0s. ii) Deermine he acceleraion a =2.0s. 15. A je lands on an aircraf carrier a a speed of 140mi/h(=63m/s). See exbook p.38-39 for soluion and answer i) Wha is he acceleraion(assumed consan) if i sops in 2.0s due o an arresing cable ha snags he je and brings i o a sop? ii) If he je ouches down a posiion x o=0, wha is is final posiion? 16. A car raveling a a consan speed of 45.0m/s passes a rooper on a moorcycle hidden behind a billboard. One second afer he speeding car passes he billboard, he rooper ses ou from he billboard o cach he car, acceleraing a a consan rae of 3.0m/s 2. How long does i ake he rooper o overake he car? See exbook p.39-40 for soluion and answer