( ) in the following way. ( ) < 2

Transcription

1 Sraigh Line Moion - Classwork Consider an obbec moving along a sraigh line eiher horizonally or verically. There are many such obbecs naural and man-made. Wrie down several of hem. Horizonal cars waer Verical rockes objecs subjeced o graviy As an obbec moves is posiion is a funcion of ime. For is posiion funcion we will denoe he variable s( ). For insance when s!! in seconds s( ) we are being old wha posiion on he horizonal or verical number line he paricle occupies a differen values of.!! Example 1) For s show is posiion on he number line for d =1 = U = = ( S( S S S1 1 ( 4 When an obbec moves is posiion changes over ime. So we can say ha he velociy funcion v of he posiion funcion over ime. We know his o be a derivaive and can hus say ha v For convenience sake we will define v Moion ( ) > ( ) in he following way. ( ) < v ( ) = v v Horizonal Line obbec moves o he righ obbec moves o he lef obbec sopped Verical Line obbec moves up obbec moves down obbec sopped ( ) is he change s-( ). Speed is no synonymous wih velociy. Speed does no indicae direcion. So we define he speed funcion: ->..<! v!. The speed of an obbec mus eiher be posiive or zero (meaning ha he obbec is sopped). = ( ) The definiion of acceleraion is he change of velociy over ime. We know his o be a derivaive and can hus say ha a( ) = v-( ) = s--( ). So given he posiion funcion s( ) we can now deermine boh he velociy and acceleraion funcion. On your cars you have wo devices o change he velociy: acceleraor brake Le us hink as somehing acceleraing he obbec o be some exernal force like wind or curren. For convenience sake le us define he acceleraion funcion like his: Moion ( ) > a( ) < a( ) = a Horizonal Line obbec acceleraing o he righ obbec acceleraing o he lef velociy no changing Verical Line obbec acceleraing upwards obbec acceleraing downwards velociy no changing Jus because an obbec s acceleraion is zero does no mean ha he obbec is sopped. I means ha he velociy is no changing. Wha device do you have on your cars ha keeps he car s acceleraion equal o zero_ cruise conrol Also bus because you have a posiive acceleraion does no mean ha you are moving o he righ. For insance suppose you were walking o he righ v when all of a sudden a large wind sared o blow o he lef [ ( ) < ] a [ ( ) > ]. Wha would ha do o your velociy_ slow you down. AB Soluions Su Schwarz

2 ! + Example ) Given ha a paricle is moving along a horizonal line wih posiion funcion s (. (. The velociy funcion v! and he acceleraion funcion a Le s complee he char for he firs 5 seconds and show where he obbec is on he number line. s( ) v ( )! v( )! a ( ) Descripion of he paricle s moion -4 4 moving lef acceleraing o he righ moving lef bu slower sill acceleraing o he righ - sopped sill acceleraing o he righ 3-1 moving o he righ acceleraing o he righ moving o he righ faser sill acceleraing o he righ moving o he righ faser ye sill acceleraing o he righ I is oo much work o do such work for complicaed funcions. We are generally ineresed when he paricle is sopped or when i has no acceleraion. We are also ineresed when he obbec is speeding up or slowing down. Realizing ha an obbec s velociy is eiher posiive (moving righ) negaive (moving lef) or zero (sopped) and an obbec s acceleraion is eiher posiive negaive or zero (consan speed) we can now use a char o deermine all he possibiliies of an obbec s moion as if you were looking a i from above. a( ) > a( ) < a( ) = ( ) > speeding up slowing down consan velociy righ ( ) < slowing down speeding up consan velociy lef sopped acceleraing righ sopped acceleraing lef sopped no acceleraion v v v! + Example 3) A paricle is moving along a horizonal line wih posiion funcion s 3 4. Do an analysis! of he paricle s direcion (righ lef) acceleraion moion (speeding up slowing down) & posiion. Sep 1: v! So v 3 Sep : Make a number line of v a d3 g h ( ) showing when v( ) i jjjjjjjjjjj he obbec is sopped and he sign and 3 direcion of he obbec a imes o he lef and righ of ha. Assume h. Sep 3: a( ) =. Does a( ) = _ No Sep 4: Make a number line of a( ) showing a( ) ijjjjjjjjjjjjjjjjjjjjjj when he obbec has a posiive and negaive acceleraion. Scale i exacly like he v( ) number line. slowing down speeding up Sep 5: Make a moion line direcly below moion i jjjjjjjjjjj he las wo puing all criical values 3 muliplying he signs and inerpreing h according o he char above. g Sep 6: Make a posiion graph o show where he posiion llllld3lllllllllldlllllllll obbec is a criical imes and how i moves AB Soluions Su Schwarz

3 ! + + Example 3) A paricle is moving along a horizonal line wih posiion funcion s 8 ( (. Do an! analysis of he paricle s direcion acceleraion moion (speeding up or slowing down) and posiion. v 1 ( v 3 v ( = U = ( a( ) = 3! 1 =!! + = (! + ) = (! )(! ) = = v( ) a( ) SSSSSSSSSSSSSSSSSSSSSSSSS !( SSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS I9F!<9F$->..<!E>!-I9F!<9F$->..<!E> 79*#9$ SSSSSSSSSSSSSSSSSSSSSSSS SSSSSSSSSSSSSS !( *= *=( *= >9-#*#9$ (!( Noe ha he posiion graph is no like he oher hree graphs. I simply shows he posiion he obbec has wih respec o he origin and criical imes of is movemen found by seing v ( ) and a( ) d.! + + ( ) is measured in fee v is he iniial velociy (velociy a d ) and s is he iniial posiion!( if s( ) is measured in meers. When an obbec is subbeced o graviy is posiion funcion is given by s 13 v s where is measured in seconds s (posiion a d ). The formula is given by s v s! + +! + v o and he acceleraion funcion a( ) =!. This is he acceleraion due o graviy on earh. When an obbec is hrown upward i is subbeced o graviy We are usually ineresed how high he paricle reaches and how fas i is going when i impacs he ground or waer. Le us analyze wha hese mean: From our original s 13 v s we can calculae he velociy funcion v When an obbec reaches is maximum heigh When an obbec his he ground wha is is wha is is velociy_ v = final posiion_ s = So o find he maximum heigh of an obbec se v So o find he velociy of an obbec when i solve for and find s( ) his he ground se s( ) d solve for and find v( ) Example 4). A probecile is launched verically upward from ground level wih an iniial velociy of 11 f/sec. a. Find he velociy and speed b. How high will he probecile c. Find he speed of he a d 3 and d 5 seconds. rise_ probecile when i his he ground.! + s v( ) =! + 11! v 13!)*%-./->..< = 13!)*%-./ v 4 (!)*%-./!->..<! = (!)*%-./! v( ) =! + 11 = = 11 = 54! ( ) + ( ) s s = 183!)*! + = s 13 11! 13! 6 = 6! ( ) + v v 6 11!)*%-./ AB Soluions Su Schwarz

4 Example 5) The equaions for free fall a he surfaces of Mars Earh and Jupier (s in meers in seconds) are: Mars: s 15 3 Earh: s ( 58 Jupier: s! 11 5 ((. How long would i ake a rock iniially a res in a space capsule over he plane o reach a velociy of 16.6 m/sec_ Mars Earh Jupier v = 1353 v = 1353 = ( 5 (3!-./ = 1538(!-./ = 563!-./ v 5 5 = 1353 Example 6) A rock hrown verically upward from he surface of he moon a a velociy of 4m/sec reaches a heigh of s = (! 5 meers in seconds. a) Find he rock s velociy and acceleraion as a b) How long did i ake he rock o reach is highes funcion of ime. (The acceleraion in his case poin_ is he acceleraion on he moon)!! v a 1. 6! = ) = v ( !-./ c) How high did he rock go_ d) How long did i ake he rock o reach half is maximum heigh_ ( )! ( ) s (! 5 = 8 s m 5! ( + 8 = ) = ( 5 8!-./ e) How long was he rock alof_ e) Find he rock s speed when hiing he moon.! = s ( 5! 51 = U =!-./ v( 3) = 4! 1. 6( 3)! v 4 speed = 4 m/sec Example 7) A ball is dropped from he op of he Washingon Monumen which is 555 fee high. a) How long will i ake for he ball o hi he b) Find he ball s speed a impac. ground_ s( ) =! = 13 = 444 ) = 458!-./! f / sec + v mph Example 8) Paul has bough a icke on a special roller coaser a an amusemen park which moves in a sraigh line. The posiion s U! 1 ( ) of he car in fee afer seconds is given by: s( ) =! +.. a) Find he velociy and acceleraion of he b) When is he roller coaser sopped_ roller coaser afer seconds_! +! + v 5 5( a 53! 5(! 5 + 5(! = / = U =!-./! c) When is Paul speeding up and slowing down_ d) Where is Paul a criical imes of his ride_! ( = ) = ( ( ) ( ) Q>..<!E>! U( U U 1!QI9F!<9F$!((U ) = s = = 8 s = 56 = 1 s = AB Soluions Su Schwarz

5 Sraigh Line Moion - Homework A paricle is moving along a horizonal line wih posiion funcion as given. Do an analysis of he paricle s direcion acceleraion moion (speeding up or slowing down) and posiion. +!! = ) =! 1. s 3 v 3 a! +!! + ) = U! ) =. s 3 8 ( v a 3 1 v( )!a ( ) -I9F#$J!<9F$!->..<#$J!E> 79*#9$ SSSSSSSSSSSSSSSSSSSSSSSSSS s( ) SSSSSSSSSSSSSSSSSSSSSSSS SSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS *= *= 11 v( )!a ( ) s( ) !SSSSSSSSSSSSSSSSSSSSSSSSS! SSSSSSSSSSSSSSSSSSSSSSSSS! I9F!<9F$!->..<!E>-I9F!<9F$->..<!E> 79*#9$ SSSSSSSSSSS! !SSSSSSSSSSS! *= *= *=1 S( 8 3. s( ) =! + 8! ( s v! + 1! ( ) = ( a ( ) =! ) 8 U = v( ) =! ) = a( ) =! ( 1) ( + 1) v( ) SSSSSSSSSSSSSS! !SSSSSSSSSSSSSS v( ) SSSSSSSSSSSSSSSSSSSSSS (!a ( ) -I9F!<9F$!->..<!E>!-I9F!<9F$->..<!E> SSSSSSSSSSSSS!++++++SSSSSSSSSSSS *#9$ ( s( ) SSSSSSSSSSSSSSSSSSSSSSSS *= S18 S14 *=( 1 *=!a( ) s( ) I9F#$J!<9F$!->..<#$J!E> 79*#9$ SSSSSSSSSSSSSSSSSSSSSSS A 45-caliber bulle fired sraigh up from he surface of he moon would reach a heigh of s =! 53 fee afer seconds. On Earh in he absence of air is heigh would be! s =! 13 fee afer seconds. How long would i ake he bulle o hi he ground in eiher case_ *= 3 1 *= Earh s (! ) = 16 = 83 ) = 5 sec Moon s (! ) =. 6 = 83 ) = 3 sec AB Soluions Su Schwarz

6 6. A ball fired downward from a heigh of 11 fee his he ground in seconds. Find is iniial velociy.! + o + = s 13 v 11! 3( + v + 11 = ) v =!( o vo =!(!)*%-./ 7. A probecile is fired verically upward (earh) from ground level wih an iniial velociy of 16 f/sec. a. How long will i ake for he probecile o hi b. How high will he probecile ge_ he ground_ o! + = s 13 13! 13! 1 = 1!-./ v( ) =! + 13 = = 13 ) = 54! ( ) + s (!)* 8. A helicoper pilo drops a package when he helicoper is f. above he ground rising a f/sec. a. How long will i ake for he package o hi b. Wha is he speed of he package a impac_ he ground_! + + =!!! s 16 ( ) ) = s sec v( ) =! 3 + v! ( ) + = f/sec 9. A man drops a quarer from a bridge. How high is he bridge if he quarer his he waer 4 seconds laer_! + = s 13 s s( ) =! 13 (() + s = s o = 43!)* 1. A probecile fired upward from ground level is o reach a maximum heigh of 16 fee. Wha is is iniial velociy_ s 16 v 16 v 3 v v = 3! 16 + ( 3) = = 16 = 1 sec v! + =! + = f/sec = AB Soluions Su Schwarz

7 11. A probecile is fired verically upward wih an iniial velociy of 96 f/sec from a ower 56 fee high. a. How long will i ake for he probecile o reach b. Wha is is maximum heigh_ is maximum heigh.! + +! + = ) = s v 83!-./! + +! + = ) = s v 83!-./ s (!)* c. How long will i ake he probecile o reach d. Wha is he velociy when i passes he saring is saring heigh on he way down_ poin on he way down_! + + = ) = s ! 13! 3 3!-./ 83!)*%-./ e. How long will i ake o hi he ground_ f. Wha will be is speed when i impacs he ground_! + + = s ) =! 16! 6! 16 8 sec v( 8) =! 3( 8) + 96 = 16 f / sec 1. John s car runs ou of gas as i goes up a hill. The car rolls o a sop hen sars rolling backwards. As i rolls is displacemen d +! ( ) in fee from he boom of he hill a seconds since he car ran ou of gas is given by: d a. When is his velociy posiive_ Wha does his b. When did he car sar o roll backwards_ mean in real world erms_ How far was i from he boom of he hill a ha ime_ v( ) = 31! >. < sec - going up he hill v( ) = 31! < sec - d f c. If John keeps his foo off he brake when will d. How far was John from he boom of he hill he be a he boom of he hill_ when he ran ou of gas_ 1 f ! = ) + ( 531!-./ d Ray is a sky-diver. When he free-falls his downward velociy v from he ime of he bump is given by: v calculaor for he firs 3 seconds of his dive.! ( ) ( ) fee per second is a funcion of seconds ( ) and a( ) on your measured in f/sec. Plo v a. Wha is Ray s acceleraion when he firs bumps_ b. Wha appears o be he erminal velociy I#7 v Why does he acceleraion decrease over ime_ /1 ( ) _! f /sec - air resisance 41!)*%-./ AB Soluions Su Schwarz