Ch4: Motion in two and thee dimensions (D and 3D) Displacement, elocity and acceleation ectos Pojectile motion Cicula motion Relatie motion 4.: Position and displacement Position of an object in D o 3D is descibed by its position ecto is dawn fom oigin to the position of the object at a gien time. It can be witten as Position ecto in 3D As paticle position changed fom duing a cetain time inteal displacement ecto that can be witten as In components fom
4.: Position and displacement Displacement ecto in D is shown in the figue below: 1 x y xiˆ + yj ˆ x y x y 1 1 4.: Position and displacement: Example1 Rabbit uns acoss a paking lot. Its position changing accoding to x and y in metes, t in seconds. Find its position ecto at t15s in unit ecto notation and in magnitude angle notation Position ecto in unit ectos can be witten as The magnitude fom the oigin is at angle
4.3: Aeage elocity and Instantaneous Velocity Since occus in time inteal t t t 1 aeage elocity 4.3: Aeage elocity and Instantaneous Velocity If t decease until t We will hae Instantaneous elocity (o elocity) The Speed + x y
4.3: Aeage elocity and Instantaneous Velocity: Example Fo the abbit in the peceding example1, find the elocity at time t 15 s. Fom Ex:1 Velocity ecto is 4.4: Aeage acceleation and Instantaneous acceleation When a paticle s elocity changes fom in a time inteal, its aeage acceleation duing is Note that, fo an object, acceleation can cause a change in the elocity magnitude o in the elocity diection, o both.
4.4: Aeage acceleation and Instantaneous acceleation Fo we will hae Instantaneous acceleation (o acceleation) 4.4: Aeage acceleation and Instantaneous acceleation: Example 3 Fo the abbit in the peceding two examples 1 and, find the acceleation at time t 15 s. a Fom Ex: we hae and Hence the acceleation ecto is
4.4: Aeage acceleation and Instantaneous acceleation: Example 4 The position of a paticle moing in the xy- plane is gien by ( t) t iˆ + 3tj ˆ whee is in mete and t is in seconds. Find a) the aeage elocity in the time inteal t s and t 4s i ˆ (4 )ˆ 3(4) j ( )ˆ i 3() ˆ 1 (4) () + j ag t t t 4 4 1 1 (3 8)ˆ i + (1 6) ˆj (1ˆ i + 3 ˆ) j m / s b) Velocity and speed of the paticle at t 3s d t tiˆ 3 ˆ ( ) 4 + j (3) 4(3)ˆ i + 3 ˆj 1ˆ i + 3 ˆj m / s dt speed 1 + 3 153 1.37 m / s d c) The acceleation at t3s a 4ˆ i m / s at any time (a is constant) dt 4.4: Aeage acceleation and Instantaneous acceleation: Example 5 A paticle stats fom the oigin at t with an initial elocity haing an x component of m/s and a y component of -15 m/s. The paticle moes in the xy plane with an acceleation in the x diection only (a x 4. m/s ). Find a) The components of the elocity ecto at any time and the total elocity ecto at any time. x x + a t ( + 4t)m/s x y y 15m/s (ay ) elocity ecto iˆ + ˆj (( + 4t)ˆ i 15 ˆ) j m/s f x y b) Calculate the elocity and speed of the paticle at t 5 s. t i 15 ˆ ( 5) ( + 4(5))ˆ j (4ˆ i 15 ˆ) j m / s speed 4 + ( 15) 185 4.7 m / s ( t 5)
4.5: Pojectile Motion Pojectile motion is an example of motion in D Conside the Two assumptions: 1) The object has an initial elocity, and is launched at an angle θ ) Fee-fall acceleation (g) is constant (neglect ai esistance) Then the tajectoy (path) of the pojectile is a paabola. The elocity changes magnitude and diection The acceleation in the y-diection is constant (-g). The acceleation in the x-diection is zeo. 4.5: Pojectile Motion The initial elocity ecto of pojectile motion is whee
We ll analyze pojectile motion as a supeposition of two independent motions (x and y diections): Constant elocity motion in the hoizontal diection (x) x 4.6: Pojectile Motion analysis x cosθ constant x x x xt xt Fom x and y (whee x and y at t ) we eliminate t equation of the pojectile path is Fee-fall motion in the etical diection (y) y y gt y y y y t 1 gt Ex: Assume the pojectile is launched with initial speed at angle θ, if the maximum hight is h and the hoizontal ange is R. Detemine a) h and b) R a) h is in y-axis a y g ;to find h we use h y y 4.6: Pojectile Motion analysis: the maximum height and hoizontal ange 1 yt A A ( t A gt t A can be found using ya yi y gt gt A A t A ; time to each max. hight h) (at max. hight, y g But ya y ) sinθ t A sinθ g sinθ 1 sinθ sin θ hence h sinθ g h g g g θ
4.6: Pojectile Motion analysis: the maximum height and hoizontal ange Assume the pojectile is launched with initial elocity i at angle θ, if the maximum hight is h and the hoizontal ange is R. Detemine a) h and b) R b) R is in x-diection a x R R x t B cosθ ( cosθ (t A ) sinθ ) g cosθ sinθ g θ R sin θ g The max ange R max is at angle θ45 (sinθsin9 1) R max g 4.6: Pojectile Motion analysis: the maximum height and hoizontal ange fom θ R sin g we can hae same ange if the pojectile is launched at two diffeent angles (fo same initial speed) so that 5 m/s θ 1 +θ 9 ex: 15 and 75 sin3 sin15
A long-jumpe leaes the gound at an angle of. aboe the hoizontal and at a speed of 11. m/s. (A) How fa does he jump in the hoizontal diection? (B) What is the maximum height eached? Solution: follow exactly same as peious fo h and R A) 4.6: Pojectile Motion analysis: example 6 sin θ 11 sin 4 R 7. 94m g 9.8 B) sin θ ymax h. 7m g 4.6: Pojectile Motion analysis: example 7 A piate ship 56 m fom a fot defending a habo entance. A defense cannon, located at sea leel, fies balls at initial speed 8 m/s. a) At what angle θ fom the hoizontal must a ball be fied to hit the ship? b) What is the maximum ange of the cannonballs? Solution: sin θ a) R g b) R max is at θ 45
4.6: Pojectile Motion analysis: example 8 Stone is thown as shown with m/s θ 3 h building 45 m a) Time t to each the gound b) Speed of stone befo it stick the gound f c) Hoizontal distance x f Solution: x cosθ cos3 17.3m / s sinθ sin 3 1m / s a) 4.6: Pojectile Motion analysis: example 8 y 1 1 y yt gt 45 1t (9.8) t 4.9t + 1t + 45 sole fo t t 4.s b) y y x x 17.3m / s (17.3ˆ i 31.4 ˆ) j m / s speed gt 1 9.8(4.) 31.4m / s B 17.3² + ( 31.4)² 35.9m / s c) x x t x 17.3(4.) 73m
4.6: Unifom cicula motion Cicula motion is an example of motion in D conside a paticle foced to moe aound a cicle of adius at constant speed unifom cicula motion only changes diection Examples: wheels, disks, motos, etc. 4.6: Unifom cicula motion Fo unifom cicula motion, The acceleation is always diected adially inwad (towads the cente). it is usually called a centipetal acceleation (meaning cente seeking acceleation ) whee is the speed and is the adius See poof at the text book
duing this acceleation at constant speed, the paticle taels the cicumfeence of the cicle (a distance of π) in time T π fo 1 cycle T π T note that ac length time π is the cicumfeence ( يطح (م of the cicle (one eolution) T is the peiod (time needed to complete one eolution (e.) Ex 9: find the acceleation at the edge of a weel of adius 35 cm spining at 6 1 e./min, 6T a c T 4.6: Unifom cicula motion but 6s 6 6.1s π T, hence we need T 6e. ~ 6s 1e. ~ T π π (.35m) m / s T.1s () ac 14m / s towads cente.35 θ t 4.6: Unifom cicula motion: Example 1 Ex: A ca taeling at 1km/h ounds a cone with a adius of 15m. Find the ca centipetal acceleation. 1m 1h 1 km / h( )( ) 33m / s km 36s (33) a c 8.7m / s² 15 towads the cente
4.8: Relatie Motion in One Dimension The elocity of a paticle depends on the efeence fame of whoee is obseing o measuing the elocity (محور اسناد ( Suppose Alex (stationay fame A) and Babaa (fame B moing at constant speed) watch ca P which is moing at speed. Both fames (A and B) will hae diffeent quantities (position, elocity, and acceleation). At the instant shown, x BA is the coodinate of B in the A fame. Also,P is at coodinate x PB in the B fame. The coodinate of P in the A fame is 4.8: Relatie Motion in One Dimension the elocity components ae the time deiatie of the position equation Diffeentiate elocity with espect to time acceleation
4.8: Relatie Motion in One Dimension: Example 11 Babaa s elocity elatie to Alex is a constant 5 km/h and ca P is moing in the negatie diection of the x axis. a) If Alex measues a constant -78 km/h fo ca P, what elocity will Babaa measue Alex (stationay fame A), Babaa (moing fame B) and ca is obseed moing object P. > BA 5 m/s and PA -78 km/h, hence b) If ca bakes to a stop elatie to Alex in time t 1 s at constant acceleation, what is its acceleation elatie to Alex and Babaa? 4.9: Relatie Motion in Two Dimension Ou two obsees ae again watching a moing paticle P fom the oigins of efeence fames A and B, while B moes at a constant elocity elatie to A Fom the aangement of the thee position ectos shown, we can elate the ectos with
4.9: Relatie Motion in Two Dimension: Example 1 A plane is to moe due east with elocity (elatie to gound) when flying in a blowing wind of elocity ( of speed 65km east of noth elatie to gound). If the plane speed is 15km at angle θ south of east elatie to wind ( ), find the plane speed elatie to gound and the angle θ Fom y-components 4.9: Relatie Motion in Two Dimension: Example 13 Ca A is moing with eocity A (3ˆ i + 7 ˆ) j m / s Ca B is moing with elocity ( ˆ i + ˆ) j m / s Find elocity of ca B with espect to A ( BA ): B by ectos A will see this ecto when looking at B BA B A ( 5ˆ i 5 ˆ) j m / s