Chapter 3 Poston, Velocty, and Acceleraton Revsted The poston vector of a partcle s a vector drawn from the orgn to the locaton of the partcle. In two dmensons: r = x ˆ+ yj ˆ (1) The dsplacement vector s the change n a partcle s poston from some tme t 1 to some tme t : r= r r As before, the average velocty of the partcle s: v avg f r = t To fnd the nstantaneous velocty, we take the lmt as the tme nterval goes to zero (as we dd n the one dmensonal case): Physcs 10 Page 1
r dr v = lm = () t 0 t dt In two dmensons: r = r r = x x ˆ + y y ˆj = x ˆ+ yj ˆ ( ) ( ) f f f r v = lm t 0 t ˆ ˆ x + yj = lm t 0 t = ˆ x y lm + ˆj lm t 0 t t 0 t = ˆdx dy + ˆj dt dt The average acceleraton s stll defned as: a avg v = t To fnd the nstantaneous acceleraton, we follow the same process that we used to fnd the nstantaneous velocty: v dv a = lm = t 0 t dt If we express the velocty vector n rectangular coordnates: v= v ˆ+ vj ˆ Now the nstantaneous acceleraton s: x y Physcs 10 Page
( x y ) d a= v ˆ+ vj ˆ dt dv dv x y = ˆ + ˆ j dt dt dx = ˆ dy + ˆj dt dt Example 1: A partcle has a poston vector gven by r = ( 30t) ˆ + ( 40t 5t ) ˆj, where r s n meters and t s n seconds. Fnd the nstantaneous velocty and nstantaneous acceleraton vectors as functons of tme. What are the poston, velocty, and acceleraton of the partcle at t =.00 s? Soluton: Physcs 10 Page 3
Projectle Moton Imagne that a partcle s launched from orgn wth an ntal speed at an angle θ above the horzontal. v The components of the partcle s ntal velocty are: v v = v cosθ x = v snθ y (3) We shall assume that the partcle experences no ar resstance. Therefore: a = 0 and a = g (4) x Substtutng these acceleratons nto our knematc equatons: vxf = vx v = v gt yf Notce that the horzontal and vertcal veloctes are ndependent of each other. Ths means that we can solve ths two dmensonal moton problem as two separate one dmensonal moton problems. The postons of our partcle are: y y Physcs 10 Page 4
x = x + v t f x 1 yf = y + vyt gt (5) We shall now consder the case where y f = y. For smplcty we wll assume that x = 0 and y = 0. f The horzontal dstance traveled by the partcle s called the range (R). The range s: R = v xt = v cosθt From the above equaton, we can see that to compute the range of the partcle we must frst compute the tme of flght. To fnd ths, we look at the vertcal moton of the partcle: 1 yf = y + vsnθt gt 1 gt = v sn θ t v t = snθ g Physcs 10 Page 5
The horzontal range of the partcle s: R = v cosθ t v = sn θ cos θ g v = sn θ g (6) Imagne that we wanted to fnd the range of our partcle at π α= θ : v R = snα g v π = sn θ g v = snπcosθ cosπsnθ g v = sn θ g Therefore the range of the partcle s the same for both angles. π R( θ ) = R θ Physcs 10 Page 6
Example : Wle E. Coyote plans to use hs new ACME super cannon to crush the Road Runner. He places the cannon at the edge of a 150 m hgh clff. The launch speed of the cannon ball s 0 m/s. How far from the base of the clff should he put the brd feed? Soluton: Physcs 10 Page 7
Example 3: Durng a tenns match, a player serves the ball at 3.6 m/s, wth the center of the ball leavng the racquet horzontally.37 m above the court surface. The net s 1 m away and 0.90 m hgh. When the ball reaches the net, (a) does the ball clear t and (b) what s the dstance between the center of the ball and the top of the net? Suppose that, nstead, the ball s served as before but now t leaves the racquet at 5.00 below the horzontal. When the ball reaches the net, (c) does the ball clear t and (d) what now s the dstance between the center of the ball and the top of the net? Soluton: Physcs 10 Page 8
Example 4: You throw a ball toward a wall at speed 5.0 m/s and at angle θ o = 40.0 above the horzontal. The wall s dstance d =.0 m from the release pont of the ball. (a) How far above the release pont does the ball ht the wall? What are the (b) horzontal and (c) vertcal components of ts velocty as t hts the wall? (d) When t hts, has t passed the hghest pont on ts trajectory? Soluton: Physcs 10 Page 9
Unform Crcular Moton When an object moves n a crcular path at a constant speed, the object s moton s called unform crcular moton. Snce the object s drecton s constantly changng, Ths acceleraton, called the centrpetal acceleraton, ponts toward the center of the crcular path and has a magntude: a c = v r (7) From the dagram (b) above, the average acceleraton s: a avg vf v v = = = a (snce acceleraton s constant) t t The two trangles (b and c) are smlar, therefore: v = r v= v r v r r Combnng these two results, our acceleraton vector s: v r v dr v v a = lm = = = t 0 v r t r dt r r If we look at the trangle (c), we see that the change n velocty vector ponts towards the center of the crcular path. Therefore the acceleraton vector wll pont toward the center of the crcular path. Physcs 10 Page 10
Curved Paths When movng along a curved path, the velocty vector changes n drecton and magntude. The partcle s movng to the rght along a partal crcular path. Ths means that the drecton of the acceleraton vector changes at every pont. How do we know? Snce the partcle s speed s changng, there s a component of the acceleraton vector tangent to the crcle. Ths s called the tangental acceleraton: a t dv = dt (8) The other component of the acceleraton vector s drected along the radus of the curved path. Ths s called the radal acceleraton: a r v = ac = r (9) Snce the radal acceleraton always ponts towards the center of the curved path and the tangental acceleraton always ponts along the drecton of moton, we know that the drecton of the acceleraton vector s always changng. The magntude of our acceleraton vector s: t r a = a + a Physcs 10 Page 11
Example 5: A boy whrls a strng n a horzontal crcle of radus 0.8 m. How many revolutons per mnute does the ball make f the magntude of the centrpetal acceleraton s g? Soluton: Physcs 10 Page 1
Example 6: An automoble whose speed s ncreasng at a rate of m 0.850 travels along a crcular road of radus 15.0 m. When the s nstantaneous speed of the automoble s 15.00 m/s, fnd (a) the tangental acceleraton component, (b) the radal acceleraton component, and (c) the magntude and drecton of the total acceleraton. Soluton: Physcs 10 Page 13
Relatve Moton We are now gong to descrbe how dfferent observatons may be made by observers n dfferent frames of reference. Imagne that you are n an arplane movng at 500 m/hr toward the east, then your velocty would also be 500 m/hr to the east. However, your velocty mght be measured wth respect to the Earth or to the wnd whch s tryng to blow you off course. What s your speed relatve to the arplane? Imagne that a partcle P moves wth velocty v PO as measured by an observer O s a frame S. An observer O n S observes that the velocty of the partcle s v PO. The observer O moves at a velocty v OO wth respect to O. These relatve veloctes are related to each other by: v = v v (10) PO PO O O Example 7: A plane fles at an arspeed of 50 km/h. There s a wnd blowng at 80 km/h n the northeast drecton at exactly 45 o to the east of north. a. In what drecton should the plane head n order to fly due north? b. What s the speed of the plane relatve to the ground? Soluton: Physcs 10 Page 14
Example 8: Two boat landngs are.0 km apart on the same bank of a stream that flows at 1.4 km/h. A motor boat makes the round trp between the two landngs n 50 mn. What s the speed of the boat relatve to the water? Soluton: Physcs 10 Page 15