Kinemtics equtions, some numbers Kinemtics equtions: x = x 0 + v 0 t + 1 2 t2, v = v 0 + t. They describe motion with constnt ccelertion. Brking exmple, = 1m/s. Initil: x 0 = 10m, v 0 = 10m/s. x(t=1s) = 10m + 10m (1s) + 1 2 ( 1m/s2 )(1s) 2 = 19.5m t = 2s : x = 10m + 10m 2 m = 28 m t = 5s : x = 10m + 50m 12.5m = 47.5m t = 10s : x = 10m + 100m 50 m = 60 m t = 15s : x = 10m + 150m 112.5m = 47.5m t = 20s : x = 10m + 200m 200 m = 10 m (huh??) A problem... the distnce should not strt decresing! Why is this mth vs. physics. A very importnt point. c2. Kinemtics equtions exmples., Ph 211. Apr 02, 2010. 1/9
For this, we cn grph the function / dt 70 Brking, with rection. 60 50 position [m] 40 30 20 10 0 2 4 6 8 10 12 14 16 time [s] c2. Kinemtics equtions exmples., Ph 211. Apr 02, 2010. 2/9
Problem nlysis / understnding (of mth & physics) The cr will stop, nd our eqution doesn t mke sense nymore. Alwys need to keep sense of our physics, when using mth. (Try the bove with different numbers initil speed, ccelertion... ) So...how fr does the cr get? How long until it stops? Slowly, this gets us into solving problems. Not just n eqution to simply put numbers in. Our equtions tools to work with: x = x 0 + v 0 t + 1 2 t2, v = v 0 + t c2. Kinemtics equtions exmples., Ph 211. Apr 02, 2010. 3/9
Stopping... Reference: x = x 0 + v 0 t + 1 2 t2, v = v 0 + t. The key observtion : the cr stops...nd so v = 0. The cr is t this prticulr spot t specific moment in time. Let me cll them... x s nd t s (for stop). Then v(t stop) = 0 0 = v 0 + t s, nd we cn directly solve for it from the bove eqution! (Importnt: t s is number, specific vlue for time.) 0 = v 0 + t s t s = v 0 Note, the ccelertion itself ws negtive. c2. Kinemtics equtions exmples., Ph 211. Apr 02, 2010. 4/9
... distnce. So we got the time it took to stop, t s = v 0 = 10m/s ( 1m/s 2 ) = 10s With time to stop, we cn go bck to the other eqution, x(t). Enter this vlue for time, nd get the position t tht time position when stopped thus. This is the distnce we wnted. (Given our coordinte system.) At time t = 10s, the position is, from x = x 0 + v 0 t + 1 2 t2 : x(t=10s) = 10m + (10m/s)(10s) + 1 2 ( 1m/s2 )(10s) 2 x s = 60m We got our distnce nd we got the time in the process. c2. Kinemtics equtions exmples., Ph 211. Apr 02, 2010. 5/9
Cn do better... How bout some other vlues? For different? Or, other v 0? Would hve to reclculte, go bck to the beginning. Or... not? Wht we got ws formul for time: t s = v 0 We then put this time into the x(t) eqution but s number. Why? We don t hve to ruin perfectly good formul! Keep the symbol: with the bove expression for t s ( x s = x 0 + v 0 t s + 1 2 t2 s x s = x 0 + v 0 v 0 ) + 1 2 ( v 0 ) 2 Crry out the lgebr... x s = x 0 1 2 v 2 0. ( is negtive!) c2. Kinemtics equtions exmples., Ph 211. Apr 02, 2010. 6/9
... nd better yet! Reference: x = x 0 + v 0 t + 1 2 t2, v = v 0 + t. The method we used: find time, substitute into position eqution. Cn we do something like tht lwys? In generl? Yes! And tht wy we ll get something relly useful. So, the quest: eliminte time from the equtions of motion. This is generl method for solving systems of equtions. It mens: solve for it from one eqution (sy, the second). Then use this expression insted of it in the other (the first). solve for time v = v 0 + t t = v v 0, nd ( ) ( ) v v0 v 2 x = x 0 + v 0 + 1 2 v0 note: time s gone! c2. Kinemtics equtions exmples., Ph 211. Apr 02, 2010. 7/9
Finlly Clering up tht lgebr (excellent exercise), we get x = x 0 1 2 (v2 0 v2 ), or: 2 (x x 0 ) = v 2 v 2 0 It is convenient to write this s v 2 = v 2 0 + 2 (x x 0 ) (x x 0 ) is simply displcement. It is often lbeled x. With this, (x x 0 ) = x, we hve equtions of kinemtics: x = x 0 + v 0 t + 1 2 t2 v = v 0 + t v 2 = v 2 0 + 2 x ( x = x x 0) Remember, the third eqution is combintion of the first two. We got it directly from them, by eliminting time. c2. Kinemtics equtions exmples., Ph 211. Apr 02, 2010. 8/9
Comments All equtions we ll need for weeks! The rest...? Problems. Relevnt mteril in the book: Ch 2 section 4, nd up to it. Fill-in on the reding. There is lot, but most of it is so bsic tht one just must hve red tht. So do it now, skip if stuck. A cr goes 60mi/hr when it strts brking t the rte of 5m/s 2. Stopping distnce? Wht is it with hrsher brking? Sy, twice hrsher? Five times? Remember the trouble with shrinking distnces in our first exmple for brking with rection...? Cn you come up with n exmple ( physicl system sitution ) tht will be described correctly nd fully by the eqution we studied! (So tht there is no need to sy only up to tht point...) c2. Kinemtics equtions exmples., Ph 211. Apr 02, 2010. 9/9