Physics 2 Lesson 5 Graphical Analysis Acceleraion I. Insananeous Velociy From our previous work wih consan speed and consan velociy, we know ha he slope of a posiion-ime graph is equal o he velociy of he objec. In addiion, he area of a velociy-ime graph is equal o he displacemen. The same principle applies for acceleraed moion, excep now velociy is also changing wih ime. Consider he following graph for an objec undergoing uniform acceleraion. 25 Posiion vs. Time 2 Posiion (m) 15 5 2 4 6 8 12 14 16 We find he insananeous velociy a a given poin by calculaing he slope of he posiion-ime graph a he poin of ineres. To find his slope, we draw a angen line a he poin of ineres (in his case a s). A angen line is a line ha ouches a curve a only one poin and is represenaive of he slope of he curve a ha poin. Once a angen line is drawn, we hen calculae he slope of he line. 25 Posiion vs. Time 2 Posiion (m) 15 d v 11m 8.4s = 1.3 m/s 5 11 m 8.4 s 2 4 6 8 12 14 16 R.L. & A.K. 5-1 217-1-27
Example 1 The following daa was colleced when a ball rolled up an inclined plane, came o a sop and hen rolled back o is saring poin. Plo displacemen-ime, velociy-ime and acceleraion-ime graphs for his moion. displacemen (m) 5 8 9 8 5 ime (s) 1 2 3 4 5 6 A. Displacemen-ime. 8 Posiion (m) 6 4 Posiion vs. Time 2 1 2 3 4 5 6 7 8 B. Velociy-ime. If we calculae insananeous velociies a hree poins (1, 3 and 5 seconds) we can hen plo he poins o make a velociy-ime graph. Posiion vs. Time 1.8 s 8 Posiion (m) 1.8 s 6 +7.6 m 4 7.6 m v 1 v 3 v 5 d 7.6m 1.8s = m/s d 7.6m 1.8s = +4.2 m/s = - 4.2 m/s 2 1 2 3 4 5 6 7 8 R.L. & A.K. 5-2 217-1-27
v (m/s) +4.2-4.2 (s) 1 3 5 +4 +2-2 - m/s -4-6 4.6 s 1 2 3 4 5 6 7 8 C. Acceleraion-ime. A sraigh line velociy-ime graph indicaes consan acceleraion. Recall ha acceleraion is he change in velociy wih ime i.e. acceleraion is he slope of a velociy-ime graph. v a For he velociy-ime graph above v m / s a = -2.2 m/s 2 4.6s Acceleraion vs. Time +2 +1 Acceleraion (m/s 2 ) -1-2 -3 1 2 3 4 5 6 7 8 R.L. & A.K. 5-3 217-1-27
II. Displacemen from Velociy Example 2 Given he following velociy-ime graph, calculae he oal displacemen from o 8 seconds. + +5-5 - -15 1 2 3 4 5 6 7 8 Displacemen equals he area of he velociy-ime graph. The key is o calculae he correc areas beween he graph and he = line. Noe ha he area of a riangle is d = ½v. + +5 d = v d =+m/s(2s) d = +2 m d = ½v d = ½(+ m/s)(2 s) d = + m -5 - d = ½v d = ½(-5 m/s)(1s) d = 2.5 m d = v = 5 m/s (3 s) d = 15 m Noe ha areas can be posiive or negaive depending on he direcion of moion. -15 1 2 3 4 5 6 7 8 To calculae he oal displacemen we sum up he individual displacemen areas. d = (+2 m) + (+ m) + (-2.5 m) + (-15 m) = +12.5 m R.L. & A.K. 5-4 217-1-27
III. Overview of Graphical Analysis In our sudy of graphical analysis we sared wih consan speed graphs. We hen exended he principles learned here ino consan velociy moion and finally ino acceleraed moion. We have learned wha he following graph shapes mean and how o properly inerpre hem. d d d saionary objec posiive velociy negaive velociy posiive acceleraion negaive acceleraion v v v a consan (+) velociy consan (-) velociy posiive acceleraion saring from v = negaive acceleraion saring from v = negaive acceleraion saring from a posiive velociy, he objec evenually sops and sars moving in he negaive direcion consan (+) acceleraion consan (-) acceleraion R.L. & A.K. 5-5 217-1-27
IV. Reference o Pearson There is an excellen discussion and more examples in pages 31 o 44 in Pearson. V. Pracice Problems 1. (+) (+) v a (-) (-) 2. (+) (+) d v (-) (-) 3. (+) (+) v a (-) (-) R.L. & A.K. 5-6 217-1-27
4. Given he posiion-ime graph below, 25 Posiion vs. Time 2 Posiion (m) 15 5 1 2 3 4 5 6 7 8 A. Wha is he insananeous velociy a 5. s? (~ -2.4 m/s) B. How far did he objec ravel from is origin afer 6. s? (-8.5 m) C. Is he acceleraion posiive or negaive? Explain. (negaive) R.L. & A.K. 5-7 217-1-27
5. Given he velociy-ime graph below, +15 + +5-5 - 2 4 6 8 12 14 16 A. Wha is he insananeous velociy a 8. s? (+ m/s) B. Wha is he acceleraion from 8 s o 14 s? (-2.5 m/s 2 ) C. Wha is he displacemen from 8 s o 16 s? (+5. m) R.L. & A.K. 5-8 217-1-27
VI. Hand-In Assignmen 1. For each of he following graphs, describe he moion involved and hen skech he missing graph. A. v a B. d v C. d v R.L. & A.K. 5-9 217-1-27
D. d a E. d v F. v a G. a d R.L. & A.K. 5-217-1-27
2. Consider he following posiion-ime graph. 25 Posiion vs. Time 2 Posiion (m) 15 5 5 15 2 25 3 35 4 A. Wha is he insananeous velociy a 15 s? (~ +.34 m/s) B. How far would he objec ravel from 5 s o 2 s? (4.1 m) C. Wha was he average speed for he inerval from s o 25 s? (.32 m/s) R.L. & A.K. 5-11 217-1-27
3. The following is a graph for an objec moving norh. 8 6 4 2 2 4 6 8 12 14 16 A. Wha is he acceleraion of he objec? (+.6 m/s 2 ) B. Wha is he displacemen from s o s? (+7 m) 4. Given he following velociy-ime graph. + +5-5 - -15 5 15 2 25 3 35 4 A. Wha was he acceleraion a 5 s and a 2 s? (, -1.7 m/s 2 ) B. Wha is he displacemen from s o 3 s? (+6 m) R.L. & A.K. 5-12 217-1-27
5. The following graph shows he moion of wo objecs raveling eas. 5 4 3 A 2 B 2 4 6 8 12 14 16 A. How much faser is objec A raveling a 8 s han objec B? (6. m/s) B. Wha is he acceleraion of each objec? (+2.7 m/s 2, +1.2 m/s 2 ) C. From s o 12 s, which objec ravelled he furhes? (A) R.L. & A.K. 5-13 217-1-27
6. Using he graph of he moion of a car ravelling in a sraigh line, deermine each of he following. a. he velociy of he car in each inerval. (+5 km/h,, -33 km/h) b. he final displacemen of he car. () Posiion vs. Time 25 2 Posiion (km) 15 5.2.4.6.8 1. 1.2 1.4 1.6 Time (h) R.L. & A.K. 5-14 217-1-27
*7. A ball rolls along he floor, up an inclined plane, and hen back down he plane and across he floor again. The graph below represens his moion. a. A wha ime is he ball a is highes poin? b. Wha is is maximum displacemen up he ramp? (1. m) c. Wha was he acceleraion when he ball was (i) rolling up he ramp, (ii) rolling down he ramp, and (iii) when he ball was insananeously a res a he op of he ramp? +1. +.5 -.5-1. -1.5 2 4 6 8 R.L. & A.K. 5-15 217-1-27