Physics 221 Fall 2008 Homework #2 Solutions Ch. 2 Due Tues, Sept 9, 2008

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1 Physics 221 Fall 28 Homework #2 Soluions Ch. 2 Due Tues, Sep 9, A paricle moving along he x-axis moves direcly from posiion x =. m a ime =. s o posiion x = 1. m by ime = 1. s, and hen moves direcly o posiion x =. m by ime = 2. s. Wha is he average velociy of he paricle (a) beween =. s and = 1. s? = 1. m = 1. m/s 1. s (b) beween = 1. s and = 2. s? = 1. m = 1. m/s 1. s (c) beween =. s and = 2. s? =. m =. m/s 2. s The average speed of a paricle over some ime inerval is he oal disance D raveled over ha ime inerval divided by he ime inerval. Disance is always posiive or zero; i is wha your odomeer reads on your car. Wha is he average speed of he paricle (d) beween =. s and = 1. s? = D = 1. m 1. s = 1. m/s (e) beween = 1. s and = 2. s? = D = 1. m 1. s = 1. m/s (f) beween =. s and = 2. s? = D = 2. m 2. s = 1. m/s 1

2 (g) Compare he resuls in pars (a)-(c) wih he respecive resuls in (d)-(f). Explain any differences. The answers in (a) and (d) are he same because he velociy and speed are boh posiive during ha ime inerval. The answers in (b) and (c) are differen from hose in (e) and (f), respecively, because of he difference in he definiions of average speed and average velociy. 2.2 A paricle moves along he x-axis. The iniial posiion x and iniial velociy v x of he paricle a ime = are boh zero. The paricle acceleraes wih an (insananeous) acceleraion given by a x = 3. m/s 2 (1. m/s 3 ) where is he ime in seconds. (a) Plo a x versus for = o = 1. s. Use a spreadshee program if you can o make his and he following plos. The hree plos for his problem were made using he program KaleidaGraph on a Macinosh. All hree requesed plos are made on he same figure below. The numerical verical scale applies o all hree plos. a x, v x, x 2 Problem x (m) a (m/s 2 ) x -1 v x (m/s) ime (s) (b) Derive an expression for he velociy v x versus. Plo v x versus for = o = 1. s. v x = v x + a x () d = + [(3. m/s 2 ) (1. m/s 3 )] d = (3. m/s 2 ) (.5 m/s 3 ) 2 (c) Derive an expression for he posiion x versus. Plo x versus for = o = 1. s. x = x + v x () d = + [(3. m/s 2 ) (.5 m/s 3 ) 2 ] d = (1.5 m/s 2 ) 2 (.167 m/s 3 ) 3 2

3 (d) Wha are he average velociy and he average acceleraion of he paricle over he ime inerval from = o = 1. s? The iniial velociy and posiion are boh zero. From pars (c) and (b), he posiion a = 1. s is x = 17 m and he velociy a = 1. s is v x = 2 m/s, so v x-av = 17 m = 1.7 m/s and a x-av = v x 1 s = 2 m/s 1 s = 2. m/s A paricle is moving along he x-axis. The posiion x of he paricle versus ime is ploed in he figure Problem x (m) ime (s) (a) A wha insans or inervals of ime is he velociy v x posiive? negaive? zero? Explain. The (insananeous) velociy is dx/d, which is he slope of he curve. Thus v x is posiive for 1. s < <. s, is negaive for. s < < 1. s, and is zero for = 1. s,. s, and 1. s. (b) A wha insans or inervals of ime is he acceleraion a x posiive? negaive? zero? Explain. The (insananeous) acceleraion is dv x /d, which is he curvaure d 2 x/d 2. The curvaure is posiive if x() is bending upwards, is negaive if x() is bending downwards, and is zero a an inflecion poin. Thus a x is posiive for 1. s <.5 s and for.5 s < 1. s, is negaive for.5 s < <.5 s, and is zero for =.5 s and.5 s. 3

4 2.4 A person hrows a ball verically upward, in he posiive y-direcion, wih an iniial speed of 29.4 m/s. The release poin is a y =. Afer release, he (consan) acceleraion of he ball due o graviy is a y = 9.8 m/s 2. The verical posiion of he ball versus ime is designaed as y(). In his problem ignore he possible influence of air resisance on he moion of he ball. (a) Wrie down an expression for he posiion y of he ball versus ime. This is a consan acceleraion problem, for which y = y + v y a y 2. Here, y =, v y = 29.4 m/s, and a y = 9.8 m/s 2 as saed in he problem. (b) Wrie down an expression for he velociy of he ball versus ime. How long does i ake for he ball o reach is maximum heigh? A he op of he rajecory, he ball is momenarily a res. We have v y = v y + a y. Seing v y = gives he ime = v y 29.4 m/s = a y 9.8 m/s = 3. s. 2 (c) How high does he ball go before i sars coming back down? From pars (a) and (b), we have y = y + v y a y 2 = + (29.4 m/s)(3. s) + ( 4.9 m/s 2 )(3. s) 2 = 44.1 m. (d) How long does i ake for he ball o reach y = again afer i is launched? Over wha ime inerval are he signs of v y and a y he same? Over wha ime inerval are he signs of v y and a y opposie o each oher? Wha are he velociy and speed of he ball when he ball reurns o y =? Using y = y + v y a y 2 from par (a), and seing y = y =, solving for he (nonzero) ime gives = 2v y a y = 4 2(29.4 m/s) 9.8 m/s 2 = 6. s. Noe ha his ime is wice he ime for he ball o reach is maximum heigh. Thus he ime for he ball o go up is he same as he ime for he ball o come down. During he ime inerval from = o = 3. s when he ball is going up, v y and a y have opposie signs: v y is posiive and a y is negaive. From = 3. s o = 6. s when he ball is coming down, v y and a y have he same negaive sign. To deermine he velociy when he ball reurns o y =, use he informaion ha he ball has zero velociy a he maximum heigh, v y =, and i akes 3. s o reurn o y = from here: v y = v y + a y = + ( 9.8 m/s 2 )(3. s) = 29.4 m/s.

5 Alernaively, one can use he ime = 6. s ha he ball akes o go up and come down: v y = v y + a y = 29.4 m/s + ( 9.8 m/s 2 )(6. s) = 29.4 m/s. Thus he speed a which he ball reurns o y = is v y = 29.4 m/s. This speed is he same as he launch speed, which we will see laer is a consequence of he law of conservaion of energy. (e) Plo y, v y and a y versus ime over he ime inerval for which he ball is a or above y =. Use a spreadshee program o make hese plos if you can. All hree plos are placed on he same figure as follows. The requesed ime inerval is 6. s. The plos were made using he program KaleidaGraph on a Macinosh. The numerical scale on he verical axis applies o all hree plos. y, v y, a y 5 Problem 2.4(e) 4 3 y (m) 2 1 v (m/s) y -1-2 a y (m/s 2 ) ime (s) Noe ha he velociy goes o zero a he op of he rajecory, he iniial and final speeds are he same, he velociy is zero and changes sign a he op of he rajecory, and he acceleraion is consan and negaive. 5

MEI Mechanics 1 General motion. Section 1: Using calculus

MEI Mechanics 1 General motion. Section 1: Using calculus Soluions o Exercise MEI Mechanics General moion Secion : Using calculus. s 4 v a 6 4 4 When =, v 4 a 6 4 6. (i) When = 0, s = -, so he iniial displacemen = - m. s v 4 When = 0, v = so he iniial velociy