PHY183-Lecture5

# 6 k m 3 6 5 2 k m p u t o r ig in o f c o o r d in a

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Unformatted text preview: = > to ta l d is ta n c e fo r th e r o u n d tr ip is 2 * 1 8 2 .6 k m , 3 6 5 .2 k m . • P u t o r ig in o f c o o r d in a te s y s te m in to D e s M o in e s ( d o e s it m a tte r where we put it?) => xD= 0 km • Position vector for Iowa City then has the value xI=+182.6 km. • D is p la c e m e n t v e c to r fo r g o in g fr o m D e s M o in e s to Io w a C ity V e lo c it y V e c t o r x ! Average velocity • Displacement divided by time interval for the displacement to take place !x Units: m!s-1 v!t = !t • Displacement: ! (Instantaneous) velocity • Obtained in the limit that the time interval for the averaging procedure approaches 0 v(t ) = lim v = lim !t " 0 !t " 0 !x ID = x I " x D = x I " 0 = +182.6 km • D is p la c e m e n t v e c to r fo r r e tu r n tr ip • T o ta l d is p la c e m e n t fo r r e tu r n tr ip is th e s u m o f b o th d is p la c e m e n ts !x DI = x D " x I = 0 " x I = "182.6 km !x dx # !t dt !xtotal = !x ID + !x DI = 182.6 km " 182.6 km = 0 August 31, 2006 Physics for Scientists&Engineers 1 9 • We need deriva...
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