hw4 - TAM212 Introductory Dynamics Spring 2005 S....

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Unformatted text preview: TAM212 Introductory Dynamics Spring 2005 S. Balachandar Homework 4 (1.45, 1.53) 1.45 (a) We are given acceleration as, a = x i = 12 it m / s 2 Integrating to get velocity v = x i = (6 t 2 + C 1 ) i m / s Using, x i = 2 i m / s at t = 1 s , 2 = 6 + C 1 C 1 =- 4 x = 6 t 2- 4 Integrating, once again, we get the displacement x = 2 t 3- 4 t + C 2 3 = 16- 8 + C 2 , using x = 3 at t = 2 s C 2 =- 5 x = 2 t 3- 4 t- 5 Now, to find the displacement vector from t = s to t = 5 s x | t = 5- x | t = = (250- 20- 5)- (0-- 5) = 230 m r 5- r = 230 i m (b) The displacement vector from t = s to t = 5 s by itself will not give us sufficient information as to the total distance traveled (Hint: the particle could have reversed its direction during this time ) We employ the fact that particle is in rectilinear motion and try and find the time at which the velocity becomes 0....
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hw4 - TAM212 Introductory Dynamics Spring 2005 S....

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