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

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TAM212 Introductory Dynamics Spring 2005 S. Balachandar Homework 4 (1.45, 1.53) 1.45 (a) We are given acceleration as, a = ¨ x ˆ i = 12 ˆ it m / s 2 Integrating to get velocity v = ˙ x ˆ i = (6 t 2 + C 1 ) ˆ i m / s Using, ˙ x ˆ i = 2 ˆ 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 = 0 s to t = 5 s x | t = 5 - x | t = 0 = (250 - 20 - 5) - (0 - 0 - 5) = 230 m r 5 - r 0 = 230 ˆ i m (b) The displacement vector from t = 0 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. ˙ x = 0 6 t 2 - 4 = 0 t = r 2 3 s 1

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So the total distance traveled is x Total = x | t = 2 3 - x | t = 0 + x | t = 5 - x | t = 2 3 = |- 7 . 18 - ( - 5) | + | 225 - ( - 7 . 18) | = 2 . 18 + 232 = 234 m 1.53 The speeder is going at a speed ˙ x s = 70 mph = 70(88 / 60) ft/s = 103 ft/s
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