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hw2solution - Homework 2 Solution CHANG 1 From the...

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Homework 2 – Solution CHANG 1 From the positions of the two particles we first obtain the velocities and accelerations for A and B for 0 t 10 s: x A t t 4 6 t 3 2 2 t 2 8 m A t 2 3 t 3 3 2 t 2 4 t m/s a A t 2 t 2 3 t 4 m/s 2 x B t t 3 3 t 2 t 1 m B t t 2 2 t 1 m/s a B t 2 t 2 m/s 2 (a) For t 1 3 s and t 2 6 s the corresponding velocities are 1 A t 3 2/3 3 3 3/2 3 2 4 3 7.5 m/s 2 A t 6 2/3 6 3 3/2 6 2 4 6 66 m/s So the average acceleration of A can be calculated to be a ave 2 1 t 2 t 1 66 7.5 6 3 24.5 m/s 2 (b) Set the accelerations of the particles equal to each other, a A a B , to find the time instant: 2 t 2 3 t 4 2 t 2 2 t 2 t 6 0 t 1.5 (discard), 2 At t 2 s, the positions of particles A and B are x A 1/6 2 4 1/2 2 3 2 2 2 8 4/3 m x B 1/3 2 3 2 2 2 1 5/3 m The distance between them is the difference in positions, and thus d | x A x B | 1/3 m. Note the absolute value makes sure there is no negative distance. CHANG 2 For the given position x t t 3 6.5 t 2 30 t 10 m for the duration 0 t 10 s, the velocity and acceleration are: t dx dt 3 t 2 13 t 30 m/s a t d dt 6 t 13 m/s 2 (a) When the particle is farthest away from the origin, it is at the extreme position on the path, and that implies the particle is momentarily at rest to change the direction of motion. So we set the velocity equal to zero to find the time instant: 0 3 t 2 13 t 30 0 t 6, 5/3 (discard) The position and acceleration can then be calculated corresponding to the time instant t 6 s: x t 6 6 3 6.5 6 2 30 6 10 188 m (This position is indeed in the x direction) a t 6 6 6 13 23 m/s 2 (b) We first need to know the starting and final locations of the particle during the time span: x 1 x t 4 4 3 6.5 4 2 30 4 10 150 m x 2 x t 8 8 3 6.5 8 2 30 8 10
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