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Unformatted text preview: Chapter 2 Kinematics of Particles 2.1 Some other stuff 2.2 Cartesian Coordinates 89 2.2.1 GOAL: Plot the path of someone following a map’s directions. DRAW: Figure 1: Paths from map SOLVE: Figure 1 shows the two cases. ( a ) shows the map’s original directions and ( b ) shows the overall change in position. ( c ) shows what we get by reversing the directions and ( d ) shows the starting position to finishing position vector. As can be seen from the figure, the overall start to finish vector is the same in each case. This illustrates how the order in which vector components are added won’t affect the final answer. 90 2.2.2 GOAL: Using the given information, find average speed and approximate instantaneous speed. GIVEN: Graph of position vs time. DRAW: FORMULATE EQUATIONS: We’ll use the graph and divide the distance traveled by the time taken in order to estimate the average speed and estimate the slope at a point in order to estimate the instantaneous speed. SOLVE: You can see from the graph of displacement versus time that the particle travels 8 meters in 20 seconds. Thus, v = 8 20 m/s = 0.4 m/s . The maximum slope occurs at t = 10 seconds. Estimating the slope with a straightedge pro vides an estimate for the maximum speed of 1 m/s . The minimum slope occurs at t = 0 and t = 20 seconds. Estimating this slope provides you with an estimate for the minumum speed of 0.1 m/s . 91 2.2.3 GOAL: Find the position of a particle at t = 3 seconds given a plot of the particle’s acceleration. GIVEN: Plot of acceleration vs. time. DRAW: FORMULATE EQUATIONS: We’ll need to integrate the acceleration to find speed and then integrate the speed to find position. SOLVE: The plot of acceleration versus time can be divided into three distinct regions on the time interval ≤ t ≤ 3 . For the region where 0 ≤ t ≤ 1, the acceleration is given by ¨ s = 10 t Integrating for velocity gives ˙ s ( t ) = 5 t 2 + v (1) Integrating again for position provides s ( t ) = 5 3 t 3 + v + s (2) Using the initial condition v = 0 in (1) and evaluating at t = 1 gives ˙ s (1) = 5. Using the initial conditions s = 1 and v = 0 in (2) yields s (1) = 8 3 . For 1 < t ≤ 2 the acceleration is constant; ¨ s = 10 Reinitialize the time counter to begin at t = 1, treating this segment as if it’s simply another initial condition problem for which the initial conditions start at t = 1. Integrating for velocity gives ˙ s ( t ) = 10 t + v 1 (3) Integrating again for position gives s ( t ) = 5 t 2 + v 1 t + s 1 (4) Applying the conditions found over the first time interval into these equations, we find ˙ s (2) = 15 and that s (2) = 38 3 . For the final interval, 2 < t ≤ 3, the acceleration equals 0, so ˙ s (3) = v 2 and s (3) = ˙ s (2) t + s 2 ....
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 Spring '06
 KAM
 Acceleration, Velocity, Sin, Cos

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