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# Position - Position Velocity and Acceleration What makes...

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Position, Velocity, and Acceleration What makes the generalization to vectors particularly simple is that the relationships between position, velocity, and acceleration stay exactly the same. Whereas before we had v ( t ) = x' ( t ) and a ( t ) = v' ( t ) = x'' ( t ) now we have v ( t ) = x â≤ ( t ) and a ( t ) = v â≤ ( t ) = x â≤â≤ ( t ). where the derivatives are taken component by component. In other words, if x ( t ) = ( x ( t ), y ( t ), z ( t )) , then x â≤ ( t ) = ( x' ( t ), y' ( t ), z' ( t )) . Therefore, all the equations derived in the previous section are valid once the scalar-valued functions are turned into vector-valued ones. As an example, consider the position function x ( t ) = a t 2 + v 0 t + x 0 , where a = (0, 0, - g ) , v 0 = ( v x , 0, v z ) , and x 0 = (0, 0, h ) . The above vector equation for position can be broken down into three one-dimensional equations: x ( t ) = v x t , y ( t ) = 0, z ( t ) = - gt 2 + v z t + h The motion in the x -direction is of constant velocity, the motion in the y -direction is non- existent (so really this is a two-dimensional problem), and the motion in the z -direction looks like that of an object moving up and down near the surface of the earth (recall that
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