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Introductory Physics Homework Help
Position from velocity time graph
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[QUOTE="Charles Link, post: 6049924, member: 583509"] These are definite integrals. See (3 lines) below. ## \\ ## The correct expression for ## v_x ## is ## v_x(t)=-4+4t ##.## \\ ## (You can write ## \frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1} ##, where you use any two points ## (x_1,y_1) ## and ## (x_2,y_2) ##. In this case ## y ## is ## v ##, and ## x ## is ## t ##). ## \\ ## The distance ## \Delta x ## it goes is ## \Delta x=\int\limits_{0}^{2} v_x(t) \, dt ##, (as previously mentioned, it's a definite integral, i.e. one that has its limits posted in the expression, and thereby there is no constant of integration to add to the result like there is with an indefinite integral). ## \\ ## Finally, ## x_{final}=x_o +\Delta x ##. ## \\ ## I'll let you perform the integral. ## \\ ## You could also write the integral as ## \Delta x=\int\limits_{0}^{1} v_x(t) \, dt +\int\limits_{1}^{2} v_x(t) \, dt ## if you want to sum these two parts separately. [/QUOTE]
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Position from velocity time graph
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