A particle undergoing constant acceleration has an x position given by x = x0 +

Question

A particle undergoing constant acceleration has an x position given by x = x0 + V0t + 1 / 2at2 where x0 = 1.0 m, v0 = 2.0 m / s, and alpha = 2.2 m / s2. Here is the graph of the particle's position versus time: On the graph sketch the tangents to the line at times t of 0 s, 1 s, and 2 s. Take the derivative of the equation of motion with respect to time. Evaluate the result for times of 0 s, 1 s, and 2 s. What are these numbers? How do they relate to the tangents you have drawn? Graph the acceleration versus time on the following axes: What are the slopes of the tangents to the line at times 0 s, 1 s, and 2 s? What are these numbers?

Explanation / Answer

slope of displacement time graph gives velocity at that instanthere v=v0+at acceleration is a so at curve is a st line parallel to X axisusually slope of at graph gives jerk here it is 0

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