Problem — Reading a three-phase x(t) graph

Problem

Graph reading: $x(t)$ in three phases. An $x(t)$ graph has three phases: a rising linear ramp for $5\;\text{s}$, then flat for $5\;\text{s}$, then a falling linear ramp for $5\;\text{s}$. Sketch $v(t)$ qualitatively.

Solution

The key is that in an $x(t)$ graph velocity is the slope: where $x$ rises $v$ is positive, where $x$ is flat $v$ is zero, where $x$ falls $v$ is negative. Let’s analyse the three phases.

Phase 1 ($0$–$5\;\text{s}$). The rising ramp has constant, positive slope: the velocity is constant and positive. From the graph $x$ rises by $2$ units in $5\;\text{s}$, so $v = +0{,}4\;\text{m/s}$ (the exact value depends on the scale; what matters is that it is constant and positive).

Phase 2 ($5$–$10\;\text{s}$). The flat stretch has zero slope: the object is stationary, $v = 0$.

Phase 3 ($10$–$15\;\text{s}$). The falling ramp has constant, negative slope, equal in magnitude to that of phase 1: the velocity is constant and negative, $v = -0{,}4\;\text{m/s}$ (the object moves back).

Resulting $v(t)$ graph: a positive step for the first $5\;\text{s}$, then zero for the next $5\;\text{s}$, then a symmetric negative step for the last $5\;\text{s}$. A step function, with vertical jumps at the points where the $x(t)$ graph changes slope abruptly.

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Topics: Cinematica Skills: Lettura dei grafici