Problem — Triangular acceleration pulse

Problem

Graph reading: $a(t)$ in an impact. An $a(t)$ graph during a collision has the shape of a triangular pulse, with $a_\text{max} = 50\;\text{m/s}^2$ and duration $0{,}05\;\text{s}$. Calculate the change in velocity of the struck object.

Solution

Principle. In an $a(t)$ graph the change in velocity is the area under the curve:

$\Delta v = \int a\,dt = \text{area under } a(t)$

Area of the triangle. The pulse is a triangle with base equal to the duration $\Delta t = 0{,}05\;\text{s}$ and height $a_\text{max} = 50\;\text{m/s}^2$:

$\Delta v = \frac{1}{2}\cdot \text{base}\cdot \text{height} = \frac{1}{2}\cdot 0{,}05 \cdot 50$

$\ev{\Delta v = 1{,}25\;\mathrm{m/s}}$

Although the peak acceleration is enormous ($50\;\text{m/s}^2$, over $5\,g$), the very short duration of the impact means the velocity changes by only $1{,}25\;\text{m/s}$: what matters for the velocity change is the area, not the height of the peak.

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Topics: Cinematica Concepts: Accelerazione Skills: Lettura dei grafici