Problem — Acceleration from the v(t) graph

Problem

Inverse: given the graph, find $a$. The $v(t)$ graph is a straight line starting from the origin and reaching $40\;\text{m/s}$ in $8\;\text{s}$. Calculate the acceleration and the distance covered.

Solution

Acceleration from the slope. In a $v(t)$ graph the acceleration is the slope of the line, i.e. the ratio between the change in speed and the time interval:

$a = \frac{\Delta v}{\Delta t} = \frac{40 - 0}{8 - 0}$

$\ev{a = 5{,}0\;\mathrm{m/s}^2}$

Distance from the area. The distance covered is the area under the $v(t)$ graph. The region is a triangle with base $8\;\text{s}$ and height $40\;\text{m/s}$:

$x = \frac{1}{2}\cdot \text{base}\cdot \text{height} = \frac{1}{2}\cdot 8 \cdot 40$

$\ev{x = 160\;\mathrm{m}}$

The two readings of the graph are complementary: the slope gives the acceleration, the area gives the distance.

Links

Topics: Cinematica Concepts: Moto uniformemente accelerato Skills: Lettura dei grafici