Three-stage motion (s, v, a graphs)

Problem

An object moves along a straight line with a position-time law represented qualitatively by three stages: (1) $0 \leq t \leq 2\;\mathrm{s}$, it accelerates from rest with $a = 3\;\mathrm{m/s}^2$; (2) $2 < t \leq 5\;\mathrm{s}$, it moves at constant speed; (3) $5 < t \leq 7\;\mathrm{s}$, it decelerates uniformly until stopping. Draw the graphs $s(t)$, $v(t)$, $a(t)$ and determine the maximum speed and the total distance covered.

Solution

Stage 1 (acceleration, $0 \leq t \leq 2\;\mathrm{s}$). The body starts from rest with constant acceleration $a_1 = 3\;\mathrm{m/s}^2$. At the end of the stage the speed is:

$v_\text{max} = a_1 \, t_1 = 3 \cdot 2 = 6\;\mathrm{m/s}$

This is the maximum speed reached in the whole motion. The distance covered is:

$s_1 = \tfrac{1}{2}\,a_1\,t_1^2 = \tfrac{1}{2}\cdot 3 \cdot 2^2 = 6\;\mathrm{m}$

Stage 2 (constant speed, $2 < t \leq 5\;\mathrm{s}$). The speed stays at $v_\text{max} = 6\;\mathrm{m/s}$ for a duration $\Delta t_2 = 3\;\mathrm{s}$, so the acceleration is zero and:

$s_2 = v_\text{max}\,\Delta t_2 = 6 \cdot 3 = 18\;\mathrm{m}$

Stage 3 (deceleration, $5 < t \leq 7\;\mathrm{s}$). The body goes from $6\;\mathrm{m/s}$ to $0$ in $\Delta t_3 = 2\;\mathrm{s}$, so the deceleration is $a_3 = -6/2 = -3\;\mathrm{m/s}^2$. The distance covered, the area of the triangle under $v(t)$, is:

$s_3 = \tfrac{1}{2}\,v_\text{max}\,\Delta t_3 = \tfrac{1}{2}\cdot 6 \cdot 2 = 6\;\mathrm{m}$

Shape of the graphs.

• $a(t)$: constant and positive ($+3\;\mathrm{m/s}^2$) from $0$ to $2\;\mathrm{s}$; zero from $2$ to $5\;\mathrm{s}$; constant and negative ($-3\;\mathrm{m/s}^2$) from $5$ to $7\;\mathrm{s}$. It is a step function.
• $v(t)$: rising straight line from $0$ to $6\;\mathrm{m/s}$ in the first stage; horizontal segment at $6\;\mathrm{m/s}$ in the second; falling straight line from $6\;\mathrm{m/s}$ to $0$ in the third. It has the shape of a trapezium.
• $s(t)$: an upward-concave parabola in the first stage; an oblique straight line (constant slope) in the second; a downward-concave parabola in the third, reaching the final value with a horizontal tangent.

Total distance. Adding the three contributions (equivalent to the area of the trapezium under $v(t)$):

$s_\text{tot} = s_1 + s_2 + s_3 = 6 + 18 + 6 = 30\;\mathrm{m}$

$\ev{v_\text{max} = 6\;\mathrm{m/s}, \quad s_\text{tot} = 30\;\mathrm{m}}$

Links

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