The cat in two phases

Problem

A cat at time $t = 5\;\text{s}$ has velocity $7\;\text{m/s}$. It accelerates to a velocity $V$ taking $3\;\text{s}$ and $10\;\text{m}$, then stops decelerating with $a = -1\;\text{m/s}^2$. Determine all the events.

Solution

Setup. We choose the axis oriented along the direction of motion, with the origin of positions at the point where the cat is at the initial instant ($s_A = 0$). The motion has two phases and three events (A, B, C). Let’s summarise the known data in the event table:

Event	$t$ (s)	$s$ (m)	$v$ (m/s)	$a$ (m/s²)
A	$5$	$0$	$7$
B	$8$	$10$	$v_B$
C	$t_C$	$s_C$	$0$	$-1$

Leg A$\to$B. In this leg the average velocity coincides with the arithmetic mean of the velocities (uniformly accelerated motion). From $\dfrac{s_B - s_A}{t_B - t_A} = \dfrac{v_A + v_B}{2}$:

$\frac{10}{3} = \frac{7 + v_B}{2} \quad\Rightarrow\quad v_B = \frac{20}{3} - 7 = -\frac{1}{3}\;\text{m/s}$

The negative result shows that the cat was in fact slowing down: the velocity at B is directed opposite to the initial motion. The acceleration in this leg is:

$a_{AB} = \frac{v_B - v_A}{t_B - t_A} = \frac{-\tfrac{1}{3} - 7}{3} = -\frac{22}{9} \approx -2{.}44\;\text{m/s}^2$

Leg B$\to$C. In the final phase the cat decelerates with $|a| = 1\;\text{m/s}^2$ until it stops ($v_C = 0$). The time needed to zero the velocity is $\Delta t = |v_B|/|a| = \tfrac{1}{3}\,\text{s}$, so:

$t_C = t_B + \frac{v_C - v_B}{a} = 8 + \frac{0 + \tfrac{1}{3}}{-1} = \frac{25}{3} \approx 8{.}33\;\text{s}$

$s_C = s_B + \frac{v_C^2 - v_B^2}{2a} = 10 + \frac{0 - \tfrac{1}{9}}{-2} = 10 + \frac{1}{18} = \frac{181}{18} \approx 10{.}06\;\text{m}$

$\ev{v_B = -\tfrac{1}{3}\;\mathrm{m/s}, \quad t_C = \tfrac{25}{3}\;\mathrm{s}, \quad s_C = \tfrac{181}{18}\;\mathrm{m}}$

Collegamenti

Argomenti: Kinematics Concetti: Uniformly accelerated motion Competenze: Choice of axes