a) The known motions are divided into rectilinear and curvilinear due to the shape of the trajectory. Due to variable velocity values - uniform rectilinear motion and non‑uniform motion. Among non‑uniform rectilinear motions, we distinguish uniformly accelerated rectilinear motion and uniformly decelerated rectilinear motionuniformly decelerated rectilinear motionuniformly decelerated rectilinear motion.
b) Acceleration describes the change of velocity values per unit of time.
c) Known motions are characterized by the time of travel, distance, average velocity and instantaneous velocity as well as acceleration.
d) AccelerationaccelerationAcceleration is calculated using the formula:

The unit of acceleration in the SI system is $\frac{m}{s^2}$.

Analysis

Given:

$v_i=72\frac{km}{h}$

$v_f=54\frac{km}{h}$

$t=15s$

Unknown:

Solution:

$a=\frac{\Delta v}{\Delta t}$

$\Delta v=v_f-v_i$

$\Delta v=72\frac{km}{h}-54\frac{km}{h}$

$\Delta v=-18\frac{km}{h}=-5\frac{m}{s}$

$a=\frac{-5\frac{m}{s}}{15s}$

$a=-\frac{1}{3}\frac{m}{s^2}$

Answer:

Acceleration is $a=-\frac{1}{3}\frac{m}{s^2}$, it is negative, so deceleration in this motion is $a_0=-a=\frac{1}{3}\frac{m}{s^2}$, it is positive.

Section I - uniformly accelerated motion, duration: 2 s.
Section II – uniform rectilinear motion, duration: 2 s.
Section III - uniformly decelerated motion, duration: 4 s.

Analysis

Given:

$v_i=600\frac{m}{s}$

$s=8cm=0,08m$

$v_f=0$

Unknown:

$a=?$

Solution:

$s=v_i\cdot t-\frac{a\cdot t^2}{2}$

$a=\frac{\Delta v}{\Delta t}$

$\Delta v=v_i-v_f$

$\Delta v=-v_i$

$t=\frac{\Delta v}{a}$

$t=\frac{v_i}{a}$

The acceleration vector in uniformly decelerated motion has the same direction and opposite sense to the velocity vector of the moving body. To the formula we substitute the acceleration as a positive value, and the fact that we are dealing with a uniformly decelerated motion is marked by inserting a minus in the right place. Therefore, the formula for the distance in uniformly decelerated motion takes the form:

$s=v_i\cdot \frac{v_i}{a}-\frac{a\cdot {\left(\frac{-v_i}{a}\right)}^2}{2}$

$s=\frac{v_i^2}{a}-\frac{v_i^2}{2a}=\frac{v_i^2}{2a}$

$a=\frac{v_i^2}{2s}[\frac{\frac{m^2}{s^2}}{m}=\frac{m^2}{s^2}\cdot \frac{1}{m}=\frac{m}{s^2}]$

$a=\frac{{600}^2}{2\cdot 0,08}$

$a=\frac{360000}{0,16}$

$a=2250000\frac{m}{s^2}$

Answer:

Deceleration is 2250000 $\frac{{\displaystyle m}}{s^2}$.

Rxpojtlj0LhwY

Ilustracja przedstawia zależności wartości prędkości od czasu. Oś pozioma opisana t, w nawiasie kwadratowym s, zaznaczone punkty 0, 4, 8 i 10. Oś pionowa opisana v, w nawiasie kwadratowym m przez s, zaznaczone punkty 24 i 20. Wykres narysowany jest jako trzy odcinki. Pierwszy zaczyna się w punkcie (0, 0) i kończy się w punkcie (4, 24). Drugi zaczyna się w punkcie (4, 24) i kończy się w punkcie (8, 24). Trzeci zaczyna się w punkcie (8, 24) i kończy się w punkcie (10, 20). Narysowane są linie przerywane od końców odcinków do osi poziomej i pionowej.

The total distance covered by the body is s = 188 m.

Uniformly decelerated motion is a uniformly accelerated motion in which the acceleration is negative.