Motion is the change in position of an object over time. There are two main types of motion we'll study today. First is uniform motion, where an object moves at constant velocity. Second is accelerated motion, where velocity changes over time. When acceleration is constant, we call it uniform acceleration. This means the rate of change of velocity remains the same throughout the motion.
In physics, we use the International System of Units, or SI units, to measure motion quantities. Displacement is measured in meters, representing the change in position. Velocity is measured in meters per second, showing how fast position changes. Acceleration uses meters per second squared, indicating how quickly velocity changes. Time is measured in seconds. These standardized units allow scientists worldwide to communicate precisely about motion.
The kinematic equations describe uniformly accelerated motion. The first equation, v equals u plus at, shows how velocity changes linearly with time. The second equation, s equals ut plus half at squared, gives displacement as a function of time. The third equation, v squared equals u squared plus 2as, relates velocity to displacement. These equations are derived from basic definitions of velocity and acceleration, and the area under a velocity-time graph represents displacement.
勻加速運動是物理學中最基本的運動類型之一。當物體的加速度保持恆定時，我們稱為勻加速運動。在這種運動中，物體的速度會以固定的速率增加或減少。今天我們將學習勻加速運動的基本方程，並了解如何應用這些方程解決實際問題。
在學習勻加速運動方程之前，我們需要先了解相關物理量的國際單位。位移用符號s表示，單位是公尺。初速度用u表示，末速度用v表示，單位都是公尺每秒。加速度用a表示，單位是公尺每秒平方。時間用t表示，單位是秒。正確理解這些單位對於解決物理問題至關重要。
勻加速運動有三個基本方程。第一個方程是v等於u加at，這描述了速度與時間的關係。第二個方程是s等於ut加二分之一at平方，這描述了位移與時間的關係。第三個方程是v平方等於u平方加2as，這是不含時間的速度位移關係。這三個方程涵蓋了勻加速運動的所有情況。
讓我們詳細了解勻加速運動方程中各個變數的物理意義。u代表初速度，即物體開始運動時的速度。v代表末速度，即物體在某一時刻的速度。a代表加速度，描述速度變化的快慢。t代表時間，即運動持續的時間。s代表位移，即物體位置的改變量。理解這些變數的意義是正確應用方程的基礎。
現在讓我們通過一個實際例題來應用勻加速運動方程。題目是：一輛汽車從靜止開始，以2公尺每秒平方的加速度加速，求5秒後汽車的速度和5秒內汽車行駛的距離。首先分析已知條件：初速度u等於0，加速度a等於2公尺每秒平方，時間t等於5秒。求速度時使用v等於u加at，代入數值得到v等於10公尺每秒。求位移時使用s等於ut加二分之一at平方，代入數值得到s等於25公尺。
Let's solve a practical example. A car accelerates from rest at 2 meters per second squared for 5 seconds. We need to find the final velocity and distance traveled. Given values are: initial velocity u equals zero, acceleration a equals 2 meters per second squared, and time t equals 5 seconds. For final velocity, we use v equals u plus at. Substituting values: v equals zero plus 2 times 5, which gives us 10 meters per second. For distance, we use s equals ut plus half at squared. Substituting: s equals zero plus half times 2 times 25, which gives us 25 meters.