When an object moves in a straight line and its acceleration is constant (uniform), we can describe its motion using three equations of motion.
These equations help us find velocity, distance, time, and acceleration.
 
Graphical derivation:
 
Velocity-time graph.svg
Velocity-time graph
 
Understanding the graph: 
  • The graph is a straight line  \(\rightarrow\)  motion has uniform acceleration
  • Initial velocity = \(u\) (starting point)
  • Final velocity = \(v\) (ending point)
  • Time taken = \(t\)
1. First equation of motion:
 
The slope of the velocity–time graph gives acceleration:
 
\(a=\frac{v-u}{t}\)
 
Rearranging,
 
\(v=u+at\)
 
This shows how velocity changes with time.
 
2. Second equation of motion:
 
\(Distance \ travelled \ = \ area \ under \ the \ velocity-time \ graph\)
 
\(Area= rectangle (AOCD) +triangle (ABD) \)
 
Rectangle area (\(AOCD\)), 
 
\(base\times height=t\times u=ut\)
 
Triangle area (\(ABD\)),
 
\(=\frac{1}{2}\times base\times height\)
 
\(\frac{1}{2}\times t\times (v-u)\)
 
Total distance:
 
\(s=ut+\frac{1}{2}(v-u)t\)
 
From first equation, \(v-u=at\), substitute
 
\(s=ut+\frac{1}{2}at^{2}\)
 
This gives the distance travelled in time \(t\)
 
3. Third equation of motion:
 
We eliminate time \(t\)
 
From first equation:  
 
\(v-u=at\Rightarrow t=\frac{v-u}{a}\)
 
Substitute in second equation:
 
\(s=ut+\frac{1}{2}at^{2}\)
 
After simplification, we get:
 
\(v^{2}=u^{2}+2as\)
 
This equation relates velocity and distance and is useful when time is not given.
 
Selecting the correct equation:
  • Use \(v=u+at\rightarrow\) when distance is not required
  • Use \(s=ut+\frac{1}{2}at^{2}\rightarrow\)  when distance is needed
  • Use \(v^{2}=u^{2}+2as \rightarrow\)  when time is not given
 
Solving numerical problems:
Steps:
  1. Write the given values (\(u\), \(v\), \(a\), \(t\), \(s\))
  2. Identify what is required
  3. Choose the correct equation
  4. Substitute values and solve