Motion • Topic 3 of 3

Equations of Motion & Graphs

When an object moves in a straight line with uniform acceleration (constant $a$), three simple equations connect its initial velocity $u$, final velocity $v$, acceleration $a$, time $t$ and displacement $s$. They are called the equations of motion and are among the most useful tools in mechanics.

The first equation, $v = u + at$, comes straight from the definition of acceleration $a = \dfrac{v-u}{t}$ rearranged. It links velocity and time. The second equation, $s = ut + \tfrac{1}{2}at^2$, gives the displacement in time $t$. The third equation, $v^2 = u^2 + 2as$, is handy when time is not given but displacement is. To use them correctly, always note the sign of $a$: it is positive for speeding up and negative for retardation, and remember that an object starting from rest has $u = 0$.

Motion can also be understood visually using graphs. A distance–time graph plots distance on the y-axis against time on the x-axis. Its key feature is that the slope of a distance–time graph gives the speed. A straight slanting line means uniform speed; a steeper line means faster motion; a horizontal line means the object is at rest; and a curved line means the speed is changing (non-uniform motion).

A velocity–time graph plots velocity on the y-axis against time. Here two things matter. First, the slope of a velocity–time graph gives the acceleration — an upward-sloping line means positive acceleration, a flat line means constant velocity (zero acceleration), and a downward line means retardation. Second, and very importantly, the area under a velocity–time graph gives the displacement. For uniform acceleration this area is a triangle or trapezium, and computing it gives exactly the same answer as $s = ut + \tfrac{1}{2}at^2$. These graphs let us read off the whole story of a journey at a glance.

  • First equation: $v = u + at$ (velocity–time).
  • Second equation: $s = ut + \tfrac{1}{2}at^2$ (displacement–time).
  • Third equation: $v^2 = u^2 + 2as$ (velocity–displacement, no $t$).
  • Distance–time graph: slope $=$ speed.
  • Velocity–time graph: slope $=$ acceleration; area under it $=$ displacement.
Velocity-time graph: slope is acceleration, shaded area is displacementTime t (s)Velocity v (m/s)uvtArea = displacement sslope = a
Solved Examples
1
Worked Example
A body starts from rest and accelerates at $2\,\text{m/s}^2$ for $6\,\text{s}$. Find its final velocity using $v = u + at$.
Solution
  1. Given $u = 0$, $a = 2\,\text{m/s}^2$, $t = 6\,\text{s}$.
  2. $v = u + at = 0 + 2 \times 6$.
  3. $= 12\,\text{m/s}$.

Answer: $v = 12\,\text{m/s}$.

2
Worked Example
A car starts from rest with acceleration $3\,\text{m/s}^2$. Find the distance covered in $4\,\text{s}$ using $s = ut + \tfrac{1}{2}at^2$.
Solution
  1. Given $u = 0$, $a = 3\,\text{m/s}^2$, $t = 4\,\text{s}$.
  2. $s = ut + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2} \times 3 \times 4^2$.
  3. $= \tfrac{1}{2} \times 3 \times 16 = 24\,\text{m}$.

Answer: $s = 24\,\text{m}$.

3
Worked Example
A vehicle moving at $u = 6\,\text{m/s}$ accelerates uniformly to $v = 14\,\text{m/s}$ over a distance of $40\,\text{m}$. Find its acceleration using $v^2 = u^2 + 2as$.
Solution
  1. $v^2 = u^2 + 2as \Rightarrow 14^2 = 6^2 + 2 \times a \times 40$.
  2. $196 = 36 + 80a$.
  3. $80a = 160 \Rightarrow a = 2\,\text{m/s}^2$.

Answer: $a = 2\,\text{m/s}^2$.

4
Worked Example
A train moving at $20\,\text{m/s}$ is brought to rest with a uniform retardation of $2\,\text{m/s}^2$. Find the distance it travels before stopping.
Solution
  1. Here $u = 20\,\text{m/s}$, $v = 0$, $a = -2\,\text{m/s}^2$.
  2. Use $v^2 = u^2 + 2as$: $0 = 20^2 + 2(-2)s$.
  3. $0 = 400 - 4s \Rightarrow 4s = 400 \Rightarrow s = 100\,\text{m}$.

Answer: $s = 100\,\text{m}$.

5
Worked Example
On a velocity–time graph, an object moves from $u = 4\,\text{m/s}$ to $v = 16\,\text{m/s}$ in $6\,\text{s}$ at uniform acceleration. Find the displacement from the area under the graph.
Solution
  1. The shape is a trapezium with parallel sides $u = 4$ and $v = 16$, height (time) $= 6$.
  2. Area $= \tfrac{1}{2}(u + v) \times t = \tfrac{1}{2}(4 + 16) \times 6$.
  3. $= \tfrac{1}{2} \times 20 \times 6 = 60\,\text{m}$.

Answer: Displacement $= 60\,\text{m}$.

6
Worked Example
A distance–time graph is a straight line passing through the origin, and the object covers $50\,\text{m}$ in $10\,\text{s}$. What does the slope represent and what is its value?
Solution
  1. The slope of a distance–time graph represents the speed.
  2. Slope $= \dfrac{\text{distance}}{\text{time}} = \dfrac{50}{10}$.
  3. $= 5\,\text{m/s}$ (uniform speed, since the line is straight).

Answer: Slope $=$ speed $= 5\,\text{m/s}$.

Key Points

  • First equation: $v = u + at$ links velocity with time.
  • Second equation: $s = ut + \tfrac{1}{2}at^2$ gives displacement in time $t$.
  • Third equation: $v^2 = u^2 + 2as$ is used when time is not given.
  • In a distance–time graph, the slope equals the speed.
  • In a velocity–time graph, the slope equals acceleration and the area under it equals displacement.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.Which equation should you use when time t is NOT given?
Explanation: v^2 = u^2 + 2as relates velocity, acceleration and displacement without involving time.
Q2.The slope of a velocity-time graph gives:
Explanation: Slope = change in velocity / change in time = acceleration.
Q3.The area under a velocity-time graph represents:
Explanation: Area under a v-t graph equals velocity x time = displacement.
Q4.A body starts from rest and accelerates at 5 m/s$^2$. Its velocity after 4 s is:
Explanation: v = u + at = 0 + 5 x 4 = 20 m/s.
Practice more MCQs → 📝 Assignment · 30 marks