Kinematics and SUVAT Equations

Kinematics and SUVAT Equations

Kinematics is the study of motion without considering the forces that cause it. The SUVAT equations describe motion with constant (uniform) acceleration in a straight line.

The Five SUVAT Variables

Variable	Meaning	SI Unit
s	Displacement	metres (m)
u	Initial velocity	m/s
v	Final velocity	m/s
a	Acceleration	m/s²
t	Time	seconds (s)

Displacement is a vector quantity — it has direction. A positive displacement typically means motion in the positive direction (e.g., upward or to the right).

The Five SUVAT Equations

Each equation connects four of the five variables (one is "missing"):

Equation	Missing variable
v = u + at	s
s = ut + ½at²	v
s = vt - ½at²	u
v² = u² + 2as	t
s = ½(u + v)t	a

How to choose: Identify the three known quantities and the one you need to find. Use the equation that contains all four.

Worked Example 1: A car accelerates from 5 m/s to 25 m/s over 8 seconds. Find the acceleration and the distance travelled.

Known: u = 5, v = 25, t = 8. Find a and s.

v = u + at: 25 = 5 + 8a, so a = 20/8 = 2.5 m/s²

s = ½(u + v)t = ½(5 + 25)(8) = ½(30)(8) = 120 m

Worked Example 2: A ball is thrown vertically upward at 20 m/s. Find the maximum height reached. (Take g = 9.8 m/s², upward positive.)

At maximum height, v = 0. Known: u = 20, v = 0, a = -9.8.

v² = u² + 2as: 0 = 400 + 2(-9.8)s

19.6s = 400, so s = 20.4 m

Vertical Motion Under Gravity

For objects moving vertically under gravity (ignoring air resistance):

• The acceleration is g = 9.8 m/s² (or 9.81, depending on the question) directed downward
• Choose a sign convention (usually upward = positive), then a = -g
• At the highest point, velocity is momentarily zero
• The time to reach maximum height equals the time to fall back down (for symmetric journeys)

Worked Example: A stone is dropped from a cliff 45 m high. Find the time to hit the ground.

Taking downward as positive: u = 0, a = 9.8, s = 45.

s = ut + ½at²: 45 = 0 + ½(9.8)t²

t² = 90/9.8 = 9.184

t = √9.184 = 3.03 seconds

Velocity-Time Graphs

A velocity-time graph is one of the most powerful tools in kinematics:

• The gradient of the graph equals the acceleration
• The area under the graph equals the displacement
• A horizontal line means constant velocity (zero acceleration)
• A straight line with positive gradient means constant positive acceleration
• A line below the time axis means motion in the negative direction

Worked Example: A vehicle's velocity-time graph shows: 0 to 10s velocity increases from 0 to 20 m/s; 10s to 25s velocity stays at 20 m/s; 25s to 30s velocity decreases from 20 to 0 m/s.

Total displacement = area under graph

= ½(10)(20) + (15)(20) + ½(5)(20)

= 100 + 300 + 50 = 450 m

Acceleration in first phase = 20/10 = 2 m/s²

Deceleration in final phase = 20/5 = 4 m/s² (magnitude)

Variable Acceleration (Calculus Methods)

When acceleration is not constant, use calculus:

• Velocity = ds/dt (differentiate displacement)
• Acceleration = dv/dt = d²s/dt² (differentiate velocity)
• Displacement = ∫v dt (integrate velocity)
• Velocity = ∫a dt (integrate acceleration)

Worked Example: A particle moves with displacement s = 2t³ - 9t² + 12t metres. Find when the particle is at rest.

v = ds/dt = 6t² - 18t + 12

At rest: v = 0, so 6t² - 18t + 12 = 0

t² - 3t + 2 = 0

(t - 1)(t - 2) = 0

The particle is at rest at t = 1 s and t = 2 s.

Exam Tips

• Always write down s, u, v, a, t and fill in what you know before choosing an equation.
• Be consistent with your sign convention throughout the problem. State it explicitly.
• In vertical motion problems, ensure you correctly assign the sign of g based on your chosen positive direction.
• When a question says "find the distance", it wants the magnitude. When it says "find the displacement", the answer may be negative.
• Velocity-time graph questions often ask for total distance (sum of absolute areas) not displacement (signed sum). Read carefully.
• With variable acceleration, the SUVAT equations do not apply — you must use calculus.