Kinematics

This lesson covers kinematics — the study of motion without considering the forces that cause it. Kinematics is one of the foundational topics in A-Level Mechanics and provides the mathematical tools needed to describe and analyse the motion of objects in one and two dimensions.

---

Key Quantities

Quantity	Symbol	Unit	Type
Displacement	$s$s	m	Vector
Velocity	$v$v	m/s	Vector
Speed	$	v	$
Acceleration	$a$a	m/s$^2$2	Vector
Time	$t$t	s	Scalar

• Displacement is the distance from a fixed point in a specified direction (a vector).
• Velocity is the rate of change of displacement with respect to time.
• Acceleration is the rate of change of velocity with respect to time.

Exam Tip: Always distinguish between distance (scalar) and displacement (vector), and between speed (scalar) and velocity (vector). Using the wrong term will lose marks.

---

The SUVAT Equations (Constant Acceleration)

When acceleration is constant, the following equations relate the five kinematic quantities:

Equation	Variables used
$v = u + at$v=u+at	$v, u, a, t$v,u,a,t
$s = ut + \frac{1}{2}at^2$s=ut+21​at2	$s, u, a, t$s,u,a,t
$s = vt - \frac{1}{2}at^2$s=vt−21​at2	$s, v, a, t$s,v,a,t
$v^2 = u^2 + 2as$v2=u2+2as	$v, u, a, s$v,u,a,s
$s = \frac{(u + v)t}{2}$s=2(u+v)t​	$s, u, v, t$s,u,v,t

Where: $s$s = displacement, $u$u = initial velocity, $v$v = final velocity, $a$a = acceleration, $t$t = time.

Example: A car accelerates from rest at 3 m/s$^2$2 for 8 seconds. Find the final velocity and distance travelled.

$u = 0$u=0, $a = 3$a=3, $t = 8$t=8

$v = u + at = 0 + 3 \times 8 = 24$v=u+at=0+3×8=24 m/s

$s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 3 \times 64 = 96$s=ut+21​at2=0+21​×3×64=96 m

---

Velocity-Time Graphs

A velocity-time graph provides a visual representation of an object's motion:

Feature	Interpretation
Gradient	Acceleration
Area under the curve	Displacement
Horizontal line	Constant velocity (zero acceleration)
Line sloping upwards	Positive acceleration
Line sloping downwards	Deceleration (negative acceleration)

Exam Tip: The area under a velocity-time graph gives displacement, not distance. If the graph goes below the time axis, the area below represents negative displacement (motion in the opposite direction).

---

Vertical Motion Under Gravity

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

• Take upwards as positive (or downwards — but state your convention clearly).
• The acceleration due to gravity is $g = 9.8$g=9.8 m/s$^2$2 (downwards).
• Use the SUVAT equations with $a = -g$a=−g (if upwards is positive) or $a = g$a=g (if downwards is positive).

Example: A ball is thrown vertically upwards with speed 20 m/s. Find the maximum height.

At maximum height, $v = 0$v=0.

$v^2 = u^2 + 2as$v2=u2+2as

$0 = 20^2 + 2(-9.8)s$0=202+2(−9.8)s

$s = \frac{400}{19.6} = 20.4$s=19.6400​=20.4 m

---

Displacement-Time Graphs

Feature	Interpretation
Gradient	Velocity
Horizontal line	Object is stationary
Straight line with positive gradient	Constant positive velocity
Curve	Changing velocity (acceleration)

---

Summary

• Kinematics describes motion using displacement, velocity, acceleration, and time.
• The SUVAT equations apply when acceleration is constant.
• Velocity-time graphs: gradient = acceleration, area = displacement.
• Displacement-time graphs: gradient = velocity.
• For vertical motion under gravity, use $a = \pm g = \pm 9.8$a=±g=±9.8 m/s$^2$2.
• Always state your sign convention and use consistent units.

Exam Tip: Before applying SUVAT equations, list all known quantities and identify which equation to use. If three quantities are known, you can find the fourth. Always check your sign convention is consistent throughout the problem.