You are viewing a free preview of this lesson.
Subscribe to unlock all 10 lessons in this course and every other course on LearningBro.
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.
| Quantity | Symbol | Unit | Type |
|---|---|---|---|
| Displacement | s | m | Vector |
| Velocity | v | m/s | Vector |
| Speed | $ | v | $ |
| Acceleration | a | m/s2 | Vector |
| Time | t | s | Scalar |
Exam Tip: Always distinguish between distance (scalar) and displacement (vector), and between speed (scalar) and velocity (vector). Using the wrong term will lose marks.
When acceleration is constant, the following equations relate the five kinematic quantities:
| Equation | Variables used |
|---|---|
| v=u+at | v,u,a,t |
| s=ut+21at2 | s,u,a,t |
| s=vt−21at2 | s,v,a,t |
| v2=u2+2as | v,u,a,s |
| s=2(u+v)t | s,u,v,t |
Where: s = displacement, u = initial velocity, v = final velocity, a = acceleration, t = time.
Example: A car accelerates from rest at 3 m/s2 for 8 seconds. Find the final velocity and distance travelled.
u=0, a=3, t=8
v=u+at=0+3×8=24 m/s
s=ut+21at2=0+21×3×64=96 m
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).
For objects moving vertically under gravity (ignoring air resistance):
Example: A ball is thrown vertically upwards with speed 20 m/s. Find the maximum height.
At maximum height, v=0.
v2=u2+2as
0=202+2(−9.8)s
s=19.6400=20.4 m
| Feature | Interpretation |
|---|---|
| Gradient | Velocity |
| Horizontal line | Object is stationary |
| Straight line with positive gradient | Constant positive velocity |
| Curve | Changing velocity (acceleration) |
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.