Kinematics – Study Notes
(KCET – Physics – 11th Grade)
1. Simple Explanation
Kinematics is the branch of physics that describes motion without considering the forces that cause it.
It focuses on the following basic concepts:
| Concept | What it describes |
|---|---|
| Displacement | Change in position (vector) |
| Velocity | Rate of change of displacement (vector) |
| Acceleration | Rate of change of velocity (vector) |
Key Idea – All motion can be broken down into position, velocity, and acceleration.
2. Key Points
| Topic | Important Equations / Definitions | Notes |
|---|---|---|
| Displacement (s) | ( \vec{s} = \vec{r_f} - \vec{r_i} ) | Always a vector; magnitude is the straight‑line distance between start and end. |
| Average Velocity ((v_{av})) | ( v_{av} = \frac{\Delta s}{\Delta t} ) | Can be a vector; sign indicates direction. |
| Instantaneous Velocity ((v)) | ( v = \lim_{\Delta t\to0}\frac{\Delta s}{\Delta t} ) | Found as the slope of the position‑time graph. |
| Average Acceleration ((a_{av})) | ( a_{av} = \frac{\Delta v}{\Delta t} ) | Vector; sign tells if speed is increasing or decreasing. |
| Instantaneous Acceleration ((a)) | ( a = \lim_{\Delta t\to0}\frac{\Delta v}{\Delta t} ) | Slope of the velocity‑time graph. |
| Uniformly Accelerated Motion | 1. ( v = u + a t ) 2. ( s = u t + \frac12 a t^2 ) 3. ( v^2 = u^2 + 2 a s ) |
(u) = initial velocity, (v) = final velocity. |
| Projectile Motion | Horizontal: ( x = u_x t ) Vertical: ( y = u_y t - \frac12 g t^2 ) |
(g = 9.8\ \text{m/s}^2) (downward). |
| Relative Velocity | ( \vec{v}_{AB} = \vec{v}_A - \vec{v}_B ) | Used when two bodies move simultaneously. |
| Graphs | • Position vs. time: slope = velocity • Velocity vs. time: slope = acceleration • Acceleration vs. time: area under curve = change in velocity |
Graphical interpretation is essential. |
| Units | Displacement & distance: m, km Velocity: m/s, km/h Acceleration: m/s² Time: s, min, h |
Always keep the units consistent. |
3. Worked Example
Problem
A car starts from rest and accelerates uniformly at (2.0\ \text{m/s}^2) for (10\ \text{s}).
Find:
- Final velocity.
- Distance covered.
- Average velocity.
- Sketch the position‑time and velocity‑time graphs.
Solution
| Step | Calculation | Result |
|---|---|---|
| 1. Initial velocity (u) | (u = 0\ \text{m/s}) | – |
| 2. Final velocity | ( v = u + a t = 0 + (2.0)(10) = 20.0\ \text{m/s} ) | (20.0\ \text{m/s}) |
| 3. Distance covered | ( s = u t + \frac12 a t^2 = 0 + \frac12(2.0)(10)^2 = 100\ \text{m} ) | (100\ \text{m}) |
| 4. Average velocity | ( v_{av} = \frac{s}{t} = \frac{100}{10} = 10\ \text{m/s} ) | (10\ \text{m/s}) |
| 5. Graphs | • Velocity‑time: straight line from (0,0) to (10,20) → slope (a=2.0\ \text{m/s}^2). • Position‑time: parabola opening upwards; equation (s = 10t^2). |
(Sketch below) |
Position‑Time Graph
s(m) | | * (10,100) | * | * | * | * |* +-------------------- t(s) 0 2 4 6 8 10Velocity‑Time Graph
v(m/s) | | * | * | * |* +-------------------- t(s) 0 2 4 6 8 10
Interpretation
• Acceleration is constant → linear velocity‑time graph.
• Distance grows faster than time (parabolic).
4. Quick Revision Summary
| Concept | Equation | What to remember |
|---|---|---|
| Displacement | (s = \vec{r_f} - \vec{r_i}) | Vector; straight‑line distance. |
| Average velocity | (v_{av} = \frac{\Delta s}{\Delta t}) | Slope of s‑t graph. |
| Instantaneous velocity | (v = \frac{ds}{dt}) | Instant slope. |
| Average acceleration | (a_{av} = \frac{\Delta v}{\Delta t}) | Slope of v‑t graph. |
| Instantaneous acceleration | (a = \frac{dv}{dt}) | Instant slope. |
| Uniform acceleration equations |