The Equations of Motion ചലനസമവാക്യങ്ങൾ
When an object moves with uniform acceleration, three short equations connect its initial velocity, final velocity, acceleration, time and displacement. Here you see where they come from (straight out of the velocity–time graph), learn which one to pick, and use them on the textbook's four examples – a car, a train leaving a station and a falling object.
1. The five symbols
ചലനസമവാക്യങ്ങൾ ഉപയോഗിക്കുമ്പോൾ സാധാരണയായി ചുവടെ കൊടുത്തിരിക്കുന്ന സൂചകങ്ങളാണ് ഉപയോഗിക്കുന്നത്. For an object moving with uniform acceleration (സമത്വരണം) we use these symbols:
പ്രവേഗവും സമയവും തമ്മിലുള്ള ബന്ധം മനസ്സിലാക്കാൻ നമുക്ക് v = u + at എന്ന സമവാക്യം ഉപയോഗിക്കാം. v = u + at links velocity and time; s = ut + ½at² links displacement and time; v² = u² + 2as links displacement and velocity.
2. Where the equations come from – the v–t graph
The textbook states the equations; here is how they come out of a velocity–time graph using only two ideas from page 10: slope = acceleration and area = displacement. Step through.
ചലന സമവാക്യങ്ങൾ: v = u + at ; s = ut + ½at² ; v² = u² + 2as. സമത്വരണത്തിലുള്ള വസ്തുക്കൾക്ക് മാത്രമേ ഈ സമവാക്യങ്ങൾ ബാധകമാകുന്നുള്ളൂ.
These equations apply only to objects moving with uniform acceleration. If the acceleration changes (like the motorbike of Table 2.7), use the graph instead.
3. Which equation should I use?
Each equation leaves out exactly one of s, v, t. List what is given and what is asked; the quantity that is neither tells you the equation.
Start values u = 6 m/s, a = 1 m/s², t = 10 s are the textbook's car example: v = 16 m/s, s = 110 m.
4. Worked examples (textbook pages 45–46)
Example 1 · Velocity 20 m/s at 2 s and 40 m/s at 6 s
Textbook: നിശ്ചലാവസ്ഥയിൽ നിന്നും പുറപ്പെട്ട ഒരു വസ്തുവിന്റെ പ്രവേഗം 2 s കൊണ്ട് 20 m/s ഉം 6 s കൊണ്ട് 40 m/s ഉം ആയി. ഈ സമയ ഇടവേളയിൽ വസ്തുവിനുണ്ടായ സ്ഥാനാന്തരമെത്ര? – An object starts from rest. Its velocity is 20 m/s at 2 s and 40 m/s at 6 s. What is its displacement during this interval?
“This interval” is from 2 s to 6 s, so u = 20 m/s (velocity at its start), v = 40 m/s, t = 6 s − 2 s = 4 s. Find a first, then use s = ut + ½at².
“Starts from rest” is not used in the calculation – only the 2 s → 6 s interval matters. In fact, if the acceleration had been 5 m/s² from t = 0, the velocity at 2 s would have been only 0 + 5 × 2 = 10 m/s, not 20 m/s. So in the first 2 s the object must have accelerated faster (20 ÷ 2 = 10 m/s²). The equations are applied only to the stretch where the acceleration is uniform.
Example 2 · A car from 6 m/s to 16 m/s in 10 s
Textbook: ഒരു കാറിന്റെ പ്രവേഗം 10 s കൊണ്ട് 6 m/s ൽ നിന്നും 16 m/s ലേക്ക് എത്തുന്നുവെങ്കിൽ, a) കാറിന്റെ ത്വരണം കണക്കാക്കുക. b) ഈ സമയം കൊണ്ട് കാറിനുണ്ടായ സ്ഥാനാന്തരം എത്ര? – A car's velocity changes from 6 m/s to 16 m/s in 10 s. (a) Find its acceleration. (b) What is its displacement in this time?
(a) a = (v − u)/t. (b) Use s = ut + ½at² with the a you just found.
Example 3 · A train leaving a station
Textbook: റെയിൽവേ സ്റ്റേഷനിൽ നിന്ന് പുറപ്പെട്ട ഒരു ട്രെയിനിന്റെ പ്രവേഗം 10 മിനിറ്റ് കൊണ്ട് 90 km/h ആയി. എങ്കിൽ, ട്രെയിനിന്റെ ത്വരണം കണക്കാക്കുക. – A train leaving a railway station reaches a velocity of 90 km/h in 10 minutes. Find its acceleration.
The train starts at the station, so u = 0. Convert first: km/h → m/s (× 5/18) and minutes → seconds (× 60). Then a = (v − u)/t.
To change km/h into m/s, multiply by 5/18; to change m/s into km/h, multiply by 18/5. Always convert before using the equations of motion.
Example 4 · An object falling from rest
Textbook: നിശ്ചലാവസ്ഥയിൽ നിന്ന് 10 m/s² സമത്വരണത്തോടെ പതിക്കുന്ന വസ്തു തറയിൽ തൊടുന്നതിന് തൊട്ടു മുമ്പ് അതിന്റെ പ്രവേഗം 20 m/s ആണെങ്കിൽ വസ്തു എത്ര ഉയരത്തിൽ നിന്നാണ് പുറപ്പെടുന്നത്? – An object falls from rest with a uniform acceleration of 10 m/s². Just before it touches the ground its velocity is 20 m/s. From what height did it start?
Given u, a and v; asked s. Time is neither given nor asked, so use v² = u² + 2as. Take downwards as positive.
📐 These four examples, with “try other numbers” labs: page 12 – In-text questions, solutions.
5. Falling coconut lab
Choose the height of the coconut, drop it and watch v grow. Air resistance is ignored, so a = 10 m/s² (downwards taken as +). At every instant v² = 2 × 10 × (distance fallen), and the area under the v–t graph equals the distance fallen.
A coconut from the top of a 20 m palm reaches the ground at about 20 m/s – that is 72 km/h, the speed of a fast bus – in only 2 s. That is why we never stand or park under coconut palms during the monsoon winds. And a Kochi Metro train pulling out of a station speeds up at roughly 1 m/s², about 24 times the acceleration of the textbook's train (1/24 m/s²).
- 📄 Physics Tutorial - 1D Kinematics - Kinematic Equations - Kinematic Equations The Physics Classroom
The equations v = u + at, s = ut + 1/2 at^2 and v^2 = u^2 + 2as with how to choose one (they write vi/vf for u/v). - ▶️ Deriving 3 equations of motion (from v-t graph) Khan Academy India - English (YouTube) · 13:04
Derives all three equations of motion from a velocity-time graph - exactly the textbook method. - ▶️ Using equations of motion (1 step numerical) Khan Academy India - English (YouTube) · 11:54
Step-by-step practice using the equations of motion in simple numericals. - 📄 Physics Tutorial - 1D Kinematics - Kinematic Equations - Sample Problems and Solutions The Physics Classroom
Extra solved problems to check your understanding. - 📄 Physics Tutorial - 1D Kinematics - Free Fall - Introduction The Physics Classroom
What 'free fall' means: an object moving only under gravity. - 📄 Physics Tutorial - 1D Kinematics - Free Fall - Acceleration of Gravity The Physics Classroom
g is about 9.8 m/s^2 downward - a ball thrown up slows by 9.8 m/s every second. - ▶️ Apollo 15 Hammer-Feather Drop NASA Solar System (YouTube) · 1:23
On the Moon (no air) an astronaut drops a hammer and a feather - they land together! - ▶️ Brian Cox visits the world's biggest vacuum | Human Universe - BBC BBC (YouTube) · 4:41
A bowling ball and feathers dropped in a huge vacuum chamber fall at the same rate.
6. Check yourself
- Symbols: u initial velocity, v final velocity, s displacement, a acceleration, t time.
- v = u + at (from the slope), s = ut + ½at² (area: rectangle + triangle), v² = u² + 2as (eliminate t).
- Valid only for uniform acceleration. Pick the equation that leaves out the quantity you neither know nor need.
- Convert first: km/h × 5/18 = m/s; minutes × 60 = s.