Motion is one of those topics where students lose easy marks by confusing two pairs: distance vs. displacement and speed vs. velocity. Get those two distinctions crystal-clear and the rest of Motion falls into place.
Here is the plain picture. Imagine you walk from your घर (home) to a shop 500 m away, buy something, and walk back. You have covered a distance of 1,000 m — that is the total path length, always positive, never zero unless you literally did not move. But your displacement is zero — you ended exactly where you started. Displacement cares only about the straight-line gap between start and end, with a direction attached.
Speed is how fast you covered that 1,000 m of path. Velocity is how fast you shifted your position — and since your final position matched your start, your average velocity for the round trip is zero even if you sprinted both ways.
Acceleration is simply the rate at which velocity changes. Constant speed in a straight line means zero acceleration. But here is the trap: a car going around a roundabout at constant speed is still accelerating, because its direction — and therefore its velocity — keeps changing.
Newton tied all of this together with three laws. Think of them as three different answers to the question "what does force do?"
F = ma). The bigger the force, the more acceleration; the heavier the object, the less acceleration for the same force.Momentum (p = mv) connects mass and velocity. Newton's second law in its most general form says force equals the rate of change of momentum. When no external force acts, momentum is conserved — this is why two skaters who push off each other move in opposite directions.
Distance is a scalar (magnitude only). Displacement is a vector (magnitude + direction). After any number of complete revolutions on a circular track, displacement is zero. This is the concept behind PYQ 6a3bc44fd136391748ad7db9 — the body completes two full circles, returns to start, displacement = 0.
For uniform acceleration along a straight line, three equations cover every scenario:
v = u + ats = ut + ½at²v² = u² + 2asWhere u = initial velocity, v = final velocity, a = acceleration, s = displacement, t = time.
Free fall is just uniform acceleration with a = g ≈ 10 m/s² (use 9.8 only if the options force you). Object starts from rest, so u = 0. Equation 2 becomes: h = ½gt², which gives t = √(2h/g).
Plug numbers into h = ½gt² and isolate t:
t = √(2h/g)
For h = 16 m, g = 10 m/s²:
t = √(2 × 16 / 10) = √(32/10) = √3.2 ≈ 1.789 ≈ 1.8 s
Always use g = 10 m/s² unless the question explicitly states 9.8. It saves calculation time and the options are set accordingly.
This is a high-yield concept. Know what each region of a v-t graph means:
| Feature | Physical Meaning |
|---|---|
| Slope of the line | Acceleration (a = Δv/Δt) |
| Area under the curve | Displacement |
| Horizontal line (slope = 0) | Constant velocity, zero acceleration |
| Line going up | Positive acceleration |
| Line going down | Deceleration (negative acceleration) |
Units check for area: velocity (ms⁻¹) × time (s) = meters (m). So area under v-t graph has units of meters — displacement, not distance (unless the graph stays above the axis throughout).
In uniform circular motion, speed is constant but velocity changes direction continuously. This means there is always an acceleration — called centripetal acceleration — directed toward the center. The force causing this is centripetal force: F = mv²/r.
After any whole number of revolutions, the object is back at the starting point. Net displacement = 0. Distance covered = number of revolutions × circumference = n × 2πr.
Friction opposes relative motion between surfaces. Two types matter for objective questions:
Friction depends on the nature of surfaces (coefficient of friction μ) and the normal force (N), but not on the area of contact or the speed of sliding. That "not on area" point is a classic MCQ trap.
First Law gives the definition of inertia. Heavier object = more inertia = harder to accelerate or stop.
Second Law (F = ma) is the calculation law. If mass doubles and force stays the same, acceleration halves.
Third Law — action and reaction act on different bodies. A book resting on a table: the book pushes the table down (action), the table pushes the book up (reaction/normal force). These two forces do not cancel each other because they act on different objects.
A projectile has horizontal velocity (constant, no air resistance) and vertical velocity (changes due to gravity). Time of flight and maximum height depend only on the vertical component; horizontal range depends on both.
45° launch angle.g downward — not zero, not changing direction.When you see "displacement" in the question, ask: did the body return to or pass through its starting point? If it completed full circles, displacement = 0. This catches the circular motion trap every time. Standard mistake approach: students calculate 40π cm (circumference × 2) thinking it is displacement. This trick takes 3 seconds vs. 30 seconds of calculation.
For free fall from rest: t = √(2h/g). With g = 10, this becomes t = √(h/5).
Quick reference: h = 5 m → t = 1 s; h = 20 m → t = 2 s; h = 45 m → t = 3 s.
For h = 16 m: t = √(16/5) = √3.2. You know √3.2 is between 1.7 (√2.89) and 1.8 (√3.24). It is closer to 1.8. Done. Standard method: full algebra — about 40 seconds. This anchor method: under 10 seconds.
Whatever is on the y-axis × whatever is on the x-axis = unit of the area. v-t graph: ms⁻¹ × s = m. acceleration-time graph: ms⁻² × s = ms⁻¹ (velocity). Always multiply the axis labels — no formula needed. Eliminates wrong answers in 5 seconds vs. 20+ seconds of unit analysis.
First Law: Is anything moving? (Inertia/equilibrium) — use when the question mentions "at rest" or "constant velocity".
Second Law: Does it say force, mass, or acceleration? — use F = ma, straight calculation.
Third Law: Does it mention pairs of objects pushing/pulling each other? — action-reaction, always equal in magnitude.
This decision splits all Newton's Law questions into the right lane in under 5 seconds. Removes the common error of applying F = ma to Third Law questions.
Any MCQ option that says friction depends on the area of contact or the speed of sliding is wrong. Static and kinetic friction depend only on μ (surface property) and normal force N. If you see "larger surface → more friction", eliminate it immediately. This single rule eliminates 2-3 options in friction MCQs instantly.
Read the question and identify the category:
Step 1 — Is it about position or path?
Step 2 — Is it a calculation question?
h = ½gt² or v² = u² + 2as with g = 10.F = ma.p = mv; conservation → m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.Step 3 — Is it a graph question?
Step 4 — Is it a conceptual Newton/friction question?
If still unsure, use elimination: identify the one or two options that are dimensionally correct, then pick the physically sensible one.
Why this question: The circular motion displacement trap is one of the most commonly recycled concepts. Students who know "circumference × revolutions = distance, not displacement" get this in 5 seconds.
Solving path: Two full revolutions → body returns exactly to starting point → straight-line gap between start and end = 0 → displacement = zero. The options 40π cm and 20π cm are the circumference-based calculations for distance, not displacement. Eliminate them. Answer: zero.
Why this question: Free fall with a number is a direct formula application. If you have h = ½gt² ready, this is a 10-second question. The trap is using g = 9.8 instead of 10.
Solving path: h = ½gt² → 16 = ½ × 10 × t² → t² = 3.2 → t = √3.2 ≈ 1.79 ≈ 1.8 s. Scan options: 1.8 sec is there. Done. If you used g = 9.8, you would get t ≈ 1.806 s — same answer, but more painful arithmetic. Always default to g = 10.
Why this question: v-t graph unit questions appear regularly and trip up students who try to remember a formula instead of multiplying axis units.
Solving path: Area under v-t graph = displacement. Unit check: velocity (ms⁻¹) × time (s) = m. Look at the options: m³, ms⁻¹, m², m. Only m is dimensionally correct. No formula recall needed — just multiply axis units. Answer: m.
Confusing distance and displacement in circular/return-journey problems. After any whole number of revolutions, displacement is zero. Distance equals n × 2πr. These are not interchangeable.
Using g = 9.8 when g = 10 works. Bihar Police Constable options are designed for g = 10 m/s². Using 9.8 creates messy arithmetic and sometimes misleads you toward the wrong option.
Thinking a body at constant speed has zero acceleration. If it is moving in a circle, its velocity vector changes direction constantly — acceleration is not zero. Uniform circular motion always has centripetal acceleration.
Applying Third Law incorrectly by saying action and reaction cancel. They act on different bodies. The book on the table: weight of book acts on the table; normal force of table acts on the book. These are on different objects, so they cannot cancel each other.
Saying friction increases with contact area. Friction = μN. Area does not appear in this equation. Any option involving larger surface area giving more friction is wrong.
Misreading v-t graph slope vs. area. Slope = acceleration (unit: ms⁻²). Area = displacement (unit: m). Swapping these two is one of the most common graph errors. A flat horizontal line has zero slope (zero acceleration) but its area still grows with time (displacement increases).