Motion in a straight line — the kinematic equations that define it, and the traps that cost marks on them.
NCERT Class 11 Physics Chapter 2 (page 1) defines the framework: an object moving along a straight line has its position described by a single coordinate. Displacement is the change in that coordinate (a signed quantity), while distance is the total path length (always non-negative). This distinction between vector displacement and scalar distance is the root of half the traps in this topic.
The three kinematic equations (NCERT Chapter 2, pages 4–6) apply strictly when acceleration is constant:
A high-frequency trap: applying these equations when acceleration varies with time or position. If a problem states acceleration changes, you must integrate — the kinematic equations are off-limits.
Galileo's odd-number rule. For an object starting from rest under constant acceleration, distances covered in successive equal time intervals follow the ratio 1 : 3 : 5 : 7 : … A common mistake is writing 1 : 2 : 3 : 4 (linear), which ignores that displacement grows as t².
Non-zero initial velocity. When a problem says an object is "thrown downward" or "projected with speed u," that u must appear in the equation. Writing v² = 2gh instead of v² = u² + 2gh drops the u² term and produces a wrong answer — a distractor that appears regularly in NEET papers.
Implicit-function kinematics. When time is given as a function of position (e.g., t = x² + x), don't try to algebraically invert for x(t). Differentiate directly: v = dx/dt = 1/(dt/dx), then use the chain rule a = v(dv/dx) for acceleration.
The v² insight. Under uniform deceleration, kinetic energy (proportional to v²) drops linearly with distance. Speed itself does not. Treating the speed ratio as the distance ratio is a trap that costs marks in multi-stage deceleration problems.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
An object starts from rest and undergoes uniform acceleration. The distances covered in the 1st, 2nd, 3rd, and 4th seconds are in the ratio:
A ball is thrown vertically downward from a tower with an initial speed of 10 m/s. It hits the ground with a speed of 30 m/s. Taking g = 10 m/s², the height of the tower is:
The kinematic equations v = v₀ + at, s = v₀t + ½at², and v² = v₀² + 2as are valid only when:
A bullet travelling at 200 m/s enters a wooden block and its speed reduces to 100 m/s after penetrating 15 cm. The further distance it travels before coming to rest (assuming uniform deceleration) is:
The position of a particle is given by x = t³ − 6t² + 9t + 4 (x in metres, t in seconds). The acceleration at t = 2 s is:
A car moving at 20 m/s decelerates uniformly at 4 m/s². The distance travelled in the 3rd second of deceleration is:
The time t (in seconds) is related to position x (in metres) of a particle by t = x² + x. The velocity of the particle when x = 1 m is:
Which of the following statements is correct for an object under uniform acceleration along a straight line?
Pattern: Multi-stage deceleration (bullet-block type)
Given
A bullet enters a wooden plank at u = 3.00 × 10² m/s. After penetrating d₁ = 4.00 × 10⁻² m (4 cm), its speed reduces to 1.00 × 10² m/s. Deceleration is uniform throughout.
Required
Find the additional distance d₂ the bullet travels before stopping.
Concept
Under constant deceleration, v² decreases linearly with distance. We apply the third kinematic equation twice: once to find the deceleration, once to find the remaining distance.
Formula
v² = u² + 2a·s (third kinematic equation, constant acceleration)
Substitution
**Stage 1:** (1.00 × 10²)² = (3.00 × 10²)² + 2a(4.00 × 10⁻²) 1.00 × 10⁴ = 9.00 × 10⁴ + 0.0800a −8.00 × 10⁴ = 0.0800a **Stage 2:** 0 = (1.00 × 10²)² + 2a·d₂
Calculation
From Stage 1: a = −8.00 × 10⁴ / 0.0800 = −1.00 × 10⁶ m/s² From Stage 2: d₂ = −(1.00 × 10⁴) / (2 × (−1.00 × 10⁶)) = 1.00 × 10⁴ / 2.00 × 10⁶ = 5.00 × 10⁻³ m = 0.500 cm **Note on exact constants:** The factor 2 in the denominator of v² = u² + 2as is a mathematical constant (exact) and does not affect significant-figure counting.
Final answer
d₂ = 5.00 × 10⁻³ m (0.500 cm). The bullet travels only 0.5 cm further after losing two-thirds of its speed over 4 cm — because it is v² (not v) that scales linearly with distance.
Common trap
The temptation is to reason: "speed went from 300 to 100 over 4 cm (a factor of 3 reduction), so it needs about 4/3 ≈ 1.3 cm more to go from 100 to 0." This linear-speed-distance assumption is wrong. Under uniform deceleration, v² drops linearly with distance: going from 300 to 100 m/s reduces v² by 80,000 (over 4 cm), while going from 100 to 0 reduces v² by only 10,000 — requiring only 1/8 of the first stage's distance.
Similar NEET-style question
A car moving at 60 m/s brakes uniformly. After 90 m, its speed is 30 m/s. How much further does it travel before stopping? (Answer: 30 m. Apply v² = u² + 2as to both stages.) ---
Motion is change in position of an object with time. Motion in a straight line (rectilinear / one-dimensional motion) is the simplest case where position can be specified by a single coordinate.
-- NCERT Class 11 Physics, Ch. 2, p. 1Choose a positive direction along the line of motion. Position, velocity, and acceleration are signed scalars on this axis. Sign carries directional meaning: positive = chosen direction, negative = opposite direction.
-- NCERT Class 11 Physics, Ch. 2, p. 1Final velocity equals initial velocity plus acceleration times the time elapsed, for motion under constant acceleration.
| Symbol | Quantity | SI Unit |
|---|---|---|
| v | Final velocity | m/s |
| v0 | Initial velocity | m/s |
| a | Constant (uniform) acceleration | m/s^2 |
| t | Elapsed time | s |
Displacement equals initial-velocity-times-time plus half of acceleration-times-time-squared. The (1/2) factor is the area of the triangle on the v-t graph.
| Symbol | Quantity | SI Unit |
|---|---|---|
| x | Final position | m |
| x0 | Initial position | m |
| v0 | Initial velocity | m/s |
| a | Constant acceleration | m/s^2 |
| t | Time elapsed | s |
Relates final velocity to initial velocity, displacement, and acceleration without using time. Most useful when t is unknown or unwanted.
| Symbol | Quantity | SI Unit |
|---|---|---|
| v | Final velocity | m/s |
| v0 | Initial velocity | m/s |
| a | Constant acceleration | m/s^2 |
| x - x0 | Displacement | m |
Maximum height attained by a projectile launched at speed v0 and angle theta0 above the horizontal, measured above the launch level.
| Symbol | Quantity | SI Unit |
|---|---|---|
| H | Maximum height (above launch) | m |
| v0 | Launch speed | m/s |
| theta0 | Launch angle | rad/deg |
| g | Gravitational acceleration | m/s^2 |
For a projectile launched from and returning to the same horizontal level with initial speed v0 at angle theta0 above the horizontal, the horizontal range R is given by this formula. R is maximised at theta0 = 45 deg.
| Symbol | Quantity | SI Unit |
|---|---|---|
| R | Horizontal range | m |
| v0 | Launch speed | m/s |
| theta0 | Launch angle above horizontal | rad (or deg with sin in deg) |
| g | Gravitational acceleration | m/s^2 |
An object moving in a circle of radius r at constant speed v has acceleration of magnitude v^2/r (or equivalently omega^2 * r) directed toward the centre. This is centripetal (radially inward), not tangential.
| Symbol | Quantity | SI Unit |
|---|---|---|
| a_c | Centripetal acceleration | m/s^2 |
| v | Tangential speed | m/s |
| r | Radius of circle | m |
| omega | Angular speed | rad/s |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Similar Terms
Student answers 1:2:3:4 for distances in successive 1-second intervals (linear) instead of 1:3:5:7 (Galileo's odd numbers).
Question asks about ratios of distances traversed in successive 1-s intervals during free fall from rest.
Distance grows quadratically (y = ½ g t²); successive interval distances are y_n - y_{n-1} = ½ g (t_n² - t_{n-1}²) = ½ g (2n-1) seconds. The factor (2n-1) gives 1, 3, 5, 7, ...
Category: Graph Interpretation
Student uses sin or cos of the angle the line makes with the time axis, instead of tan, to extract velocity.
Question gives an angle the x-t line makes with the t-axis (often 30°, 45°, 60°) and asks for velocity or its ratio.
Velocity = dx/dt = slope of x-t line = tan(angle), where the angle is measured from the time axis. Always tan, not sin or cos.
Category: Overthinking
Student attempts to invert t(x) algebraically before differentiating, getting tangled in messy algebra; misses chain rule.
Question gives t as function of x (instead of x as function of t), e.g. t = x² + x.
Differentiate the given relation directly: dt/dx = (function of x). Then v = dx/dt = 1/(dt/dx). For acceleration use chain rule: a = dv/dt = (dv/dx)(dx/dt) = v dv/dx.
Category: Sign Convention
Student treats a 'thrown vertically downward' problem as if the object were dropped (u = 0). The result is wrong by an additive u² term in v² = u² + 2gh. When the question explicitly states a launch speed, that speed is non-zero and CANNOT be ignored.
Question phrases: 'thrown vertically downward', 'projected with initial velocity', 'launched with speed u'. If u is given numerically, it MUST appear in the equation.
Always parse the launch verbal cue and write down u with its sign before reaching for v² = 2gh. Use the full v² = u² + 2gh (or u² - 2gh for upward motion).
Category: Similar Terms
Student plugs into v₀² sin(2θ)/g (range) when asked for maximum height, or vice versa. The two share v₀ and θ but have different sin-vs-sin² and 2g-vs-g terms.
Question mentions launch speed and angle and asks for max height (H) or range (R). Distractors include the wrong formula's answer.
Memorise BOTH formulas explicitly: H = v₀² sin² θ / (2g) (note sin²); R = v₀² sin(2θ) / g (note sin of doubled angle). Check by setting θ = 45°: max range, half max height.
Category: Sign Convention
Student plugs angle θ into v cos θ when the question states 'angle with the vertical' (which makes the horizontal component v sin θ).
Question phrases like 'thrown at angle θ with the vertical direction' or 'with horizontal'.
Always identify reference axis explicitly. From horizontal: vx = v cos θ, vy = v sin θ. From vertical: vx = v sin θ, vy = v cos θ. The two are complementary (θ_h + θ_v = 90°).
Category: Sign Convention
Student fails to distinguish between same-direction and opposite-direction relative velocities, treating both as magnitudes.
Question describes two objects moving on the same line; observer somewhere between or alongside.
Relative velocity is a VECTOR. Same direction: v_rel = v_a - v_b (smaller magnitude). Opposite direction: v_rel = v_a + v_b (larger magnitude). Use sign convention consistently along chosen axis.
Category: Overthinking
Student assumes proportionality of speed to remaining distance under uniform deceleration. In fact, KE drops linearly with distance (v² is the linear quantity, not v): v² = u² - 2as. Speed-vs-distance is a sqrt-curve, not a line.
Question describes a body decelerating through stages with given speed at one stage; asks for distance to stop or speed at another stage.
Always work with v², not v, when uniform deceleration is in play. The work-energy theorem gives the same answer faster: ½ m v² = work done against constant force over distance.
Category: Overthinking
Student uses the radius R as the projectile launch height or fails to compute the UCM speed from period.
Question describes a particle in UCM with given (R, T) then says 'now launched vertically up with same speed; find max height'.
Step 1: speed v = 2πR/T (from UCM). Step 2: max projectile height H = v²/(2g) = (2πR/T)² / (2g). Don't shortcut by setting H = R.
Category: Similar Terms
Student claims velocity is constant in uniform circular motion (it's not — direction changes).
Question asks 'in uniform circular motion at constant speed, which is also constant?'
In UCM: SPEED constant; KE constant. VELOCITY (vector) NOT constant. ACCELERATION (centripetal, magnitude v²/r) constant in MAGNITUDE but NOT in direction.
Root cause: concept gap
Average velocity = total displacement / total time (vector). Average speed = total path length / total time (scalar, always >= |average velocity|). For round-trip motion, average velocity is zero; average speed is not.
Distractor offers (v1+v2)/2 instead of total-distance/total-time.
Root cause: formula misuse
The three kinematic equations require CONSTANT acceleration. For variable acceleration, use a = dv/dt and integrate, or use v dv = a dx for position-dependent acceleration. Verify constant-a before applying these formulas.
Distractor uses constant-acceleration kinematic equations on a problem where the question explicitly says acceleration changes with time or position.
Root cause: concept gap
Standard range R = v0^2 sin(2*theta)/g and H = v0^2 sin^2(theta)/(2g) assume (i) launch and landing at the same height, (ii) negligible air drag, and (iii) constant g. For asymmetric trajectories, use the full kinematic decomposition along x and y.
Distractor applies R = v0^2 sin(2*theta)/g to a projectile launched from a cliff.
Root cause: concept gap
Acceleration is the rate of change of VELOCITY (a vector), not speed. In uniform circular motion, the speed is constant but the velocity direction changes continuously, giving a centripetal acceleration of magnitude v^2/r toward the centre.
Distractor option says 'a = 0 because speed is constant'.
A particle moving with uniform speed in a circular path maintains:
The ratio of the distances travelled by a freely falling body in the 1st, 2nd, 3rd and 4th second
uses arithmetic progression 1 2 3 4
Linear-time intuition
treats t x as explicit
Default to t-as-input thinking
treats speed ratio as distance ratio
Linear thinking: 'speed went from u to u/3, so distance traversed should be 3× the original'
uses sin or cos instead of tan
Trig confusion under time pressure
uses radius as height
Surface confusion of geometric R and projectile H
uses sin instead of sin squared
Confusing the height formula H = v²sin²θ/(2g) with the range 2v sin θ / g
uses sin instead of cos
Sign-of-angle-axis confusion
speed zero at top
True only for vertical motion (no horizontal v₀)
ignores direction relative to observer
Treating bus-passing time as absolute
treats velocity as scalar
Conflates speed (scalar) with velocity (vector)
drops initial velocity term
Student conflates 'thrown' with 'dropped' and uses v² = 2gh
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