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 13) 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 16–18) 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.
A car travels 40 m east along a straight road and then 30 m west along the same road. Its total distance travelled and its displacement are:
Answer: A. A is correct. Distance adds the path lengths regardless of direction: 40 + 30 = 70 m. Displacement is the change in position along the line: +40 − 30 = +10 m, i.e. 10 m east. Distance can never be smaller than the magnitude of displacement, and here it is much larger because the car doubled back.
Why B is wrong: B is wrong because it swaps the two quantities. The 70 m is how far the car went; the 10 m is where it ended up.
Why C is wrong: C is wrong because it treats both legs as if they were in the same direction. The westward 30 m cancels part of the eastward 40 m in the displacement.
Why D is wrong: D is wrong because 50 m is √(40² + 30²), which would be right only if the two legs were at right angles. On a straight road they are along the same line, so they add as signed numbers.
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:
Answer: A. A is correct. Using v² = u² + 2gh with the downward direction positive: 30² = 10² + 2(10)h, so 900 − 100 = 20h and h = 40 m. The u² term is what carries the information that the ball was thrown rather than released. (NCERT Class 11 Physics Chapter 2, page 18.)
Why B is wrong: B is wrong because 45 m comes from v² = 2gh, dropping the u² = 100 term: h = 900/20 = 45 m. That is the classic reading of 'thrown downward' as 'dropped from rest'. (trap: initial velocity dropped)
Why C is wrong: C is wrong because 35 m comes from subtracting 2u² = 200 instead of u² = 100: (900 − 200)/20 = 35 m. Only u² appears in v² = u² + 2gh.
Why D is wrong: D is wrong because 25 m comes from subtracting (2u)² = 400 instead of u² = 100: (900 − 400)/20 = 25 m. Squaring the doubled speed is not the u² term of v² = u² + 2gh.
The kinematic equations v = v₀ + at, s = v₀t + ½at², and v² = v₀² + 2as are valid only when:
Answer: A. All three kinematic equations are derived under the assumption that acceleration a is constant (uniform) in both magnitude and direction. (NCERT Class 11 Physics Chapter 2, pages 16–18.)
Why B is wrong: B is wrong because if velocity is constant, acceleration is zero and there is no need for kinematic equations beyond s = vt. The equations are designed for the case where velocity changes uniformly.
Why C is wrong: C is wrong because when acceleration is proportional to time (a = kt), the motion is non-uniformly accelerated and requires integration: v = v₀ + ½kt², not v = v₀ + at. (mistake: applying constant-a formulas to variable-a problems)
Why D is wrong: D is wrong because the equations work for both speeding up and slowing down, as long as acceleration is constant. Deceleration is simply a negative value of a in the same sign convention.
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:
Answer: C. Stage 1: v² = u² + 2as₁ → (100)² = (200)² + 2a(0.15) → a = (10000 − 40000)/0.30 = −100000 m/s². Stage 2: 0 = (100)² + 2(−100000)s₂ → s₂ = 10000/200000 = 0.05 m = 5 cm. Note that v² (not v) scales linearly with distance under constant deceleration. (NCERT Class 11 Physics Chapter 2, page 18.)
Why A is wrong: A is wrong because 15 cm comes from assuming that since speed halved over 15 cm, it takes another 15 cm to go from 100 m/s to 0. This treats speed as linearly proportional to distance — but it is v² that is linear in distance under uniform deceleration. (trap: linear speed-distance ratio)
Why B is wrong: B is wrong because 10 cm may result from incorrectly assuming v scales linearly with distance and that the '100 m/s to 0' drop is half the '200 to 100' drop, so the distance is 15/1.5 = 10 cm. The correct approach uses v² = u² + 2as.
Why D is wrong: D is wrong because 7.5 cm may come from halving the first-stage distance (15/2), assuming that halving the speed means halving the remaining penetration. Under constant deceleration, the remaining distance scales with v², not v.
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:
Answer: D. v = dx/dt = 3t² − 12t + 9. a = dv/dt = 6t − 12. At t = 2 s: a = 6(2) − 12 = 0 m/s². This is a variable-acceleration problem solved by differentiation, not by kinematic equations.
Why A is wrong: A is wrong because 12 m/s² might come from computing 6t at t = 2 and ignoring the constant term −12 entirely, or from confusing the coefficient of the t³ term with the acceleration expression.
Why B is wrong: B is wrong because 6 m/s² would result from evaluating 6t at t = 2 without subtracting the constant −12, i.e., computing 6(2) = 12 and then dividing by 2, or from a differentiation error where the −6t² term is handled incorrectly.
Why C is wrong: C is wrong because −6 m/s² could arise from evaluating a = 6t − 12 at t = 1 s instead of t = 2 s, giving 6(1) − 12 = −6. Always substitute the correct time.
A car moving at 20 m/s decelerates uniformly at 4 m/s². The distance it travels during the 3rd second of its deceleration is:
Answer: A. A is correct. The distance in the 3rd second is the position at t = 3 s minus the position at t = 2 s. s₂ = 20(2) + ½(−4)(2)² = 40 − 8 = 32 m; s₃ = 20(3) + ½(−4)(3)² = 60 − 18 = 42 m; the difference is 10 m. Check first that the car is still moving: v = 20 − 4t reaches zero at t = 5 s, so the whole of the 3rd second is real motion. (NCERT Class 11 Physics Chapter 2, page 17.)
Why B is wrong: B is wrong because 12 m is the speed at the START of the 3rd second, v(2 s) = 20 − 4(2) = 12 m/s, multiplied by one second. Speed is changing through that second, so its opening value overstates the distance.
Why C is wrong: C is wrong because 8 m is the speed at the END of the 3rd second, v(3 s) = 20 − 4(3) = 8 m/s, multiplied by one second. That is the same error as B with the other endpoint, and it understates the distance. The true answer, 10 m, is the average of these two.
Why D is wrong: D is wrong because 14 m is the distance travelled in the 2nd second, s₂ − s₁ = 32 − 18 = 14 m — an off-by-one in counting which second is the 3rd.
The time t (in seconds) is related to the position x (in metres) of a particle by t = x² + x. The velocity of the particle when x = 1 m is:
Answer: A. A is correct. The relation gives t as a function of x, so differentiate it that way: dt/dx = 2x + 1. Velocity is dx/dt, which is the reciprocal: v = 1/(2x + 1). At x = 1 m, v = 1/3 m/s. There is no need to invert t = x² + x into x(t). (trap: implicit differentiation kinematics)
Why B is wrong: B is wrong because 3 m/s is dt/dx evaluated at x = 1, reported as if it were the velocity. dt/dx has units of s/m; the velocity is its reciprocal. (trap: implicit differentiation confusion)
Why C is wrong: C is wrong because 1/2 m/s comes from differentiating only the x² term, dt/dx = 2x, and dropping the +1. At x = 1 that gives 1/2 instead of 1/3.
Why D is wrong: D is wrong because 2 m/s is 2x at x = 1 — differentiating x² and then treating the result as dx/dt directly, so both the +1 and the reciprocal are lost.
Which of the following statements is correct for an object under uniform acceleration along a straight line?
Answer: B. Uniform acceleration means a = Δv/Δt = constant, so velocity changes by the same amount (aΔt) in every equal time interval. This is the defining property. (NCERT Class 11 Physics Chapter 2, page 16.)
Why A is wrong: A is wrong because under uniform acceleration, distance is proportional to t² (not t). Distance proportional to time holds only for uniform velocity (zero acceleration).
Why C is wrong: C is wrong because uniform acceleration can be negative (deceleration). In that case, speed decreases until the object stops or reverses. 'Uniform acceleration' describes constant a, not necessarily positive a.
Why D is wrong: D is wrong because 'uniform acceleration' by definition means acceleration does NOT change with time. If acceleration increased with time, the motion would be non-uniformly accelerated.
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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.)
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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. 13Choose 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. 23Final 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 |
More in Kinematics: 14 exam traps and mistakes · 3 formulas · 10 question patterns from its other lessons.
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