Chapter 2: Motion in a Straight Line

Rule 1: Displacement and path length — Δx = x₂ − x₁. The sign of Δx gives the direction. Displacement depends only on the end points; path length depends on the whole path and is never less than |Δx|.

• Easy : Position changes from 3 m to 11 m ⇒ Δx = 11 − 3 = +8 m.
• Medium : A boy walks 40 m east then 15 m west. Path length = 55 m; Δx = 40 − 15 = +25 m.
• Exam : A scooter goes 12 km east in 20 min, then 4 km west in 10 min. Path = 16 km; Δx = +8 km; total time = 30 min = 0.5 h.

Rule 2: Average velocity and average speed — average velocity = Δx ÷ Δt (a vector quantity, can be positive, negative or zero); average speed = path length ÷ Δt (never negative). Average speed is always greater than or equal to the magnitude of average velocity.

• Easy : 100 m in 20 s along a straight line ⇒ 5 m s⁻¹.
• Medium : For the boy above in 50 s : average speed = 55 ÷ 50 = 1.1 m s⁻¹; average velocity = 25 ÷ 50 = 0.5 m s⁻¹.
• Exam : For the scooter : average speed = 16 ÷ 0.5 = 32 km h⁻¹; average velocity = 8 ÷ 0.5 = 16 km h⁻¹ (east).

Rule 3: The particle idea — If the size of an object is very small compared with the distances it travels, it can be treated as a point (a particle). A 200 m train on a 400 km journey is a particle; the same train at a platform is not.

• Easy : A bus travelling from Pune to Mumbai can be treated as a particle.
• Medium : Returning to the starting point gives Δx = 0 and average velocity 0, even though the path length and average speed are not zero.
• Exam : A runner completes one full lap of a 400 m track in 80 s. Average speed = 400 ÷ 80 = 5 m s⁻¹; average velocity = 0.

Spot the difference

In the figure : The dashed tangent shows how steeply the position is changing at the marked instant.

Rule 1: Reading an x–t graph — A horizontal line means rest; a straight sloping line means uniform velocity; a steeper line means a larger velocity; a curve means the velocity is changing; a line sloping downwards means motion in the negative direction.

• Easy : x goes from 0 to 30 m in 6 s along a straight line ⇒ v = 30 ÷ 6 = 5 m s⁻¹.
• Medium : For x = 4t² (x in m, t in s) : v = dx ÷ dt = 8t. At t = 3 s, v = 24 m s⁻¹.
• Exam : For x = 2t³ − 3t : v = 6t² − 3. At t = 2 s, v = 21 m s⁻¹; v = 0 when 6t² = 3, i.e. t = 0.71 s.

Rule 2: Average and instantaneous values — In uniform motion the average and instantaneous velocities are equal. In non-uniform motion they differ, and the average over an interval is the slope of the chord between the two points.

• Easy : x = 4t² : x(1) = 4 m, x(3) = 36 m.
• Medium : Average velocity from t = 1 s to t = 3 s = (36 − 4) ÷ 2 = 16 m s⁻¹, while v at t = 3 s is 24 m s⁻¹.
• Exam : Because v = 8t increases uniformly, the average velocity over 1 s to 3 s equals the instantaneous velocity at the middle instant t = 2 s : 8 × 2 = 16 m s⁻¹.

Rule 3: Instantaneous speed — speed = |v|. For an instant (unlike a long interval) speed always equals the magnitude of the velocity.

• Easy : v = −12 m s⁻¹ means speed 12 m s⁻¹ moving in the negative direction.
• Medium : A speedometer shows speed only; it does not show direction.
• Exam : A car moves with v = 15 m s⁻¹ to the east and later v = −15 m s⁻¹ ; both have the same speed but opposite velocities.

Spot the difference

In the figure : The car speeds up for 6 s, moves steadily for 4 s and stops in 2 s; the area under the line is the displacement.

Rule 1: Computing acceleration — a = (v − v₀) ÷ t for uniform acceleration. If v and a have the same sign the object speeds up; if opposite, it slows down.

• Easy : 0 to 20 m s⁻¹ in 5 s ⇒ a = 4 m s⁻².
• Medium : 25 to 5 m s⁻¹ in 4 s ⇒ a = (5 − 25) ÷ 4 = −5 m s⁻² (a slowing down).
• Exam : In the figure, the last 2 s take the velocity from 12 m s⁻¹ to 0 ⇒ a = −6 m s⁻².

Rule 2: Area under the v–t graph — The area between the curve and the time axis equals the displacement; areas below the axis are negative. For distance travelled add the areas without signs.

• Easy : Constant 10 m s⁻¹ for 8 s ⇒ area = 80 m.
• Medium : Velocity rises uniformly from 0 to 12 m s⁻¹ in 6 s ⇒ area = ½ × 6 × 12 = 36 m.
• Exam : The full figure : 36 + (4 × 12) + (½ × 2 × 12) = 36 + 48 + 12 = 96 m.

Rule 3: Zero velocity does not mean zero acceleration — The sign of a depends on the direction you call positive; compare it with the sign of v to know whether the object speeds up or slows down. A ball thrown upwards has v = 0 at the top but a = g downwards.

• Easy : Taking up as positive, a = −9.8 m s⁻² for both the rising and the falling ball.
• Medium : Rising : v positive, a negative ⇒ slowing down. Falling : v negative, a negative ⇒ speeding up.
• Exam : For uniform motion a = 0 and the v–t graph is a horizontal line; for uniform acceleration the v–t graph is a sloping straight line and the x–t graph is a parabola.

Spot the difference

Rule 1: The three equations — v = v₀ + at; x = x₀ + v₀t + ½at² (displacement x − x₀); v² = v₀² + 2a(x − x₀). Each equation misses out one quantity; pick the one that does not contain the unknown you do not need. All quantities carry signs.

• Easy : From rest with a = 3 m s⁻² for 8 s ⇒ v = 24 m s⁻¹; displacement = ½ × 3 × 64 = 96 m.
• Medium : A car at 30 m s⁻¹ brakes with a = −6 m s⁻². Time to stop = 30 ÷ 6 = 5 s; distance = 30² ÷ (2 × 6) = 75 m.
• Exam : A train slows from 72 km h⁻¹ to 36 km h⁻¹ in 200 m. v₀ = 20 m s⁻¹, v = 10 m s⁻¹ ⇒ a = (10² − 20²) ÷ (2 × 200) = −0.75 m s⁻²; time = (10 − 20) ÷ (−0.75) = 13.3 s.

Rule 2: Free fall — Falling under gravity alone, every object has the same acceleration g = 9.8 m s⁻² downwards. Dropped from rest : v = gt, y = ½gt², v² = 2gy. Thrown upwards (up positive, a = −g) : H = v₀² ÷ 2g, time to the top = v₀ ÷ g, total time = 2v₀ ÷ g, and it returns with the same speed.

• Easy : Dropped from 19.6 m : t = √(2 × 19.6 ÷ 9.8) = 2 s; v = 9.8 × 2 = 19.6 m s⁻¹.
• Medium : Thrown up at 24.5 m s⁻¹ : H = 24.5² ÷ (2 × 9.8) = 30.6 m; time up = 2.5 s; total time = 5 s.
• Exam : A stone dropped from a tower reaches the ground in 3 s. Height = ½ × 9.8 × 9 = 44.1 m. In the last second it covers 44.1 − ½ × 9.8 × 4 = 44.1 − 19.6 = 24.5 m.

Rule 3: Galileo's odd-number rule and stopping distance — From rest, the distances covered in successive equal seconds are in the ratio 1 : 3 : 5 : 7 … The stopping distance of a vehicle is d = v₀² ÷ (2|a|), so doubling the speed makes it 4 times larger.

• Easy : Distances in successive seconds from rest : 4.9 m, 14.7 m, 24.5 m — ratio 1 : 3 : 5.
• Medium : At 15 m s⁻¹ (54 km h⁻¹) with a = −5 m s⁻² : d = 225 ÷ 10 = 22.5 m.
• Exam : At 30 m s⁻¹ (108 km h⁻¹) with the same braking : d = 900 ÷ 10 = 90 m, four times the distance for double the speed.

Derivation : v² = v₀² + 2ax

• Step 1 : From v = v₀ + at, t = (v − v₀) ÷ a.
• Step 2 : For uniform a the average velocity is (v₀ + v) ÷ 2, so x = (v₀ + v) ÷ 2 × t.
• Step 3 : Substitute t : x = (v₀ + v)(v − v₀) ÷ (2a) = (v² − v₀²) ÷ (2a).
• Step 4 : Rearranging gives v² = v₀² + 2ax.

Spot the difference

Rule 1: Relative velocity on a line — Take one direction as positive and subtract the velocity of the reference object. If both move the same way, the relative speed is the difference of speeds; if they move in opposite directions, it is the sum.

• Easy : A moves at +20 m s⁻¹ and B at +15 m s⁻¹ ⇒ v_AB = +5 m s⁻¹ and v_BA = −5 m s⁻¹.
• Medium : Two trains approach each other, A at +60 km h⁻¹ and B at −40 km h⁻¹ ⇒ v_AB = 60 − (−40) = 100 km h⁻¹. If they are 500 m apart, they meet after 0.5 ÷ 100 = 0.005 h = 18 s.
• Exam : A car at 90 km h⁻¹ overtakes a truck at 54 km h⁻¹. Relative speed = 36 km h⁻¹ = 10 m s⁻¹. To pass completely it must cover the truck length 18 m plus its own 4 m = 22 m, so the time = 22 ÷ 10 = 2.2 s.

Rule 2: Equal velocities — If two objects have equal velocities, the relative velocity is zero; each seems at rest to the other.

• Easy : Two cars both at 60 km h⁻¹ in the same direction ⇒ v_AB = 0.
• Medium : Passengers inside a moving train are at rest relative to the train but move at the train's speed relative to the ground.
• Exam : x–t graphs of two objects with equal slopes are parallel lines; the gap between them stays constant.

Spot the difference