Babu Sir’s marked questions, fully worked

• Total track length s = 60 km
• First leg: s₁ = 30 km at v₁ = 30 km h⁻¹
• Second leg: s₂ = 30 km at v₂ = ?
• Required average speed for the whole trip v̄ = 40 km h⁻¹
• Implicit: both speeds uniform, motion in one straight line with no reversal, so distance = displacement

• The speed v₂ needed on the second 30 km

Average speed is total distance divided by total time — never the average of the two speeds. Since the two legs cover the same distance but at different speeds, they take different times, so the slower leg carries more weight.

The trick is to stop thinking about speed and think about a time budget. Fixing the average speed for the whole journey fixes the total time allowed. Subtract the time already spent, and whatever remains is all the time the second leg gets.

Strategy: convert the target average into a total time, subtract the time used by leg 1, then divide the remaining distance by the remaining time.

1. Total time allowed. t_total = s/v̄ = 60/40 = 1.5 h.
2. Time used by the first leg. t₁ = s₁/v₁ = 30/30 = 1 h.
3. Time left for the second leg. t₂ = 1.5 − 1 = 0.5 h.
4. Speed needed. v₂ = s₂/t₂ = 30/0.5 = 60 km h⁻¹.
5. Check. Total distance 60 km, total time 1 + 0.5 = 1.5 h, so v̄ = 60/1.5 = 40 km h⁻¹. ✓

• Harmonic-mean shortcut. For two equal distances, v̄ = 2v₁v₂/(v₁+v₂). Put 40 = 2(30)v₂/(30+v₂) ⇒ 1200 + 40v₂ = 60v₂ ⇒ v₂ = 60. One line.
• Sanity bound. The harmonic mean always lies between the two speeds and below their arithmetic mean. Since the target 40 sits above 30, the answer must exceed 40 — and it does.
• The half-time danger. If the target average had been 60 km h⁻¹, the whole 1 h budget would already be spent and no speed could work. Always check the time budget is positive before hunting for a number.
• Never do this: (30 + v₂)/2 = 40 ⇒ v₂ = 50. That is the answer for equal times, not equal distances.

• x = 4t² − 15t + 25 (x in metre, t in second)
• Coefficients carry units: 4 m s⁻², −15 m s⁻¹, 25 m
• Implicit: the negative middle term means the particle sets off in the negative x-direction

• Position, velocity and acceleration at t = 0
• The instant at which v = 0
• Whether the acceleration is uniform

Same three-rung ladder as before: differentiate for velocity, differentiate again for acceleration.

What makes this question interesting is the minus sign on the 15t. It means the particle starts off moving backwards while the acceleration pushes forwards. So the particle slows down, stops for an instant, and then reverses — exactly like a ball thrown upward.

‘Uniform acceleration’ simply means a does not depend on t. Differentiate twice: if the answer still contains a t, the acceleration is not uniform. If it is a bare number, it is.

Strategy: build v(t) and a(t) first, then read off everything the question asks by substituting or solving.

1. Position at t = 0. x(0) = 4(0) − 15(0) + 25 = 25 m.
2. Differentiate for velocity. v = dx/dt = 8t − 15 m s⁻¹.
3. Velocity at t = 0. v(0) = 8(0) − 15 = −15 m s⁻¹. The minus sign says it is moving in the −x direction at 15 m s⁻¹.
4. Differentiate again for acceleration. a = dv/dt = 8 m s⁻², positive and with no t in it.
5. Find when v = 0. 8t − 15 = 0 ⇒ t = 15/8 = 1.875 s.
6. Is a uniform? Yes. a = 8 m s⁻² is a constant, so the motion is uniformly accelerated — the three kinematic equations are legal here.
7. Where does it turn? (not asked, but worth knowing) x(1.875) = 4(3.5156) − 15(1.875) + 25 = 175/16 = 10.94 m.

• Read all three off the pattern. For x = At² + Bt + C: initial position = C, initial velocity = B, acceleration = 2A. Here that is 25, −15 and 8 — all three answers with no calculus written down.
• Turning time from the vertex formula. The particle stops at the vertex of the parabola, t = −B/2A = 15/8. Same answer, no differentiation.
• Sign reading. B and 2A have opposite signs here, which is the algebraic fingerprint of ‘goes one way, comes back’. If they had matched signs, the particle would never turn.
• Uniform-acceleration test in one glance. Highest power of t is 2 → acceleration constant. Highest power 3 or more → not constant. Problem 2 had t³, which is why its acceleration was 15t.
• Beware of the word ‘displacement’ here. The question calls x a displacement but gives it a non-zero value at t = 0, so it is really a position measured from a fixed origin. Treat it as a position.

• t = √x + 3, with x in metres and t in seconds
• The force is constant, so the acceleration is constant
• Implicit: √x ≥ 0, so the relation only describes times t ≥ 3 s — before that the particle is on the incoming branch

• The displacement x at the instant when v = 0

The equation is written the wrong way round. It gives t in terms of x, but v = dx/dt needs x in terms of t. So the first move is not calculus at all — it is algebra: make x the subject.

To do that, isolate the square root and then square both sides. Squaring is legal here because both sides are non-negative once the 3 has been moved across.

After that it is the ordinary ladder: differentiate, set the velocity to zero, solve for t, and put that t back into x.

Strategy: rearrange → differentiate → set v = 0 → solve for t → substitute back into x.

1. Isolate the root. t = √x + 3 ⇒ √x = t − 3.
2. Square both sides. x = (t − 3)².
3. Expand (optional but useful). x = t² − 6t + 9.
4. Differentiate. v = dx/dt = 2(t − 3) = 2t − 6 m s⁻¹.
5. Set v = 0. 2t − 6 = 0 ⇒ t = 3 s.
6. Substitute back. x = (3 − 3)² = 0² = 0.
7. Interpret. The particle is at the origin at the moment it stops, so its displacement measured from the start of the motion is zero.

• Read the vertex, skip the calculus. x = (t − 3)² is a parabola with its minimum at t = 3, and a perfect square has minimum value 0. Both answers are visible in the bracket without differentiating at all.
• Confirm the acceleration is constant. a = dv/dt = 2 m s⁻², which agrees with the ‘constant force’ in the question. If your a had come out with a t in it, you would know the rearrangement went wrong.
• Never differentiate the equation as printed. Writing dt/dx = 1/(2√x) and then flipping it does work, but it is far slower and invites a reciprocal error. Rearrange first, every time.
• Careful with the word displacement. The answer is 0 m, which is not the same as saying the particle never moved. It travelled in from x = 9 m (at t = 0) and arrived at the origin — the distance covered is 9 m, but the displacement asked for is 0.

• Four position–time relations, each a quadratic in t
• All are one-dimensional motions along a straight line
• Implicit: every one of them therefore has constant acceleration, because the highest power of t is 2

• Q1: v(t) and a in symbols
• Q2: the initial velocity, i.e. v at t = 0
• Q3: v and a at t = 2 s
• Q4: initial velocity, velocity at t = 4 s, and acceleration

All four are the same question wearing different clothes. Any quadratic position law can be written x = C + Bt + At², and differentiating it twice gives:

v = B + 2At    a = 2A

Compare that with the kinematic equation x = x₀ + ut + ½at² and the three coefficients name themselves: C is the starting position, B is the initial velocity, and 2A is the acceleration.

So you never actually have to differentiate these. You read the answers off the coefficients.

Strategy: match each expression to C + Bt + At², then quote v = B + 2At and a = 2A and substitute any time required.

1. Q1. Here C = α, B = −β, A = γ.
   v = dx/dt = −β + 2γt and a = dv/dt = 2γ.
2. Q2. v = dx/dt = −5 + 12t. At t = 0: v = −5 units. The minus sign means it sets off in the negative x-direction.
3. Q3, velocity. v = 8 + 14t; at t = 2: v = 8 + 28 = 36 m s⁻¹.
4. Q3, acceleration. a = 14 m s⁻² — constant, so the value at t = 2 s is the value at every instant.
5. Q4 (i). v = 180 + 100t; at t = 0: v = 180 m s⁻¹.
6. Q4 (ii). At t = 4: v = 180 + 400 = 580 m s⁻¹.
7. Q4 (iii). a = 100 m s⁻².

• The coefficient trick. Initial velocity = coefficient of t. Acceleration = twice the coefficient of t². That answers Q2, and both parts of Q4 (i) and (iii), with no working at all.
• Sign discipline. In Q1 and Q2 the t-coefficient is negative, so the initial velocity is negative. Carry the minus sign into your answer — ‘5 units’ would be marked wrong; the answer is −5 units.
• Watch the units in Q1. Since x is a length, α is in m, β in m s⁻¹ and γ in m s⁻². Then a = 2γ comes out in m s⁻² automatically — a free dimensional check.
• Constant acceleration means one answer for all times. In Q3 the question says ‘at t = 2 s’ for both parts, but the acceleration does not depend on t. Do not waste time substituting.
• Recognise the disguise. Q4 is simply a body with u = 180 m s⁻¹ and a = 100 m s⁻². Written as x = ut + ½at² it would look completely familiar.

• Distance in the 2nd second: S₂ = 12 m
• Distance in the 4th second: S₄ = 20 m
• Implicit: uniform acceleration throughout (otherwise the n-th-second formula would not apply), motion in a straight line with no reversal
• Unknowns to find first: initial velocity u and acceleration a

• The distance covered in the 5th second, S₅
• The distance covered during the 4 seconds that follow the 5th second, i.e. from t = 5 s to t = 9 s

‘Distance in the n-th second’ means the one-second slice between t = n − 1 and t = n — not the total up to n seconds. Subtracting the two totals gives the standard formula:

S_n = u + (a/2)(2n − 1)

Notice this is linear in n. So the successive slices form an arithmetic progression whose common difference is exactly a. Two given slices are therefore enough to pin down both u and a.

The second part asks for something different: a four-second stretch. For that you go back to total distances and subtract.

Strategy: write the n-th-second formula for n = 2 and n = 4, solve the pair simultaneously for u and a, then answer both parts.

1. Write the formula for the 2nd second. S₂ = u + (a/2)(2×2 − 1) = u + 3a/2 = 12  …(i)
2. Write it for the 4th second. S₄ = u + (a/2)(2×4 − 1) = u + 7a/2 = 20  …(ii)
3. Subtract (i) from (ii). (7a/2 − 3a/2) = 20 − 12 ⇒ 2a = 8 ⇒ a = 4 m s⁻².
4. Back-substitute into (i). u + 3(4)/2 = 12 ⇒ u + 6 = 12 ⇒ u = 6 m s⁻¹.
5. Answer part one. S₅ = u + (a/2)(2×5 − 1) = 6 + 2(9) = 6 + 18 = 24 m.
6. Set up part two. The 4 seconds after the 5th second run from t = 5 s to t = 9 s, so the answer is s₉ − s₅ using total distances.
7. Compute the two totals.
   s₉ = 6(9) + ½(4)(81) = 54 + 162 = 216 m
   s₅ = 6(5) + ½(4)(25) = 30 + 50 = 80 m
8. Subtract. 216 − 80 = 136 m.

• The gap between slices is the acceleration. The 2nd and 4th seconds are two apart, and the distances differ by 8 m, so a = 8/2 = 4 m s⁻² immediately — no simultaneous equations needed.
• Then step the AP forward. Slices go 8, 12, 16, 20, 24 … Once you know the 4th second is 20 m and the common difference is 4, the 5th second is 20 + 4 = 24 m in your head.
• Part two by the AP as well. The 6th, 7th, 8th and 9th seconds are 28, 32, 36 and 40 m. Their sum is 28 + 32 + 36 + 40 = 136 m. ✓ Same answer, and it doubles as a check on the subtraction method.
• Sum-of-an-AP shortcut. Four terms, first 28 and last 40, so the sum is 4 × (28 + 40)/2 = 4 × 34 = 136.
• The classic trap. ‘In the 5th second’ (24 m) is not ‘in 5 seconds’ (80 m). Both numbers will be in the options. Circle the words in the before you start.

• Speed–time graph, straight lines, triangular shape
• v = 0 at t = 0
• v = 12 m s⁻¹ at t = 5 s (the peak)
• v = 0 at t = 10 s
• Crucially: the particle moves along a fixed direction, so it never reverses and distance = displacement

• (i) distance travelled from t = 0 to t = 10 s
• (ii) average speed over that interval
• (iii) the time at which the speed was minimum
• (iv) the time at which the speed was maximum

On a speed–time graph the area underneath is the distance travelled. Here the area is a triangle, so you need nothing more than ½ × base × height.

The phrase along a fixed direction is doing important work. It tells you the particle never turns round, so there is no negative area to worry about and the average speed and the magnitude of the average velocity agree.

Parts (iii) and (iv) need no calculation at all — you read them off the shape. The lowest points of the graph are where the speed is least; the peak is where it is greatest.

Strategy: take the area for (i), divide by the total time for (ii), and simply look at the graph for (iii) and (iv).

1. (i) Identify the triangle. Base along the time axis from t = 0 to t = 10 s, so base = 10 s. Height is the peak speed, 12 m s⁻¹.
2. Take the area. s = ½ × base × height = ½ × 10 × 12.
3. Compute. s = 60 m.
4. (ii) Average speed. = total distance / total time = 60/10 = 6 m s⁻¹.
5. (iii) Minimum speed. The graph touches zero at both ends, so the speed is least (v = 0) at t = 0 and at t = 10 s.
6. (iv) Maximum speed. The apex of the triangle is at t = 5 s, where v = 12 m s⁻¹.

• Triangle average is always half the peak. For any speed–time triangle that starts and ends at zero, the average speed is exactly half the maximum — here 12/2 = 6. You can answer (ii) before you answer (i).
• Cross-check with the two accelerations. Rising: a = 12/5 = +2.4 m s⁻². Falling: a = −12/5 = −2.4 m s⁻². Equal magnitudes, so the triangle is symmetric and each half contributes 30 m.
• Split it if the triangle is not symmetric. Two right triangles, ½(5)(12) + ½(5)(12) = 30 + 30 = 60. Slower, but it never fails.
• Watch the axis label. This graph says speed, not velocity, so nothing dips below the axis and no signed areas arise. If it had said velocity and gone negative, distance and displacement would part company (see Problem 12).
• ‘Minimum’ has two answers here. Both endpoints qualify. Quoting only t = 0 loses half the mark.

• OA: velocity rises from 0 to 20 m s⁻¹ between t = 0 and t = 2 s
• AB: velocity constant at 20 m s⁻¹ from t = 2 s to t = 5 s
• BC: velocity falls from 20 m s⁻¹ to 0 between t = 5 s and t = 10 s
• Implicit: velocity is positive throughout, so the object never reverses and distance = displacement

• The acceleration during each of the three phases
• The total distance travelled in the 10 s
• The average velocity over the 10 s

Two rules do everything on a velocity–time graph:

Slope = acceleration. A rising line means speeding up; a flat line means zero acceleration; a falling line means slowing down (negative acceleration).

Area = displacement. Chop the shape into rectangles and triangles and add.

Notice that during AB the object is not at rest — it is moving fast and steadily. A horizontal line on a v–t graph means constant velocity; a horizontal line on an x–t graph would mean at rest. Reading the wrong axis is the classic error here.

Strategy: take the slope of each of the three segments, then take the area of each and add them up.

1. Acceleration OA. a₁ = Δv/Δt = (20 − 0)/(2 − 0) = +10 m s⁻².
2. Acceleration AB. The line is flat, so Δv = 0 and a₂ = 0. Constant velocity.
3. Acceleration BC. a₃ = (0 − 20)/(10 − 5) = −20/5 = −4 m s⁻² — a retardation of 4 m s⁻².
4. Area OA (triangle). ½ × 2 × 20 = 20 m.
5. Area AB (rectangle). 3 × 20 = 60 m.
6. Area BC (triangle). ½ × 5 × 20 = 50 m.
7. Total distance. 20 + 60 + 50 = 130 m.
8. Average velocity. v̄ = 130/10 = 13 m s⁻¹.

• Trapezium in one shot. The whole shape is a trapezium with parallel sides 10 s (the base OC) and 3 s (the top AB), and height 20 m s⁻¹. So s = ½(10 + 3)(20) = ½(13)(20) = 130 m. One line instead of three areas.
• Average velocity without the distance. Once you spot the trapezium, the average velocity is ½(10 + 3)(20)/10 = 13 directly — and notice 13 sits sensibly between 0 and the peak 20.
• Sign the retardation properly. Write a₃ = −4 m s⁻², or say ‘a retardation of 4 m s⁻²’. Writing ‘acceleration = 4’ without the sign is the commonest slip on this question.
• Steepness tells you the story. OA is steeper than BC, so the object speeds up faster than it slows down — which is why the braking phase needs 5 s against the 2 s of acceleration.
• Check the units of the area. m s⁻¹ multiplied by s gives m. If your answer came out in m s⁻¹, you took a slope where you needed an area.

• 0–4 s: v rises linearly from 0 to +20 m s⁻¹
• 4–8 s: v falls linearly from +20 to −20 m s⁻¹, crossing zero at t = 6 s
• 8–10 s: v constant at +10 m s⁻¹
• Implicit: the sign change at t = 6 s means the object reverses direction, so distance and displacement will differ

• The net distance covered from t = 0 to t = 10 s
• The displacement over the same interval

Both answers come from the area under the graph, but they treat the signs differently.

Displacement = signed area. Area above the time axis counts as positive, area below counts as negative, and they can cancel.

Distance = total area with every sign made positive. The odometer never runs backwards.

So the first and most important step is to find where the graph crosses the axis, because that is the instant the object turns round. Here the line from (4, +20) to (8, −20) crosses zero exactly halfway, at t = 6 s. Splitting there is the whole technique.

Strategy: find the zero crossing, split the graph into four simple pieces, take each area with its sign, then add twice — once keeping signs, once ignoring them.

1. Find the crossing. The segment runs from +20 at t = 4 s to −20 at t = 8 s. It is symmetric, so it passes through zero at t = 6 s.
2. Piece 1: 0 to 4 s (triangle above the axis). ½ × 4 × 20 = +40 m.
3. Piece 2: 4 to 6 s (triangle above the axis). ½ × 2 × 20 = +20 m.
4. Piece 3: 6 to 8 s (triangle below the axis). ½ × 2 × 20 = 20 m of area, counted as −20 m of displacement.
5. Piece 4: 8 to 10 s (rectangle above the axis). 2 × 10 = +20 m.
6. Displacement — keep the signs. 40 + 20 − 20 + 20 = 60 m.
7. Distance — drop the signs. 40 + 20 + 20 + 20 = 100 m.

• Merge pieces 1 and 2. They are both above the axis and adjacent, so together they form one triangle from t = 0 to t = 6 s with height 20: ½ × 6 × 20 = 60 m. That saves a step.
• Distance minus displacement = twice the negative area. Here 100 − 60 = 40 = 2 × 20. ✓ A fast consistency check on any question of this type.
• Always find the zero crossing first. If you integrate straight through from 0 to 10 s you get 60 m, which is the displacement — the reversal has already silently cancelled part of the journey. Splitting at v = 0 is the entire technique.
• The jump at t = 8 s is not physical. Velocity leaping from −20 to +10 instantly would need infinite acceleration; it is a simplification of the textbook figure. Do not let it distract you — the areas are still perfectly well defined.
• Sanity bound. Distance can never be smaller than |displacement|. If your two answers come out the other way round, you have mixed up which one keeps the signs.

• Initial velocity u = 0 (from rest)
• Final velocity v = 180 km h⁻¹ — must be converted to m s⁻¹
• Time t = 25 s
• Acceleration uniform throughout, motion along a straight road

• The distance covered in the 25 s

The moment a speed arrives in km h⁻¹, convert it. Multiply by 5/18, because a kilometre is 1000 m and an hour is 3600 s, and 1000/3600 tidies to 5/18.

After that, pick the kinematic equation by asking what is missing. Acceleration is not given, so use the version without a: s = [(u + v)/2] t. Or find a first and use s = ut + ½at² — both are one line.

Because the body starts from rest, the average velocity is simply half the final velocity, and the distance is that average multiplied by the time.

Strategy: convert → find the average velocity → multiply by time. (The acceleration is a bonus you get almost free.)

1. Convert the speed. v = 180 × 5/18 = 50 m s⁻¹.
2. Find the acceleration. a = (v − u)/t = (50 − 0)/25 = 2 m s⁻².
3. Choose the distance equation. s = ut + ½at²; with u = 0 this is s = ½at².
4. Square the time. 25² = 625 s².
5. Substitute. s = ½ × 2 × 625 = 625 m.
6. Check the other way. Average velocity = (0 + 50)/2 = 25 m s⁻¹, so s = 25 × 25 = 625 m. ✓

• Half-the-final-speed trick. From rest under uniform acceleration, s = (v/2) × t. Here 25 × 25 = 625 — and you never need the acceleration at all.
• Handy coincidence. With a = 2 m s⁻² the formula s = ½at² collapses to s = t². So 25² = 625 straight away.
• Common conversions worth memorising. 36 → 10, 54 → 15, 72 → 20, 90 → 25, 108 → 30, 180 → 50 km h⁻¹ to m s⁻¹.
• The forgotten-conversion alarm. Using 180 raw would give a = 7.2 and s = 2250 — a car covering 2.25 km in 25 s. Absurd answers are your signal that the 5/18 was skipped.
• Reality check. 625 m in 25 s is an average of 90 km h⁻¹, which is exactly half of 180. Consistent with starting from rest.

• Initial velocity u = 16 m s⁻¹
• Final velocity v = 0 (comes to rest)
• Distance inside the trunk s = 0.4 m
• Implicit: the retardation is assumed uniform over that 0.4 m, and the motion is in a straight line

• The time taken during the retardation
• (and, as an intermediate, the retardation itself)

Two unknowns lurk here — the acceleration and the time — and you are given u, v and s. So attack in two steps, choosing each equation by what is missing.

Step one: time is missing, so use v² = u² + 2as to get a.

Step two: distance is now spare, so use v = u + at to get t.

Because the bullet is slowing down, a will come out negative. That negative value is correct and must be carried through; the word ‘retardation’ refers to its magnitude.

Strategy: the no-t equation first for the acceleration, then the no-s equation for the time.

1. Write the no-time equation. v² = u² + 2as.
2. Substitute. 0² = 16² + 2a(0.4), i.e. 0 = 256 + 0.8a.
3. Solve for a. 0.8a = −256 ⇒ a = −320 m s⁻². The retardation is 320 m s⁻².
4. Write the no-distance equation. v = u + at.
5. Substitute. 0 = 16 + (−320)t.
6. Solve for t. 320t = 16 ⇒ t = 16/320 = 0.05 s.
7. Check with the average-velocity route. Average velocity = (16 + 0)/2 = 8 m s⁻¹, so t = s/v̄ = 0.4/8 = 0.05 s. ✓

• Skip the acceleration entirely. For a body brought to rest uniformly, t = 2s/u, because the average velocity is exactly u/2. Here t = 2(0.4)/16 = 0.8/16 = 0.05 s in one step. This is the fastest route and it is worth memorising.
• And the matching one for retardation. a = u²/2s = 256/0.8 = 320 m s⁻² in magnitude.
• Decimal care. 2 × 0.4 = 0.8, not 8. Dividing 256 by 8 gives 32 and an answer ten times too small — a very common slip. Convert 0.4 m to 40 cm only if you convert everything else too.
• Feel the size of the number. 320/9.8 ≈ 33, so the bullet decelerates at about 33 g. That is why the stopping time is only five hundredths of a second.
• Sign discipline in the answer. Write a = −320 m s⁻² in the working but describe it as ‘a retardation of 320 m s⁻²’ in the final sentence.

• v_A = 10 m s⁻¹
• v_B = 6 m s⁻¹
• Both cars move in the same direction along a straight road
• A is ahead of B
• Convention: take the common direction of travel as positive

• v_AB, the velocity of A as seen from B
• v_BA, the velocity of B as seen from A

Relative velocity is a subtraction: v_AB = v_A − v_B. Read it as ‘velocity of A as seen from B’.

Because both cars travel the same way, both velocities carry the same sign, and subtracting them gives a small number. If they had been travelling in opposite directions, one velocity would be negative and subtracting would have added the speeds.

Swapping the two labels flips the sign but never changes the size. That is not a coincidence — the two drivers must agree on how fast the gap between them is changing, and only disagree on which way round it is.

Strategy: fix the positive direction, subtract in the order the subscripts tell you, and then say in words what the sign means.

1. Fix the sign convention. Take the direction both cars are travelling as positive, so v_A = +10 and v_B = +6 m s⁻¹.
2. Velocity of A relative to B. v_AB = v_A − v_B = 10 − 6 = +4 m s⁻¹.
3. Say what the sign means. Positive, so the driver of B sees car A moving ahead of him at 4 m s⁻¹ — the gap opens by 4 m every second.
4. Velocity of B relative to A. v_BA = v_B − v_A = 6 − 10 = −4 m s⁻¹.
5. Say what that sign means. Negative, so the driver of A (looking back) sees car B lagging behind at 4 m s⁻¹.
6. Confirm the reversal rule. v_BA = −v_AB, as it always must be.

• Same direction subtract, opposite directions add. If B had been coming the other way, v_AB = 10 − (−6) = 16 m s⁻¹ — four times larger. That contrast is what NEET tests.
• Read the subscripts as a sentence. ‘AB’ means A first, B second, so it is v_A − v_B in that order. Getting the order backwards is a recorded recurring slip — it flips the sign and loses the mark.
• The magnitudes must match. If your two answers had different sizes, something went wrong. Both drivers watch the same gap change at the same rate.
• Free extra result. Since neither car accelerates, a_AB = 0, so in B’s frame car A simply drifts away at a steady 4 m s⁻¹ forever. If they started 100 m apart, that gap becomes 140 m after 10 s.
• The fact that A is ahead does not enter the arithmetic. It only matters for describing the picture, not for computing the relative velocity.

• Radius of the circular track R = 100 m
• Time for one complete revolution T = 4 minutes
• Implicit: one complete revolution, so the finishing point is the starting point

• (i) The average speed over one revolution
• (ii) The average velocity over one revolution

This question exists to separate two words that sound the same in everyday English.

Average speed = total path length ÷ time. The path here is the full circumference, 2πR — every metre of track the cyclist actually rode.

Average velocity = displacement ÷ time. Displacement is the straight-line arrow from start to finish. After one complete lap that arrow has zero length, because she is back exactly where she started.

So the average velocity is zero — and this says nothing at all about whether she was moving. Her speed was never zero for a single instant.

Strategy: compute the circumference for the path, note that the displacement is zero, and divide each by the same time.

1. Find the path length. One revolution means one full circumference: 2πR = 2π(100) = 200π m.
2. (i) Average speed. = 200π/4 = 50π metre/minute.
3. Put a number on it. 50π ≈ 157 m/min. In SI that is 200π/240 ≈ 2.62 m s⁻¹.
4. Find the displacement. The cyclist ends at the same point she started, so the straight-line separation is 0.
5. (ii) Average velocity. = 0/4 = 0.
6. Interpret. Zero average velocity, but a very much non-zero average speed — the two answers together are the point of the question.

• Any closed path gives zero average velocity. Circle, square, figure-of-eight — if you finish where you started, the average velocity is zero. Ask ‘did she come back?’ before calculating anything.
• Watch the units the book wants. The printed answer is in metre/minute, so leaving the time as 4 minutes is correct here. Only convert to seconds if the question asks for SI units.
• Leave π in the answer. 50π is exact; 157 is rounded. Options in NEET are usually written with π still in them.
• Half a lap is the useful variant. Then the displacement is the diameter 2R = 200 m while the path is πR = 100π m, giving a ratio of π/2 ≈ 1.57. Papers alternate between the full lap and the half lap.
• Do not answer ‘zero and zero’. Average speed can only be zero if the body never moved at all.