Chapter 2 · Solved paper40 marks

Model Test – Solved ഉത്തരങ്ങൾ

Every question of the model paper, in the same order, with the full answer in exam layout (Given → Formula → Substitution → Answer with unit), the marking scheme, a step-by-step animation for every numerical and graph question and a link to the page where the idea is taught. Try each one again before you open the answer, and mark your own paper honestly.

➕ Sign convention used in every answer

East, forward and upward are positive; west, backward and downward are negative (പോസിറ്റീവ് / നെഗറ്റീവ്). A negative acceleration is a retardation (മന്ദീകരണം): a = −5 m/s² means a retardation of 5 m/s².

SECTION AAnswer all · 8 × 1 = 8 marks

The SI unit of acceleration is .

(1 mark)

💡 Unit of velocity ÷ unit of time.

✅ Answer: m/s²

Acceleration (ത്വരണം) a = (v − u)/t = change in velocity ÷ time, so its unit is (m/s) ÷ s = m/s².

Marking: 1 mark. “m/s” gets no mark.

📖 Review this on Page 8 · Unit of acceleration

Which one of the following is a scalar quantity?

(a) displacement(b) velocity(c) speed(d) acceleration

(1 mark)

💡 Which one does not need a direction?

✅ Answer: (c) speed

Displacement, velocity and acceleration need a direction – they are vector quantities (സദിശ അളവുകൾ). Speed (വേഗം) has only a magnitude – a scalar quantity (അദിശ അളവുകൾ).

Marking: 1 mark for (c).

📖 Review this on Page 5 · Vectors and scalars · Page 6 · Speed and velocity

72 km/h = m/s.

(1 mark)

💡 Multiply by 5/18.

✅ Answer: 20 m/s

1 km/h = 1000 m ÷ 3600 s = 5/18 m/s, so 72 km/h = 72 × 5/18 = 20 m/s.

Marking: 1 mark.

📖 Review this on Page 2 · km/h → m/s · Page 11 · Key box: 5/18

A car goes round a circular track at a constant speed. Its velocity is (uniform / non-uniform).

(1 mark)

💡 Does the direction of motion change?

✅ Answer: non-uniform

The speed stays the same, but the direction of motion (ചലനദിശ) keeps changing on a circular path (വർത്തുളപാത). A change of direction is a change of velocity, so the velocity is non-uniform (അസമപ്രവേഗം).

Marking: 1 mark.

📖 Review this on Page 7 · Car C on the circular track

In a velocity–time graph, the area under the graph for a time interval gives the in that interval.

(1 mark)

💡 Height × width of the shaded region = velocity × time.

✅ Answer: displacement

Area under the velocity–time graph (പ്രവേഗ-സമയ ഗ്രാഫ്) = velocity × time = displacement (സ്ഥാനാന്തരം).

Marking: 1 mark (“distance” for motion in one direction: ½ mark).

📖 Review this on Page 10 · Area under the v–t graph

Find the relation and fill in: v = u + at : velocity–time relation :: v² = u² + 2as :

(1 mark)

💡 Which quantities appear in v² = u² + 2as? Is t one of them?

✅ Answer: displacement–velocity relation

v = u + at relates velocity and time; s = ut + ½at² relates displacement and time; v² = u² + 2as relates displacement and velocity (it has no t).

Marking: 1 mark.

📖 Review this on Page 11 · The three equations

A road sign with a red triangular border, such as “School ahead”, is a sign.

(1 mark)

💡 Red circle, red triangle or blue rectangle – which one warns?

✅ Answer: cautionary sign (കോഷനറി സൈനുകൾ)

Red circle – mandatory (മാൻഡേറ്ററി സൈനുകൾ, must obey); red triangle – cautionary (കോഷനറി സൈനുകൾ, a warning of what is ahead); blue rectangle – informatory (ഇൻഫോമേറ്ററി സൈനുകൾ).

Marking: 1 mark.

📖 Review this on Page 9 · Road signs

An athlete runs exactly one full round of a circular track and stops at her starting point. Her displacement is

(a) the circumference(b) the diameter(c) zero(d) twice the circumference

(1 mark)

💡 Where does she start and where does she stop?

✅ Answer: (c) zero

Displacement is the straight-line distance from the initial position to the final position. They are the same point, so the displacement is zero – even though the distance travelled is one full circumference (Let us assess Q5: 440 m, displacement 0).

Marking: 1 mark for (c).

📖 Review this on Page 5 · Displacement · Page 13 · Let us assess Q5

SECTION BAnswer all · 5 × 2 = 10 marks

Write two differences between distance and displacement.

(2 marks)

💡 Think of the child walking B → D → C on the posts.

✅ Answer: any two differences

Distance ദൂരം	Displacement സ്ഥാനാന്തരം
Total length of the path travelled	Shortest straight-line distance from the initial to the final position, with direction
Scalar – magnitude only	Vector – magnitude and direction
Depends on the path	Does not depend on the path
Never zero once the object moves; always positive	Can be positive, negative or zero

They are equal in magnitude only when the object moves along a straight line in one direction.

Marking: 1 mark for each correct difference (any two rows of the table).

📖 Review this on Page 5 · Table 2.2

10.

A KSRTC bus starts from rest and reaches a velocity of 54 km/h in 10 s. Find its acceleration.

(2 marks)

💡 Convert km/h to m/s first: multiply by 5/18.

🎬 Step by step

✅ Answer: 1.5 m/s²

Given: u = 0 (starts from rest); v = 54 km/h = 54 × 5/18 m/s = 15 m/s; t = 10 s

Formula: a = (v − u)/t

Substitution: a = (15 − 0)/10

Answer: a = 1.5 m/s²

Marking: 1 mark for converting 54 km/h to 15 m/s, 1 mark for the answer with unit.

📖 Review this on Page 8 · a = (v − u)/t · Page 2 · km/h → m/s

11.

Many accidents to pedestrians happen at night and in the monsoon. (a) Write two rules a pedestrian must follow while crossing a road. (b) Why should pedestrians wear light-coloured clothes at night?

(2 marks)

💡 Think about what the driver can see in the dark.

✅ Answer: (a) any two rules (b) drivers can see them

(a) Any two: cross only at the zebra crossing (സീബ്രാ ക്രോസിങ്), following the signal; look right, left and right again before crossing; walk on the right side of the road, facing the traffic; use footpaths and foot-over-bridges; do not run across or use a mobile phone while crossing.

(b) In poor light, drivers cannot easily see people in dark clothes. Light or bright colours (and reflective strips) are seen from far away, so the driver gets time to slow down and stop.

Marking: ½ mark for each rule in (a), 1 mark for the reason in (b).

📖 Review this on Page 9 · Pedestrian rules · Page 14 · Let us assess Q10

12.

The position–time graph of an autorickshaw moving along a straight road is shown. (a) What kind of velocity does it have? Give a reason. (b) Find its velocity.

(2 marks)

💡 Pick two points on the line and divide the change in position by the change in time.

🎬 Step by step

✅ Answer: (a) uniform velocity (b) 10 m/s

(a) The graph is a sloping straight line (ചരിവുള്ള നേർരേഖ): the autorickshaw covers equal displacements (20 m) in equal intervals of time (2 s), so it has a uniform velocity (സമപ്രവേഗം).

Given: from the graph: at t₁ = 2 s, s₁ = 20 m; at t₂ = 8 s, s₂ = 80 m

Formula: v = (s₂ − s₁)/(t₂ − t₁)

Substitution: v = (80 − 20)/(8 − 2) = 60/6

Answer: v = 10 m/s

Marking: 1 mark for (a) with the reason, 1 mark for (b) with unit.

📖 Review this on Page 10 · Graph 2.4 (slope)

13.

A train 150 m long, moving at 54 km/h, crosses a platform 300 m long. How long does it take to cross the platform completely?

(2 marks)

💡 Do not forget the length of the train itself.

🎬 Step by step

✅ Answer: 30 s

To cross completely, the engine must travel the platform’s length plus the train’s own length, until the last coach leaves the platform.

Given: length of train = 150 m; length of platform = 300 m; v = 54 km/h = 54 × 5/18 = 15 m/s; s = 300 + 150 = 450 m

Formula: t = s/v

Substitution: t = 450/15

Answer: t = 30 s

Marking: 1 mark for s = 450 m (adding the train’s length) and the conversion, 1 mark for the answer.

📖 Review this on Page 6 · Train and bridge · Page 15 · Train calculator

SECTION CAnswer all · 4 × 3 = 12 marks

14.

Four electric posts P, Q, R and S stand 50 m apart, in this order, along a straight road running from west to east. A cyclist starts at P, rides to S, turns back and stops at Q. The whole trip takes 50 s. Find (a) the distance travelled and the displacement (b) the speed (c) the velocity of the cyclist.

(3 marks)

💡 Distance adds every metre ridden; displacement is only from P to Q.

🎬 Step by step

✅ Answer: (a) 250 m; 50 m east (b) 5 m/s (c) 1 m/s east

Take east as positive. PS = 3 × 50 = 150 m; S back to Q = 2 × 50 = 100 m.

Given: PS = 150 m, SQ = 100 m, PQ = 50 m (east), t = 50 s

Formula: speed = distance/time; velocity v = s/t (displacement/time)

Substitution: (a) distance = 150 + 100 = 250 m; displacement = +150 − 100 = +50 m (b) speed = 250/50 (c) v = 50/50

Answer: (a) distance 250 m, displacement 50 m towards the east (P → Q) (b) speed 5 m/s (c) velocity 1 m/s towards the east

The velocity is much smaller than the speed because the cyclist came back part of the way – the displacement is only 50 m.

Marking: 1 mark for each part; in (a) ½ for each value. Velocity without its direction: ½ mark.

📖 Review this on Page 5 · Posts and Table 2.1 · Page 6 · Speed vs velocity

15.

The velocity–time graph of a car is shown. Find (a) the acceleration in the first 4 s (b) the displacement between 4 s and 10 s (c) the retardation between 10 s and 12 s.

(3 marks)

💡 Slope of the v–t graph = acceleration; area under it = displacement.

🎬 Step by step

✅ Answer: (a) 5 m/s² (b) 120 m (c) 10 m/s²

Read the graph: O (0 s, 0) → A (4 s, 20 m/s) → B (10 s, 20 m/s) → C (12 s, 0).

Given: (a) u = 0, v = 20 m/s, t = 4 s (b) v = 20 m/s for t = 10 − 4 = 6 s (c) u = 20 m/s, v = 0, t = 2 s

Formula: (a) a = (v − u)/t (b) displacement = area under the graph = velocity × time (c) a = (v − u)/t

Substitution: (a) a = (20 − 0)/4 (b) area of the rectangle = 20 × 6 (c) a = (0 − 20)/2 = −10 m/s²

Answer: (a) 5 m/s² (b) 120 m (c) acceleration −10 m/s², so the retardation is 10 m/s²

Extra: the total displacement in 12 s = area of the whole figure = ½ × 4 × 20 + 120 + ½ × 2 × 20 = 40 + 120 + 20 = 180 m.

Marking: 1 mark for each part (½ for the working, ½ for the answer with unit). Writing “retardation = −10 m/s²”: lose ½ mark.

📖 Review this on Page 10 · Graph 2.5 · Page 13 · Let us assess Q9

16.

A car moving with a velocity of 10 m/s accelerates uniformly at 2 m/s² for 5 s. Find (a) its final velocity (b) the distance it covers in these 5 s. (c) Check your answers using v² = u² + 2as.

(3 marks)

💡 List u, a, t first; then choose the equation that has the unknown.

🎬 Step by step

✅ Answer: (a) 20 m/s (b) 75 m (c) 400 = 400 ✓

Given: u = 10 m/s, a = 2 m/s², t = 5 s

Formula: (a) v = u + at (b) s = ut + ½at² (c) v² = u² + 2as

Substitution: (a) v = 10 + 2 × 5 (b) s = 10 × 5 + ½ × 2 × 5² = 50 + 25 (c) v² = 20² = 400; u² + 2as = 10² + 2 × 2 × 75 = 100 + 300 = 400

Answer: (a) v = 20 m/s (b) s = 75 m (c) both sides are 400, so the answers agree ✓

Marking: 1 mark for each part (formula + substitution ½, answer with unit ½).

📖 Review this on Page 11 · Equations of motion · Page 12 · Worked examples

17.

The velocities of a motorbike and a scooter, both starting from rest, were noted every 2 seconds:

Time (s)	2	4	6
Motorbike velocity (m/s)	6	12	18
Scooter velocity (m/s)	4	10	12

(a) Find the acceleration of the motorbike in each 2 s interval. (b) Find the acceleration of the scooter in each 2 s interval. (c) Which vehicle has uniform acceleration? What will the shape of its velocity–time graph be?

(3 marks)

💡 Find the change in velocity in each interval first.

🎬 Step by step

✅ Answer: (a) 3, 3, 3 m/s² (b) 2, 3, 1 m/s² (c) motorbike; a straight line

Use a = (v − u)/t with t = 2 s for each interval.

(a) Motorbike: (6 − 0)/2 = 3 m/s², (12 − 6)/2 = 3 m/s², (18 − 12)/2 = 3 m/s².

(b) Scooter: (4 − 0)/2 = 2 m/s², (10 − 4)/2 = 3 m/s², (12 − 10)/2 = 1 m/s².

(c) The motorbike: its velocity changes equally in equal intervals of time – uniform acceleration (സമത്വരണം). Its v–t graph is a sloping straight line through the origin. (The scooter has non-uniform acceleration (അസമത്വരണം); its graph bends.)

Marking: 1 mark for each part.

📖 Review this on Page 9 · Uniform and non-uniform acceleration · Page 8 · Table 2.6

SECTION DAnswer all · 2 × 5 = 10 marks

18.

A car is moving at 72 km/h. The driver sees a child on a zebra crossing 60 m ahead. His reaction time is 0.5 s (the car keeps its speed during this time); then the brakes give a uniform retardation of 5 m/s². (a) Find the distance travelled during the reaction time. (b) Find the braking distance. (c) How long does the car take to stop after the brakes are applied? (d) Does the car stop before the zebra crossing? (e) If the car had been moving at 36 km/h, what would its total stopping distance be? What lesson does this give about speed near schools?

(5 marks)

💡 Two parts: constant speed during the reaction time, then uniform retardation.

🎬 Step by step

✅ Answer: (a) 10 m (b) 40 m (c) 4 s (d) yes, 10 m before it (e) 15 m

Given: u = 72 km/h = 72 × 5/18 = 20 m/s; reaction time = 0.5 s; then a = −5 m/s² (retardation 5 m/s²), v = 0; distance to the crossing = 60 m

Formula: (a) distance = speed × time (b) v² = u² + 2as (c) v = u + at

Substitution: (a) 20 × 0.5 (b) 0 = 20² + 2 × (−5) × s, so s = 400/10 (c) 0 = 20 + (−5)t, so t = 20/5

Answer: (a) 10 m (b) 40 m (c) 4 s

(d) Total stopping distance = 10 + 40 = 50 m < 60 m. Yes – the car stops 10 m before the crossing.

(e) u = 36 km/h = 10 m/s: reaction distance = 10 × 0.5 = 5 m; braking distance = 10²/(2 × 5) = 10 m; total = 15 m. Halving the speed cut the braking distance to a quarter (it depends on u²). Lesson: drive slowly near schools and crossings – a slow vehicle can stop in a short distance; obey the speed-limit boards.

Extra: a driver looking at a phone may take 1.5 s to react. Then the car covers 30 m before braking and needs 70 m in all – it reaches the crossing still moving at 10 m/s (36 km/h).

Marking: 1 mark for each part. In (d) the comparison 50 m < 60 m must be shown. Not converting km/h: lose 1 mark in total.

📖 Review this on Page 15 · Stopping-distance lab · Page 11 · v² = u² + 2as · Page 9 · Road safety

19.

A ball is thrown vertically upwards with a velocity of 30 m/s and falls back to the thrower’s hand. Take the upward direction as positive and the acceleration as −10 m/s². (a) How long does the ball take to reach its maximum height? (b) Find the maximum height. (c) Draw the velocity–time graph of the ball for 6 s. (d) Using the graph, find the displacement of the ball after 6 s and the total distance travelled. (e) Find the velocity of the ball after 4 s and say what the sign means.

(5 marks)

💡 At the highest point the velocity is zero for an instant.

🎬 Step by step

✅ Answer: (a) 3 s (b) 45 m (c) straight line 30 → −30 m/s (d) 0; 90 m (e) −10 m/s (downwards)

Given: u = +30 m/s, a = −10 m/s²; at the highest point v = 0

Formula: (a) v = u + at (b) v² = u² + 2as

Substitution: (a) 0 = 30 + (−10)t, so t = 30/10 (b) 0 = 30² + 2 × (−10) × s, so s = 900/20

Answer: (a) t = 3 s (b) maximum height 45 m

(c) v = 30 − 10t is a sloping straight line from (0 s, +30 m/s) through (3 s, 0) to (6 s, −30 m/s): the same slope (−10 m/s²) all the time.

(d) Area above the time axis (0–3 s) = ½ × 3 × 30 = +45 m (going up); area below (3–6 s) = −45 m (coming down). Displacement = 45 + (−45) = 0 – the ball is back in the hand. Distance = 45 + 45 = 90 m.

(e) v = u + at = 30 + (−10) × 4 = −10 m/s. The minus sign means the ball is moving downwards at 10 m/s.

Marking: 1 mark for each part; in (c) the line must cross the time axis at 3 s and reach −30 m/s at 6 s.

📖 Review this on Page 14 · Extended activity 2 · Page 15 · Ball-thrown-up lab · Page 11 · Equations of motion

Marks summary

Section	Questions	Marks each	Total	Topics
A	1–8	1	8	unit of acceleration, scalars, km/h → m/s, circular motion, area under a v–t graph, the three equations, road signs, displacement on a round track
B	9–13	2	10	distance vs displacement, acceleration of a bus, pedestrian safety, position–time graph, train crossing a platform
C	14–17	3	12	speed and velocity of a cyclist, reading a v–t graph, equations of motion, uniform and non-uniform acceleration
D	18–19	5	10	stopping distance at a zebra crossing (road safety), a ball thrown up and its v–t graph
Total			40

⏱️ Time management (15 min cool-off + 90 minutes)

Cool-off (15 min) – read the whole paper. Mark every quantity given in km/h or minutes – they must be converted (× 5/18, × 60). Note which equation each numerical needs.
Section A – about 10 minutes. One-word answers; always write the unit (m/s, m/s²).
Section B – about 20 minutes. For numericals write Given, Formula, Substitution, Answer – the formula line alone often earns half a mark. For a graph, show the two points you read.
Section C – about 25 minutes. Give the direction with every displacement and velocity. Slope of a v–t graph = acceleration; area under it = displacement.
Section D – about 25 minutes. Split each question into its parts (a)–(e). Draw graphs neatly with a scale written in the corner.
Checking – keep 10 minutes: units? directions? km/h converted? “retardation” written without a minus sign?

🎉 Finished the chapter?

Count your marks: 32 or more – excellent; 24–31 – good, revise the pages linked under the questions you missed; below 24 – go back through pages 5, 6, 8, 10 and 11 with their labs and do the scored quiz again. Then return to the chapter home and tick off what you have learned.