Chapter 2 · Let us assessTextbook pages 49–52

Let Us Assess, Q10–Q17 & Extended Activities വിലയിരുത്താം · തുടർപ്രവർത്തനങ്ങൾ

The last eight end-of-chapter questions – dark uniforms and road safety, the trip round Fig 2.13, a stone thrown up, stopping from 40 m/s, speeding up over 120 m, a bullet in a log, plotting Table 2.12 and the two cars of Graph 2.10 – followed by the three extended activities. Try each one in your notebook first, then step through the solution.

📖 What you need (pages 5, 6, 8, 10 and 11 of this guide)

Question 10 · Dark-coloured uniforms?

10. രാത്രിയിലും വെളിച്ചം കുറവുള്ള സമയത്തും കാൽനടയാത്രക്കാർ കടും നിറത്തിലുള്ള വസ്ത്രങ്ങൾ ധരിക്കുന്നത് സുരക്ഷിതമല്ല എന്നറിയാമല്ലോ. നിങ്ങളുടെ സ്കൂളിൽ കടും നിറത്തിലുള്ള യൂണിഫോം തിരഞ്ഞെടുക്കുവാൻ സ്കൂൾ അധികൃതർ തീരുമാനിച്ചു. ഈ തീരുമാനത്തോട് നിങ്ങളുടെ പ്രതികരണം രേഖപ്പെടുത്തുക. റോഡ് സുരക്ഷയുമായി ബന്ധപ്പെട്ട് ഉത്തരം സാധൂകരിക്കുക. We know that it is not safe for pedestrians to wear dark-coloured clothes at night and when the light is poor. The authorities of your school have decided to choose a dark-coloured uniform. Record your response to this decision. Justify your answer with reference to road safety.

Think of a driver on a dark, rainy evening: which clothes does the headlight show early, and which late? Link “seen late” with the distance a vehicle needs to stop. Then suggest a practical change instead of only saying “no”.

🌙 Night road – when does the driver first see the pedestrian?

Illustrative, approximate distances often used in road-safety classes; real values depend on the headlights, rain and the driver. The stopping distance assumes 50 km/h, a reaction time of 1 s and braking at 7 m/s².

Model response: I respectfully disagree with choosing a dark-coloured uniform, and suggest a change.
Many of us walk to and from school along busy roads in the early morning, in the evening after special classes, and on dark monsoon days. In low light a driver sees a pedestrian in dark clothes only when he is very close – often closer than the distance the vehicle needs to stop. A car at 50 km/h travels about 14 m during the driver's reaction time of one second and needs about 14 m more to brake, roughly 28 m in all. Light colours and reflective material are seen from much farther away, so the driver gets time to slow down.
Suggestions: choose a light or bright uniform (white, light blue, yellow), or, if the dark colour must stay, add reflective strips to the uniform and school bags; also walk on the right side facing traffic and use zebra crossings. Road safety of students should come before the convenience of a colour that hides dirt.
⚠ At night or in heavy rain, wear something light or reflective, carry a torch, walk facing the traffic, and never assume a driver has seen you.
💡 Why this matters

A vehicle cannot stop instantly: it needs a reaction distance plus a braking distance (v²/2a). Being seen early gives the driver that distance (try Extended activity 1).

Question 11 · Round the path P → Q → R → S → P (Fig 2.13)

11. ഒരു കുട്ടി P യിൽ നിന്ന് യാത്ര തിരിച്ച് Q, R എന്നീ ഭാഗങ്ങളിലൂടെ S ലും തുടർന്ന് നേരെ P യിൽ തിരിച്ചെത്തിയതും ചിത്രീകരിച്ചിരിക്കുന്നു. ചിത്രം വിശകലനം ചെയ്ത് താഴെ കൊടുത്ത പട്ടിക പൂർത്തിയാക്കുക. The figure shows a child who starts from P, goes through Q and R to S and then comes straight back to P. Analyse the figure and complete the table (Table 2.11): the speed വേഗം and velocity പ്രവേഗം on reaching Q, R, S and P again.

Always measure from the start P. Speed uses the total distance along the path and the total time; velocity uses the straight-line displacement from P (PQ = 60 m, PR = 125 m, PS = 136 m) and the same total time.

🎬 Solution animation – the trip, with Table 2.11 filling itself
Travelling from P P യിൽ നിന്ന് സഞ്ചരിച്ച്Speed വേഗംVelocity പ്രവേഗം
on reaching Q Q വരെ എത്തുമ്പോൾ
on reaching R R വരെ എത്തുമ്പോൾ
on reaching S S വരെ എത്തുമ്പോൾ
back at P P യിൽ തിരികെ എത്തുമ്പോൾ
Answer (Table 2.11):
Q (10 s): speed = 60 m ÷ 10 s = 6 m/s; velocity = 60 m ÷ 10 s = 6 m/s from P towards Q.
R (10 + 15 = 25 s, path 60 + 90 = 150 m): speed = 150 ÷ 25 = 6 m/s; velocity = 125 m ÷ 25 s = 5 m/s from P towards R.
S (25 + 9 = 34 s, path 150 + 54 = 204 m): speed = 204 ÷ 34 = 6 m/s; velocity = 136 m ÷ 34 s = 4 m/s from P towards S.
P again (34 + 16 = 50 s, path 204 + 136 = 340 m): speed = 340 ÷ 50 = 6.8 m/s; displacement = 0, so velocity = 0.
⚠️ Common trap

Using the time of one leg only (e.g. 125 ÷ 15 for R). Velocity is the displacement from the start divided by the whole time taken from the start – 25 s for R. And back at P the velocity is 0 even though the child ran 340 m.

Question 12 · A stone thrown straight up at 20 m/s

12. ഒരു കല്ല് 20 m/s പ്രവേഗത്തിൽ മുകളിലേക്ക് കുത്തനെ എറിയുന്നു. (a = −10 m/s²) a) ഈ കല്ല് പരമാവധി എത്ര ഉയരത്തിലെത്തും? b) പുറപ്പെട്ട് 3 s കൊണ്ട് ഈ കല്ല് എത്ര ദൂരം സഞ്ചരിക്കും? c) പുറപ്പെട്ട് 3 s കഴിയുമ്പോൾ ഈ കല്ല് തറയിൽ നിന്നും എത്ര ഉയരത്തിലായിരിക്കും? A stone is thrown vertically upwards with a velocity of 20 m/s (a = −10 m/s²).

  1. What is the maximum height the stone reaches?
  2. What distance does the stone travel in 3 s after it is thrown?
  3. How high above the ground will the stone be 3 s after it is thrown?

Take up as positive, so u = +20 m/s and a = −10 m/s². At the highest point the velocity is momentarily zero. For (b), the stone goes up for 2 s and then comes down for 1 s – distance adds both parts. For (c) use s = ut + ½at² with t = 3 s.

🎬 Solution animation – the stone, its height–time and velocity–time graphs
🧪 Try other numbers – throw speed

Double the throw speed → twice the time to the top but four times the height (h = u²/2g).

Answer (up = +): u = 20 m/s, a = −10 m/s².
(a) At the top v = 0: v² = u² + 2as ⇒ 0 = 400 + 2(−10)s ⇒ s = 400/20 = 20 m (reached at t = (0 − 20)/(−10) = 2 s).
(b) In the 3rd second it falls: ½ × 10 × 1² = 5 m. Distance travelled = 20 m + 5 m = 25 m.
(c) s = ut + ½at² = 20 × 3 + ½ × (−10) × 3² = 60 − 45 = 15 m above the ground.
⚠️ Common trap

Answering (b) with the formula: s = ut + ½at² gives the displacement (15 m), not the distance. The stone went up 20 m and came back 5 m – it travelled 25 m. Whenever the motion turns back, split it at the turning point.

Question 13 · From 40 m/s to rest with retardation 8 m/s²

13. ഒരു വസ്തു 40 m/s വേഗത്തിൽ സഞ്ചരിക്കുന്നു. ഇതിന് 8 m/s² മന്ദീകരണം നൽകിയാൽ a) ഈ വസ്തു നിശ്ചലമാകാൻ എത്ര സമയമെടുക്കും? b) ഇത്രയും സമയം കൊണ്ട് വസ്തുവിനുണ്ടാകുന്ന സ്ഥാനാന്തരമെത്ര? An object is moving at 40 m/s. If a retardation of 8 m/s² is given to it,

  1. how long will it take to come to rest?
  2. what is its displacement in this time?

Direction of motion = positive. Retardation 8 m/s² means a = −8 m/s². “Comes to rest” means v = 0. Use v = u + at for the time and v² = u² + 2as for the displacement.

🎬 Solution animation – v–t graph and the area under it
🧪 Try other numbers – speed and retardation
Answer (direction of motion = +): u = 40 m/s, v = 0, a = −8 m/s².
(a) v = u + at ⇒ 0 = 40 + (−8)t ⇒ t = 40/8 = 5 s.
(b) v² = u² + 2as ⇒ 0 = 1600 + 2(−8)s ⇒ s = 1600/16 = 100 m (in the direction of motion). Check: area of the v–t triangle = ½ × 5 × 40 = 100 m ✓.
⚠️ Common trap

Putting a = +8 m/s² into v² = u² + 2as gives s = −100 m – a negative distance that makes no sense. The word “retardation” already tells you the acceleration is negative.

Question 14 · 20 m/s, a = 5 m/s², after 120 m

14. 20 m/s പ്രവേഗത്തിൽ ഒരു വസ്തു ചലിച്ചുകൊണ്ടിരിക്കുന്നു. ഈ വസ്തുവിന് 5 m/s² ത്വരണം നൽകുന്നു. സ്ഥാനാന്തരം 120 m ആകുന്ന അവസരത്തിലെ പ്രവേഗമെത്ര? An object is moving with a velocity of 20 m/s. An acceleration of 5 m/s² is given to it. What is its velocity when its displacement becomes 120 m?

You know u, a and s and want v; time is not given and not asked. Which equation has no t in it?

🎬 Solution animation – speeding up over 120 m
Answer (direction of motion = +): u = 20 m/s, a = 5 m/s², s = 120 m, v = ?
v² = u² + 2as = 20² + 2 × 5 × 120 = 400 + 1200 = 1600 ⇒ v = √1600 = 40 m/s (in the direction of motion).
Check: t = (v − u)/a = (40 − 20)/5 = 4 s, and s = 20 × 4 + ½ × 5 × 4² = 80 + 40 = 120 m ✓.
⚠️ Common trap

Stopping at v² = 1600 and writing “v = 1600 m/s”. Take the square root as the last step: √1600 = 40.

Question 15 · A bullet stopped by a log in 2 cm

15. 60 m/s പ്രവേഗത്തിൽ സഞ്ചരിച്ച വെടിയുണ്ട ഒരു മരത്തടിയിൽ 2 cm തുളച്ചു കയറിയപ്പോൾ നിശ്ചലമായി. ഈ വെടിയുണ്ടയുടെ ത്വരണമെത്ര? മന്ദീകരണമെത്ര? A bullet travelling at 60 m/s penetrates 2 cm into a log of wood and comes to rest. What is the acceleration of the bullet? What is its retardation?

First change 2 cm into metres. u = 60 m/s, v = 0, s = 0.02 m; no time is given, so use v² = u² + 2as and solve for a.

🎬 Solution animation – into the log (magnified)
🧪 Try other numbers – speed and depth

Half the depth → twice the retardation. Double the speed → four times the retardation.

Answer (direction of the bullet's motion = +): u = 60 m/s, v = 0, s = 2 cm = 0.02 m.
v² = u² + 2as ⇒ 0 = 3600 + 2 × a × 0.02 ⇒ 0.04a = −3600 ⇒ a = −90 000 m/s² (= −9 × 104 m/s²).
Acceleration = −9 × 104 m/s²; retardation = 9 × 104 m/s² (90 000 m/s², no minus sign).
⚠️ Common trap

Using s = 2 instead of 0.02 gives a = −900 m/s² – one hundred times too small. Change cm to m before substituting. And do not write “retardation = −9 × 104 m/s²”: the minus belongs to the acceleration.

Question 16 · Plot Table 2.12 and read it

16. ഒരു വസ്തുവിന്റെ ചലനവുമായി ബന്ധപ്പെട്ട പട്ടിക നിരീക്ഷിച്ച് അനുയോജ്യമായ സ്കെയിൽ തിരഞ്ഞെടുത്ത് ഗ്രാഫ് വരയ്ക്കുക. ഗ്രാഫ് വിശകലനം ചെയ്ത് ചുവടെ കൊടുത്ത ചോദ്യങ്ങൾക്ക് ഉത്തരമെഴുതുക. Observe the table related to the motion of an object, choose a suitable scale and draw the graph. Analyse the graph and answer the questions below.

Time സമയം s051015202530
Velocity പ്രവേഗം m/s20253030302520

Table 2.12

  1. ത്വരണമില്ലാത്ത സമയ ഇടവേള ഏത്? During which time interval is there no acceleration?
  2. മന്ദീകരണം ഉണ്ടായിരുന്ന സമയ ഇടവേള ഏത്? During which time interval was there retardation?
  3. ഈ വസ്തുവിന് 30 s ൽ ഉണ്ടായ സ്ഥാനാന്തരം കണക്കാക്കുക. Calculate the displacement of the object in 30 s.

Scale X 1 cm = 5 s, Y 1 cm = 5 m/s fits the paper. A horizontal part means constant velocity (a = 0); a part sloping down means retardation. For (c), split the area under the graph into a trapezium, a rectangle and a trapezium.

🎬 Solution animation – plotting step by step, then the area
Answer (scale X 1 cm = 5 s, Y 1 cm = 5 m/s):
(a) The graph is horizontal from 10 s to 20 s – velocity constant at 30 m/s, so no acceleration.
(b) It slopes down from 20 s to 30 s: a = (20 − 30)/10 = −1 m/s², a retardation of 1 m/s². (From 0 to 10 s it accelerates at +1 m/s².)
(c) Displacement = area under the graph = ½(20 + 30) × 10 + 30 × 10 + ½(30 + 20) × 10 = 250 + 300 + 250 = 800 m.
⚠️ Common trap

Taking “velocity × time” with one velocity (30 × 30 = 900 m) – the velocity is not 30 m/s all the time. Or forgetting the strip under the graph down to the time axis: the area goes all the way to v = 0, not just from 20 m/s.

Question 17 · Cars A and B (Graph 2.10)

17. A, B എന്നീ കാറുകളുടെ ചലനവുമായി ബന്ധപ്പെട്ട പ്രവേഗ-സമയഗ്രാഫ് നൽകിയിരിക്കുന്നു. a) ഏത് കാറാണ് ആദ്യം യാത്ര തിരിച്ചത്? b) രണ്ട് കാറുകൾക്കും ഒരേ പ്രവേഗമെത്താൻ എത്ര സമയം വീതം എടുത്തു? c) ഏത് കാറിനാണ് കൂടുതൽ ത്വരണം ഉള്ളത്? d) ഏത് കാറിനാണ് സ്ഥാനാന്തരം കൂടുതൽ? The velocity–time graph of the motion of two cars A and B is given (Graph 2.10).

  1. Which car started first?
  2. How much time did each car take to reach the same velocity?
  3. Which car has the greater acceleration?
  4. Which car has the greater displacement?

Where does each line leave the time axis? Where do the two lines cross? Steeper line = bigger acceleration. Displacement = area under each line.

🎬 Solution animation – Graph 2.10, both cars running
Answer:
(a) Car A – its line starts at t = 0; car B's starts at t = 8 s.
(b) The lines cross at t = 14 s, v = 40 m/s. Car A took 14 − 0 = 14 s; car B took 14 − 8 = 6 s.
(c) aA = 40/14 ≈ 2.9 m/s²; aB = 40/6 ≈ 6.7 m/s². Car B has the greater acceleration (steeper line).
(d) Up to 14 s: area under A = ½ × 14 × 40 = 280 m; under B = ½ × 6 × 40 = 120 m. Car A has the greater displacement (it has also been moving for longer).
💡 Why this matters

A v–t graph answers three different questions from three different features: where it starts (when), its slope (acceleration) and the area under it (displacement). Car B is quicker to speed up, yet car A has gone farther.

Extended activities തുടർപ്രവർത്തനങ്ങൾ

1. Awareness class – “vehicle accidents due to overspeeding”

'അമിതവേഗം മൂലമുണ്ടാകുന്ന വാഹനാപകടങ്ങൾ' എന്ന വിഷയത്തിൽ സയൻസ് ക്ലബ് ഒരു ബോധവൽക്കരണ ക്ലാസ് സംഘടിപ്പിക്കുന്നു. ഈ ക്ലാസിൽ അവതരിപ്പിക്കുന്നതിനാവശ്യമായ സ്ലൈഡുകൾ തയ്യാറാക്കൂ. (സൂചന - ട്രാഫിക് നിയമങ്ങൾ, സൈൻ ബോർഡുകൾ, കാൽനടക്കാർ പാലിക്കേണ്ട റോഡ് നിയമങ്ങൾ തുടങ്ങിയവ) The science club is organising an awareness class on “vehicle accidents caused by overspeeding”. Prepare the slides needed for this class. (Hints: traffic rules, sign boards, road rules for pedestrians, etc.)
🛠️ Guidance card
🚗 Stopping-distance lab (use these numbers on slide 3)

Stopping distance = v × reaction time + v²/2a (from v² = u² + 2as with v = 0). Retardations are rough model values.

Slide outline (8 slides) – one idea per slide, big text, a drawing or graph on each:

  1. Title + Graph 2.1 (accidents in Kerala, 2016–2022)
  2. Speed, velocity, retardation – with units
  3. Reaction + braking distance (lab table)
  4. Double speed → 4 × braking distance; wet roads
  5. Traffic rules – speed limits, lane lines, helmets, belts
  6. Sign boards – mandatory, cautionary, informatory
  7. Pedestrian rules – right side, zebra crossing, light clothes
  8. Our pledge + a quiz for the audience

Use the lab's numbers: 50 km/h on a dry road ≈ 28 m to stop, 100 km/h ≈ 83 m. Add a short role-play – “the driver who saw the child too late”.

⚠ Collect photos and data from newspapers, the police or the Motor Vehicles Department – never by standing close to speeding traffic.
🌐 Explore online needs internet · opens in a new tab

2. Velocity–time graph of an object thrown up at 30 m/s

ഒരു വസ്തുവിനെ 30 m/s പ്രവേഗത്തോടു കൂടി മുകളിലേക്കെറിയുന്നു. ഒരു നിശ്ചിത സമയത്തിനുശേഷം വസ്തു അതേ സ്ഥലത്ത് തിരിച്ചെത്തി. ഈ വസ്തുവിന്റെ പ്രവേഗ - സമയഗ്രാഫ് വരച്ച് ക്ലാസിൽ പ്രദർശിപ്പിക്കുക (മന്ദീകരണം 10 m/s² ആയി പരിഗണിക്കുക). An object is thrown upwards with a velocity of 30 m/s. After a certain time it returns to the same place. Draw its velocity–time graph and display it in the class (take the retardation as 10 m/s²).
🛠️ Guidance card – up = positive, a = −10 m/s²
🎬 Throw it – the v–t graph draws itself

Points to plot (v = 30 − 10t; X 1 cm = 1 s, Y 1 cm = 10 m/s): (0, 30), (1, 20), (2, 10), (3, 0), (4, −10), (5, −20), (6, −30) – one straight line of slope −10 m/s², going up and coming down. At 3 s it crosses the time axis (at rest at the top). Maximum height = upper triangle = ½ × 3 × 30 = 45 m; the lower triangle is −45 m, so the total displacement after 6 s = 0.

3. Project – are the road-safety measures in our area enough?

നിങ്ങളുടെ പ്രദേശത്ത് റോഡപകടങ്ങൾ കുറയ്ക്കുന്നതിനു വേണ്ടി നടപ്പിലാക്കിയിട്ടുള്ള സുരക്ഷാ സംവിധാനങ്ങൾ പര്യാപ്തമാണോ എന്നതിനെക്കുറിച്ച് ഒരു പ്രോജക്ട് തയാറാക്കൂ. പ്രോജക്ട് ആസൂത്രണം ടീച്ചറുടെ സഹായത്തോടെ ചെയ്യുമല്ലോ. പ്രസക്തമായ കണ്ടെത്തലുകൾ റോഡ് സുരക്ഷാ അതോറിറ്റിയുടെ ശ്രദ്ധയിൽപ്പെടുത്തൂ. Prepare a project on whether the safety measures put in place in your area to reduce road accidents are adequate. Plan the project with your teacher's help, and bring the relevant findings to the attention of the road safety authority. (For more information, the services of the National Transportation Planning and Research Centre (NATPAC) and the Motor Vehicles Department can be used.)
🛠️ Guidance card – what the report must have റിപ്പോർട്ടിൽ അത്യാവശ്യമായി ഉണ്ടാകേണ്ടവ
  1. Introduction ആമുഖം – the road or junction studied and why.
  2. Hypothesis പരികല്പന – “The safety measures near our school are not enough for pedestrians.”
  3. Objectives ഉദ്ദേശ്യലക്ഷ്യങ്ങൾ – list the measures, find the gaps, suggest improvements.
  4. Methodology പഠനരീതി – checklist (zebra crossings, signals, sign boards, footpaths, speed breakers, street lights), a short questionnaire, a 10-minute vehicle count from a safe place, talks with police/MVD officials.
  5. Analysis അപഗ്രഥനം – tables and bar graphs of what you found.
  6. Findings പഠനഫലങ്ങൾ – e.g. faded zebra crossing, no footpath, street lights off.
  7. Conclusion നിഗമനം – is the hypothesis accepted or rejected?
  8. Suggestions നിർദേശങ്ങൾ – repaint markings, school-zone sign, reflective strips; send them to the road safety authority.
⚠ Do all survey work in a group, with your teacher's permission, standing well away from the road edge.

Check yourself

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