Let Us Assess, Q1–Q9 – Solutions വിലയിരുത്താം
The first nine end-of-chapter questions: a car that covers 200 m from rest, a velocity after 10 s, reading three velocity–time graphs, two cars on Graph 2.7, a runner on a circular track, a train reaching 72 km/h, plotting Table 2.10, matching graphs to motions, and Graph 2.9. Try each one in your notebook first, then step through the solution.
- Positive direction: in every question here the direction of motion is taken as positive (+). A minus sign on an acceleration then means the velocity is falling – a retardation മന്ദീകരണം.
- Speed = distance ÷ time; velocity പ്രവേഗം = displacement ÷ time. Acceleration ത്വരണം a = (v − u)/t, unit m/s².
- Equations of motion (uniform acceleration only): v = u + at, s = ut + ½at², v² = u² + 2as. km/h × 5/18 = m/s; minutes × 60 = seconds.
- On a velocity–time graph പ്രവേഗ-സമയ ഗ്രാഫ്: the slope is the acceleration and the area under the graph പരപ്പളവ് is the displacement.
Question 1 · From rest, 200 m in 20 s
1. നിശ്ചലാവസ്ഥയിൽ നിന്ന് ഒരു കാർ സമത്വരണത്തോടെ ചലിക്കുന്നു. ചലനം ആരംഭിച്ച് 20 s കൊണ്ട് 200 m ദൂരം എത്തിയെങ്കിൽ കാറിന്റെ ത്വരണം കണക്കാക്കുക. A car starts from rest and moves with uniform acceleration. If it covers 200 m in 20 s after starting, calculate the acceleration of the car.
“From rest” means u = 0. You know s and t, and you want a – which equation of motion has s, u, t and a but not v? (s = ut + ½at².)
Double the time for the same distance and the acceleration becomes a quarter – the time is squared.
s = ut + ½at² ⇒ 200 = 0 × 20 + ½ × a × 20² = 200a
a = 200 ÷ 200 = 1 m/s².
Dividing 200 m by 20 s gives 10 m/s – that is the average velocity, not the acceleration. The car started at 0 and ended at 20 m/s; its average was 10 m/s.
Question 2 · 2 m/s² for 10 s
2. നിശ്ചലാവസ്ഥയിൽ നിന്ന് നീങ്ങിത്തുടങ്ങുന്ന ഒരു വസ്തു 2 m/s² ത്വരണത്തോടെ സഞ്ചരിക്കുന്നു. എങ്കിൽ 10 s കഴിയുമ്പോൾ വസ്തുവിന്റെ പ്രവേഗം എത്രയായിരിക്കും? An object starting from rest moves with an acceleration of 2 m/s². What will its velocity be after 10 s?
An acceleration of 2 m/s² means the velocity grows by 2 m/s every second. Use v = u + at with u = 0.
v = u + at = 0 + 2 × 10 = 20 m/s (in the direction of motion).
A Kochi Metro train leaving a station speeds up by roughly this much: in 10 s it is already at 20 m/s = 72 km/h. That is why standing passengers must hold the handles.
Question 3 · Reading the three graphs of Graph 2.6
3. ഒരു വാഹനത്തിന്റെ ചലനവുമായി ബന്ധപ്പെട്ട 3 വ്യത്യസ്ത ഗ്രാഫുകൾ ചുവടെ നൽകിയിരിക്കുന്നു. ഗ്രാഫുകൾ വിശകലനം ചെയ്ത് ചലനത്തിന്റെ പ്രത്യേകതകൾ കണ്ടെത്തുക. Three different graphs related to the motion of a vehicle are given below (Graph 2.6). Analyse the graphs and find the features of the motion.
All three are velocity–time graphs (velocity on Y, time on X). Ask for each: where does it start (at zero or not)? Is it a straight line? Does it go up, down, or both?
(ii) A straight line falling to the time axis: the vehicle has a velocity at the start, which decreases by equal amounts in equal times until it comes to rest – uniform retardation (uniform negative acceleration).
(iii) A zig-zag line not starting at zero: the vehicle was already moving; its velocity increases, decreases, increases again and then decreases – unequal changes, so non-uniform acceleration അസമത്വരണം (acceleration and retardation one after another).
A v–t graph tells the whole story without words: up = speeding up, flat = steady velocity, down = slowing down, straight = uniform, bent or zig-zag = non-uniform. Graph (iii) is a KSRTC bus in town traffic.
Question 4 · Car A and car B (Graph 2.7)
4. കാർ A, കാർ B ഇവയുടെ സഞ്ചാരത്തിന്റെ ഗ്രാഫ് തന്നിരിക്കുന്നു. a) ഏത് കാറിനാണ് ത്വരണം കൂടുതൽ? എന്തുകൊണ്ട്? b) തോത് മാറ്റി വരച്ച് ഗ്രാഫുകൾ താരതമ്യം ചെയ്യുക. The graph of the motion of car A and car B is given (Graph 2.7; scale X-axis 1 cm = 1 s, Y-axis 1 cm = 1 m/s).
- Which car has the greater acceleration? Why?
- Redraw the graphs with a different scale തോത് and compare them.
Read where each line leaves the time axis (the car starts from rest there) and where it reaches 5 m/s. Acceleration = change in velocity ÷ time taken. Both cars gain the same 5 m/s – which one does it faster?
(a) Car A: u = 0 at 5 s, v = 5 m/s at 7 s ⇒ a = (5 − 0)/(7 − 5) = 5/2 = 2.5 m/s².
Car B: u = 0 at 3 s, v = 5 m/s at 7 s ⇒ a = (5 − 0)/(7 − 3) = 5/4 = 1.25 m/s².
Car A has the greater acceleration: it gains the same 5 m/s in less time (2 s instead of 4 s), so its line is steeper.
(b) Redrawn with Y-axis 1 cm = 2 m/s, both lines become less steep, and car A's line is still steeper than car B's. The accelerations read from the graph are unchanged (2.5 m/s² and 1.25 m/s²): changing the scale changes the look of the graph, not the value of the physical quantity ഭൗതിക അളവ്.
Comparing steepness between two graphs drawn to different scales. Steepness can be compared only on the same graph paper with the same scale – always calculate Δv ÷ Δt with the numbers.
Question 5 · A runner on a circular track (Fig 2.12)
5. ചിത്രം നിരീക്ഷിക്കൂ. ഒരു കുട്ടി 440 m ചുറ്റളവുള്ള വൃത്തപാതയിൽ കൂടി സമവേഗത്തിൽ ഓടുന്നു. വൃത്തപാതയുടെ ആരം 70 m ആണ്. A യിൽ നിന്ന് B യിലൂടെ തിരികെ A യിൽ ഓടിയെത്താനെടുക്കുന്ന സമയം 80 s ആണ്. പട്ടികയിൽ നൽകിയ ഓരോ സന്ദർഭത്തിലും ഉള്ള ദൂരം, സ്ഥാനാന്തരം, വേഗം, പ്രവേഗം ഇവ കണ്ടെത്തുക. Observe the figure (Fig 2.12). A child runs at uniform speed along a circular path of circumference 440 m. The radius of the path is 70 m. The time taken to run from A through B and back to A is 80 s. Find the distance, displacement, speed and velocity in each situation given in Table 2.9 (on reaching B; on reaching A).
A and B are the two ends of a diameter, so B is half a round away (40 s, 220 m along the track). Displacement is the straight line from the start A to where the child is – at B that is the diameter, 2 × 70 m. Speed uses distance; velocity uses displacement.
| Situation സന്ദർഭം | Distance ദൂരം | Displacement സ്ഥാനാന്തരം | Speed വേഗം | Velocity പ്രവേഗം |
|---|---|---|---|---|
| On reaching B (40 s) | ? | ? | ? | ? |
| On reaching A again (80 s) | ? | ? | ? | ? |
The table fills itself when the runner reaches B and A. In between, watch the displacement arrow: it grows to 140 m at B and shrinks back to 0.
At B (t = 40 s): distance = 440 ÷ 2 = 220 m; displacement = AB = 2 × 70 = 140 m (A → B); speed = 220 ÷ 40 = 5.5 m/s; velocity = 140 ÷ 40 = 3.5 m/s (A → B).
At A again (t = 80 s): distance = 440 m; displacement = 0 (start = end); speed = 440 ÷ 80 = 5.5 m/s; velocity = 0 ÷ 80 = 0.
Writing the displacement at B as 220 m (half the circumference). Displacement is the straight-line distance from start to end – the diameter, 140 m. (The radius 70 m and 440 m agree: 2 × 22/7 × 70 = 440.)
Question 6 · A train reaches 72 km/h in 5 minutes
6. നിശ്ചലാവസ്ഥയിൽ നിന്നും പുറപ്പെട്ട ഒരു ട്രെയിനിന്റെ പ്രവേഗം 5 മിനിറ്റ് കൊണ്ട് 72 km/h ആകുന്നു. ട്രെയിനിന്റെ ത്വരണവും ഈ സമയം കൊണ്ട് ട്രെയിനിനുണ്ടായ സ്ഥാനാന്തരവും കണ്ടുപിടിക്കുക. A train starting from rest reaches a velocity of 72 km/h in 5 minutes. Find the acceleration of the train and the displacement of the train in this time.
Convert first: km/h × 5/18 → m/s, minutes × 60 → s. Then a = (v − u)/t and s = ut + ½at² (or the area of the triangle under the v–t graph).
a = (v − u)/t = (20 − 0)/300 = 1/15 m/s² ≈ 0.067 m/s².
s = ut + ½at² = 0 × 300 + ½ × (1/15) × 300² = 90000/30 = 3000 m (3 km).
Check: average velocity = (0 + 20)/2 = 10 m/s; 10 m/s × 300 s = 3000 m ✓.
Using 72 and 5 directly: 72/5 = 14.4 “m/s²” is wrong by a factor of 216. Units must be m and s before they go into the formula.
Question 7 · Plot Table 2.10
7. പട്ടിക 2.10 വിശകലനം ചെയ്ത് പ്രവേഗ - സമയ ഗ്രാഫ് വരയ്ക്കുക. Analyse Table 2.10 and draw the velocity–time graph.
| X time (s) | 0 | 2 | 4 | 6 | 8 | 10 |
|---|---|---|---|---|---|---|
| Y velocity (m/s) | 10 | 15 | 20 | 20 | 20 | 15 |
- From the graph, find the time interval in which there is no acceleration.
- Find the time interval in which there is retardation.
- Find the displacement between the 4th second and the 8th second.
Choose a scale that fits: X 1 cm = 2 s, Y 1 cm = 5 m/s. A flat part = no acceleration; a part going down = retardation. Displacement = area under the graph between 4 s and 8 s.
(a) No acceleration from 4 s to 8 s – the graph is flat at 20 m/s (uniform velocity).
(b) Retardation from 8 s to 10 s: a = (15 − 20)/(10 − 8) = −2.5 m/s², i.e. a retardation of 2.5 m/s².
(c) Displacement from 4 s to 8 s = area of the rectangle = 20 m/s × 4 s = 80 m.
(Also: from 0 to 4 s, a = (20 − 10)/4 = 2.5 m/s² – uniform acceleration.)
Starting the line at the origin. At t = 0 the velocity is already 10 m/s, so the first point is (0, 10) on the Y-axis.
Question 8 · Match the graphs (Graph 2.8)
8. ഗ്രാഫുകൾ നിരീക്ഷിക്കുക. ഏത് ഗ്രാഫിലാണ് വസ്തുവിന് a) സമത്വരണമുള്ളത്? b) സമപ്രവേഗമുള്ളത്? c) ത്വരണവും മന്ദീകരണവുമുള്ളത്? Observe the graphs (Graph 2.8). In which graph does the object have
- uniform acceleration?
- uniform velocity?
- acceleration and retardation?
Flat line → velocity not changing. Straight line going up → velocity rising by equal amounts each second. A curve that rises and then comes down → speeding up, then slowing down.
(b) Uniform velocity → graph (i): a horizontal line; the velocity stays the same, acceleration = 0.
(c) Acceleration and retardation → graph (iii): the curve rises (velocity increases – acceleration) and then comes down after the highest point (velocity decreases – retardation). Being a curve, the acceleration is non-uniform.
Question 9 · Graph 2.9 – accelerate, cruise, brake
9. ഒരു വസ്തുവിന്റെ നേർരേഖാചലനവുമായി ബന്ധപ്പെട്ട പ്രവേഗ-സമയ ഗ്രാഫ് നൽകിയിരിക്കുന്നു. ഗ്രാഫിൽ ഏത് അവസരത്തിലാണ് വസ്തുവിന് a) ത്വരണമുള്ളത്? b) സമപ്രവേഗമുള്ളത്? c) മന്ദീകരണമുള്ളത്? The velocity–time graph of an object moving in a straight line is given (Graph 2.9). At which stage in the graph does the object have
- acceleration?
- uniform velocity?
- retardation?
Read the corners: O (0, 0), A (4 s, 50 m/s), B (8 s, 50 m/s), C (10 s, 0). Going up = acceleration, flat = uniform velocity, going down = retardation.
(a) Acceleration from 0 to 4 s (O → A): a = (50 − 0)/4 = 12.5 m/s².
(b) Uniform velocity from 4 s to 8 s (A → B): steady 50 m/s, a = 0.
(c) Retardation from 8 s to 10 s (B → C): a = (0 − 50)/(10 − 8) = −25 m/s², a retardation of 25 m/s².
Extra: total displacement = area = ½ × 4 × 50 + 50 × 4 + ½ × 2 × 50 = 100 + 200 + 50 = 350 m.
This is the shape of every Kochi Metro run between two stations: speed up, cruise, brake. The braking part (2 s) is steeper than the speeding-up part (4 s), so the retardation (25 m/s²) is bigger than the acceleration (12.5 m/s²).