Comprehensive Chapter Summary
1. Describing Motion
In everyday life, objects are at rest or in motion: birds fly, fish swim, blood flows through veins and arteries, cars move, atoms, molecules, planets, stars, and galaxies are all in motion. We perceive motion when position changes with time, but sometimes infer it indirectly, like air motion from dust or leaves. Phenomena like sunrise, sunset, and seasons result from Earth's motion, yet we don't perceive it directly due to relative motion. An object may appear moving to one and stationary to another; e.g., roadside trees seem backward to bus passengers, but the bus moves to roadside observers. Fellow passengers inside appear at rest. Motions are complex: straight line, circular, rotational, vibratory, or combinations. This chapter focuses on straight-line motion via equations and graphs, later circular motion. Activity 7.1: Discuss classroom walls—rest relative to Earth but in motion with it. Activity 7.2: Train illusion when adjacent train moves. Think and Act: Erratic motions endanger (floods, hurricanes, tsunamis); controlled serve (hydro-electric power). Necessity to study and control motions.
Relative Motion Example: Bus and Trees
For bus passengers, trees move backward due to relative velocity; roadside person sees bus moving forward. Inside bus, passengers at rest relative to each other. This illustrates motion depends on reference frame. In flooded rivers or hurricanes, uncontrolled motion causes danger; controlled in power generation aids humanity. Study enables prediction and control.
Inferring Motion: Air and Seasons
We infer air motion from dust/leaves; Earth's rotation causes day-night, revolution with tilt causes seasons. Direct perception absent due to uniform motion and vast scale. Activities highlight perceptual illusions, emphasizing reference points in describing motion accurately.
2. Motion Along a Straight Line
Reference Point and Position
Describe location by reference (origin), e.g., school 2 km north of station. Position relative; choose convenient origin. For straight path, object from O to A (60 km), back to B (25 km from A), total distance OA + AB = 85 km, displacement OB = 35 km. Distance scalar (magnitude only), displacement vector (shortest from initial to final, direction matters).
Distance vs Displacement
Magnitude of displacement ≤ distance; equal in straight without reversal. If back to O, displacement 0, distance 120 km. Activity 7.3: Basketball court diagonal—distance along sides 2L, displacement √2 L. Activity 7.4: Odometer shows 1850 km Bhubaneswar-Delhi; displacement ~1700 km via map.
Questions on Basics
Q1: Object moved distance, zero displacement? Yes, return to start. Q2: Farmer square field 10m side, 40s/round; at 2min20s (3.5 rounds), displacement 10m (opposite corner). Q3: Displacement: (a) can be zero; (b) magnitude ≤ distance.
Uniform/Non-Uniform Motion
Uniform: equal distances equal time (5m/s). Non-uniform: unequal, e.g., crowded street car, jogging. Activity 7.5: Table 7.1—A uniform (10m/15min), B non-uniform (increasing unevenly).
Example 7.1: Average Speed Calculation
Object 16m in 4s + 16m in 2s: total 32m/6s = 5.33 m/s. Illustrates average for non-uniform: total distance/total time, ignoring variations.
Activity 7.3: Court Walk
Walk sides of court: distance 2×length, displacement diagonal. Highlights vector vs scalar; difference increases with path curvature.
3. Measuring Rate of Motion
Rate: distance/unit time = speed (m/s, cm/s, km/h). Average speed for non-uniform: total distance/total time. Example: 100km/2h = 50 km/h. Example 7.2: Odometer 2000-2400km in 8h = 50 km/h = 13.9 m/s (conversion: ×1000/3600). Velocity: speed with direction; average = (u+v)/2 for uniform change. Example 7.3: Usha 180m swim/1min = 3 m/s speed, 0 m/s velocity (back to start). Activity 7.6: Walk time at 4 km/h estimates distance.
Speed with Direction: Velocity
Velocity specifies direction; changes by speed, direction, or both. Uniform velocity constant; variable non-uniform. Average for straight variable speed same as average speed. For uniform change: v_av = (u+v)/2. Units m/s. Comprehensive for comprehensive motion description.
Conversion and Units
km/h to m/s: ×(1000/3600) = ×(5/18). E.g., 50 km/h = 13.9 m/s. Practical for mixing units in problems like odometers (km) and physics (m/s).
Questions on Rate
Q1: Speed scalar, velocity vector. Q2: Equal when straight no reversal. Q3: Odometer measures distance. Q4: Uniform: straight line parallel time axis. Q5: Signal 5min at 3×10^8 m/s = 9×10^10 m.
4. Rate of Change of Velocity
Uniform motion: constant velocity, zero change. Non-uniform: varies, non-zero change. Acceleration a = (v-u)/t; positive same direction, negative opposite. SI m/s². Uniform: equal velocity changes equal time (free fall). Non-uniform: unequal changes. Activity 7.7: Thunder delay measures lightning distance (sound 346 m/s). Activity 7.8: Examples—(a) accelerating car forward; (b) braking opposite; (c) free fall uniform; (d) speeding car non-uniform.
Example 7.4: Bicycle Acceleration
Rahul: 0 to 6 m/s in 30s = 0.2 m/s²; brakes 6 to 4 m/s in 5s = -0.4 m/s². Shows positive/negative.
Uniform Acceleration: Free Fall
Body falls with constant g (~10 m/s²), velocity increases linearly. Example of straight-line uniform acceleration; contrasts non-uniform like varying car speed.
5. Graphical Representation of Motion
Line graphs: distance/velocity vs time. Distance-time: uniform straight line (Fig. 7.3), slope = speed v=(s2-s1)/(t2-t1). Non-uniform curved (Fig. 7.4, Table 7.2: 0-36m in 12s). Velocity-time: uniform horizontal (Fig. 7.5), area = displacement. Uniform acceleration straight line (Fig. 7.6, Table 7.3: 0-15 m/s in 30s). Area under v-t = distance (rectangle + triangle). Non-uniform any shape (Fig. 7.7). Activity 7.9: Train distances/times (Table 7.4), plot uniform segments. Activity 7.10: Feroz/Sania distances (Table 7.5), plot—Feroz uniform steeper, Sania non-uniform.
Distance-Time Graphs
Straight proportional for uniform; curved non-linear for accelerated. Use slope for instantaneous speed. E.g., small AB segment slope gives average over interval, approximates instantaneous.
Velocity-Time Graphs
Slope = acceleration; area = displacement. For uniform accel, straight line from origin. Non-uniform: varying slope. Practical: Speedometer data plots engine test velocity.
Questions on Graphs
Q1: Uniform straight, non-uniform curved. Q2: At rest (parallel time axis). Q3: Uniform speed (parallel time axis). Q4: Area below v-t = displacement.
6. Equations of Motion
For uniform acceleration straight line: v = u + at; s = ut + ½at²; v² = u² + 2as. Relate velocity, accel, distance, time. Derived graphically. Example 7.5: Train 0-72 km/h (20 m/s) in 5min (300s): a=1/15 m/s², s=3km. Example 7.6: Car 18-36 km/h (5-10 m/s) in 5s: a=1 m/s², s=37.5m. Example 7.7: Brakes -6 m/s², stop in 2s: u=12 m/s, s=12m (maintain distance caution).
Derivation Insight
From v-t graph: slope a=(v-u)/t → v=u+at. Area s= average v × t = (u+v)/2 × t → s=ut + ½at². Eliminate t: v²=u²+2as. Useful for unknown variables.
Practical: Braking Distance
Example 7.7 shows why spacing vehicles: even short time covers distance at speed. Applies to safety in roads, trains.
7. Uniform Circular Motion
Velocity changes direction: accelerated despite constant speed. Athlete rectangular track (Fig. 7.8a): 4 direction changes/round. Hexagonal (b): 6; octagonal (c): 8; limit circle (d). Speed v=2πr/t. Uniform circular: constant speed circular path. Activity 7.11: Thread stone circle, release tangential straight (direction at instant). Hammer throw: circular to release tangential. Examples: Moon/Earth orbit, satellite, cyclist track.
Direction Change Implication
Even constant magnitude, direction change = velocity change = acceleration (centripetal). Tangential release shows instantaneous direction preserved post-release, per inertia.
Everyday Circular Motions
Planets/stars, galaxies; controlled for orbits. Uncontrolled: erratic spins in storms. Study enables satellite tech, amusement rides safety.
8. Exercises and Applications
Exercises: Athlete 200m diameter track 40s/round; 2min20s: distance 2.2km, displacement 0 (back). Joseph jogs: A-B 300m/2.5min=2 km/h speed, 2 km/h velocity; A-C: 400m/3.5min≈1.14 km/h speed, -1.14 km/h velocity. Abdul round: 24 km/h average. Boat 3 m/s² 8s: 96m. Driver 52 km/h brakes. Graphs: Shade areas, uniform parts. Fig 7.10: A fastest; yes same point; C 100m when B passes A; B 200m passing C. Ball 20m 10 m/s²: 20 m/s velocity, 2s time. Speed-time Fig 7.11: 4s distance 40m; uniform middle. Possibles: (a) at top projectile; (b) uniform circular; (c) circular. Satellite 42250km 24h: ~3.07 km/s.
What You Have Learnt
Motion: position change, distance/displacement. Uniform/non-uniform via velocity constant/changing. Speed distance/time, velocity displacement/time. Acceleration velocity change/time. Graphs show motions. Equations: v=u+at, s=ut+½at², v²=u²+2as for uniform accel. Uniform circular: constant speed, direction change.
Broader Implications
Motion study foundational physics; applies engineering (vehicles), astronomy (orbits), safety (braking). Graphs/equations predict, control motions in tech, sports.
Questions and Answers from Chapter
Short Questions
Q1. An object has moved through a distance. Can it have zero displacement?
Answer: Yes.
Q2. What is the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds?
Answer: 10 m.
Q3. Which is true for displacement? (a) It cannot be zero.
Answer: False.
Q4. Its magnitude is greater than the distance travelled.
Answer: False.
Q5. Distinguish between speed and velocity.
Answer: Speed scalar, velocity vector.
Q6. Under what condition is magnitude of average velocity equal to average speed?
Answer: Straight line no reversal.
Q7. What does odometer measure?
Answer: Distance.
Q8. Path for uniform motion?
Answer: Straight line parallel time axis.
Q9. Signal from spaceship in 5 min at 3×10^8 m/s distance?
Answer: 9×10^10 m.
Q10. When is body in uniform acceleration?
Answer: Equal velocity changes equal time.
Q11. Non-uniform acceleration?
Answer: Unequal velocity changes.
Q12. Bus 80-60 km/h in 5s acceleration?
Answer: -2 m/s².
Q13. Train 40 km/h in 10 min acceleration?
Answer: 1/90 m/s².
Q14. Nature of distance-time for uniform motion?
Answer: Straight line.
Q15. For non-uniform?
Answer: Curved.
Q16. Motion if distance-time straight parallel time axis?
Answer: At rest.
Q17. Speed-time straight parallel time axis?
Answer: Uniform speed.
Q18. Quantity by area below velocity-time graph?
Answer: Displacement.
Q19. Athlete 200m diameter 40s round; distance 2min20s?
Answer: 2200 m.
Q20. Displacement at end?
Answer: 0 m.
Medium Questions
Q1. A farmer moves along boundary of square field 10m side in 40s. Magnitude of displacement at end of 2min20s?
Answer: Farmer completes 3 full rounds (120s) back to start, then half round (70s, but 140s total wait—2min20s=140s, 3.5 rounds: after 3 at start, 0.5 to opposite corner 10m displacement. (3 marks)
Q2. Distinguish speed and velocity.
Answer: Speed is scalar (magnitude, distance/time), velocity vector (magnitude+direction, displacement/time). Speed no direction, velocity changes if direction alters even constant speed. Units same m/s. (3 marks)
Q3. Condition for average velocity magnitude = average speed?
Answer: When displacement = distance, i.e., straight path without reversal. Path length equals shortest distance. In curves/reversals, displacement < distance. (3 marks)
Q4. What does odometer measure? Path for uniform motion?
Answer: Odometer: total distance travelled. Uniform: straight line on distance-time graph parallel? No, inclined straight; slope constant. Parallel time axis: rest. (3 marks)
Q5. Signal from spaceship 5min at light speed distance?
Answer: Time 300s, speed 3×10^8 m/s, distance = speed×time = 9×10^10 m. Illustrates vast space scales. (3 marks)
Q6. When uniform/non-uniform acceleration?
Answer: Uniform: equal velocity changes equal intervals (e.g., free fall). Non-uniform: unequal changes (e.g., car varying speed). (3 marks)
Q7. Bus 80-60 km/h in 5s acceleration?
Answer: u=80/3.6≈22.22 m/s, v=60/3.6≈16.67 m/s, t=5s, a=(v-u)/t≈ -1.11 m/s² (deceleration). (3 marks)
Q8. Train 40 km/h in 10min acceleration?
Answer: u=0, v=40/3.6≈11.11 m/s, t=600s, a=v/t≈0.0185 m/s² or 1/54 m/s² approx. (3 marks)
Q9. Nature of distance-time graphs uniform/non-uniform?
Answer: Uniform: straight line (proportional). Non-uniform: curved (non-linear variation). Slope gives speed. (3 marks)
Q10. Motion if distance-time straight parallel time axis?
Answer: Object at rest; no distance change with time. Horizontal line indicates zero speed. (3 marks)
Q11. Speed-time straight parallel time axis?
Answer: Uniform speed; constant velocity magnitude, horizontal line slope zero acceleration. (3 marks)
Q12. Quantity by area below velocity-time graph?
Answer: Displacement (or distance if straight positive). Integral of velocity over time. (3 marks)
Q13. Joseph jogs A to B 300m in 2.5min, back 100m in 1min; average speeds/velocities A-B and A-C?
Answer: A-B: speed 300m/150s=2 m/s, velocity +2 m/s. A-C: total 400m/210s≈1.9 m/s speed, velocity -400m/210s≈ -1.9 m/s (direction back). (3 marks)
Q14. Abdul 20 km/h to school, 30 km/h back; average for trip?
Answer: Assume equal distance d: total 2d/(d/20 + d/30)=2d/(3d/60 + 2d/60)=2d/(5d/60)=24 km/h. Harmonic mean for equal distances. (3 marks)
Q15. Motorboat 3 m/s² 8s distance?
Answer: u=0, a=3, t=8, s=½at²=½×3×64=96m. Uniform acceleration from rest. (3 marks)
Q16. Driver 52 km/h brakes; shade graph distance, uniform part?
Answer: Assume v-t decreasing; shade trapezoid area. Uniform: constant velocity segment before brakes. (3 marks)
Q17. Fig 7.10: Which fastest? Same point?
Answer: Steepest slope: A fastest. Intersect: yes, all meet at points. (3 marks)
Q18. How far C when B passes A? B when passes C?
Answer: From graph: C ~100m at intersection; B ~200m at next. (3 marks)
Q19. Ball dropped 20m at 10 m/s² velocity/time to ground?
Answer: u=0, a=10, s=20; v=√(u²+2as)=√400=20 m/s; t=√(2s/a)=2s. (3 marks)
Q20. Speed-time graph car first 4s distance? Uniform part?
Answer: Shade triangle area ~ (1/2×4×10)=20m assume; uniform horizontal segment. (3 marks)
Long Questions
Q1. Describe motion along straight line with distance/displacement example. When magnitude displacement = distance?
Answer: Simplest motion straight path. Reference O, positions A/B/C. Object O to A 60km, back to C via B 35km total: distance 95km (scalar path), displacement 35km (vector shortest O to C). Equals when no reversal, straight to end (e.g., O to A 60km both). Zero displacement if back to start, distance non-zero. Activities: Court walk distance sides, displacement diagonal; odometer vs map. Vectors include direction; scalars not. Essential for navigation, physics. (6 marks)
Q2. Explain uniform/non-uniform motion with examples. Data Table 7.1 analysis.
Answer: Uniform: equal distances equal time intervals (small), e.g., 5m each second. Non-uniform: unequal, e.g., car crowded street varying speed, jogger park. Table 7.1: A 10m/15min constant= uniform; B 12-44m uneven= non-uniform. Implications: Uniform predictable; non-uniform average used. Daily: Bus uniform highway, non-uniform traffic. Study controls erratic motions. (6 marks)
Q3. Define speed, average speed, velocity. Calculate car 100km/2h, object 16m/4s +16m/2s.
Answer: Speed: distance/time scalar. Average: total distance/total time non-uniform. Velocity: displacement/time vector. Car: 50 km/h average (may vary). Object: 32m/6s=5.33 m/s. Usha swim: 180m/60s=3 m/s speed, 0 velocity. Conversions: km/h to m/s ×5/18. Units m/s. Velocity includes direction for full description. (6 marks)
Q4. Explain acceleration, uniform/non-uniform. Bicycle example 7.4 details.
Answer: a=(v-u)/t; positive same direction velocity, negative opposite. Uniform: equal changes equal time (free fall). Non-uniform: unequal (car road). Rahul: 0-6 m/s/30s=0.2 m/s² accel; 6-4 m/s/5s=-0.4 m/s² decel. Units m/s². Activity: Thunder delay distance sound speed. Examples: Forward accel car, brake opposite, uniform fall, non-uniform speed up. (6 marks)
Q5. Describe distance-time graphs uniform/non-uniform. How find speed? Table 7.2 plot.
Answer: Uniform: straight proportional time, slope speed. Non-uniform: curved. Speed: slope (s2-s1)/(t2-t1) small segment. Table 7.2: 0-36m/12s curved accel. Fig 7.3 uniform line; 7.4 non. Activity 7.9 train uniform segments plot. Interprets motion visually, predicts positions. (6 marks)
Q6. Velocity-time graphs: uniform, accelerated. Area meaning? Table 7.3 example.
Answer: Uniform: horizontal, area rectangle displacement. Accelerated uniform: straight slope a, area rectangle+triangle. Table 7.3: 0-15 m/s/30s line, area ~135 m. Non-uniform curved. Fig 7.5 uniform 40 km/h; 7.6 accel. Activity 7.10 bicycles: Feroz uniform, Sania non. Slope acceleration, area distance. (6 marks)
Q7. Derive equations of motion. Train example 7.5 application.
Answer: Uniform accel: v=u+at (slope v-t). s=ut+½at² (area). v²=u²+2as (elim t). Train: 0-20 m/s/300s a=1/15 m/s²; v²=2as=400, s=3000m=3km. Car 7.6:1 m/s²,37.5m. Brakes 7.7:12m stop. Solves unknowns graphically derived. (6 marks)
Q8. Explain uniform circular motion. Why accelerated? Activity 7.11 observation.
Answer: Constant speed circular path, v=2πr/t. Direction changes each point= velocity change= acceleration (centripetal). Tracks: rectangular 4 changes, hexagonal 6, octagonal 8, circle infinite small. Activity: Stone thread circle, release tangential straight; direction at instant. Hammer/discus release tangential. Moon orbit example. (6 marks)
Q9. Bus 0.1 m/s² 2min: speed acquired, distance?
Answer: t=120s, v=u+at=12 m/s. s=ut+½at²=½×0.1×14400=720m. Applies equations growth from rest. (6 marks)
Q10. Train 90 km/h brakes -0.5 m/s² stop distance?
Answer: u=25 m/s, v=0, v²=u²+2as, 0=625-1s, s=312.5m. Safety calc. (6 marks)
Q11. Trolley incline 2 cm/s² velocity 3s start?
Answer: u=0, a=0.02 m/s², v=at=0.06 m/s. Uniform accel down. (6 marks)
Q12. Racing car 4 m/s² 10s distance?
Answer: u=0, s=½at²=½×4×100=200m. From rest. (6 marks)
Q13. Stone upward 5 m/s, a=-10 m/s² height/time?
Answer: v=0 top, 0=25-20h, h=1.25m; t=v/a=0.5s. Projectile top. (6 marks)
Q14. Situations possible? Examples: constant a zero v; a uniform speed; a perpendicular direction.
Answer: (a) Yes, projectile top momentary zero v constant a. (b) Yes, uniform circular constant speed a centripetal. (c) Yes, circular motion tangential v, radial a. (6 marks)
Q15. Satellite 42250 km orbit 24h speed?
Answer: r=42250×10^3 + Earth radius approx, but given r, circumference 2πr≈2.65×10^8 m, t=86400s, v≈3070 m/s. Uniform circular. (6 marks)
Q16. Explain relative motion with bus example. Implications daily life.
Answer: Bus passengers: trees backward, co-passengers rest; roadside: bus forward. Relative to frame. Implications: Illusions like stationary train moving; safety judging speeds; astronomy Earth motion inferred. (6 marks)
Q17. Differentiate distance/displacement with 60km forth 25km back example.
Answer: Distance 85km total path scalar. Displacement 35km net vector. Zero if full back. Used navigation (GPS displacement), travel logs (odometer distance). (6 marks)
Q18. Describe average speed/velocity. Usha swim example.
Answer: Average speed total distance/time=180m/60s=3 m/s. Velocity displacement/time=0/60s=0 m/s. For non-straight, differ. (6 marks)
Q19. Bus decel 80-60 km/h 5s find a.
Answer: Convert u=22.22 m/s v=16.67 m/s t=5 a=(16.67-22.22)/5=-1.11 m/s². Negative decel. (6 marks)
Q20. Train uniform accel 40 km/h 10min a and distance.
Answer: v=11.11 m/s t=600s a=0.0185 m/s²; s=½vt=3330m approx. Equations apply. (6 marks)
