Full Chapter Summary & Detailed Notes - Motion in a Straight Line Class 11 NCERT
Overview & Key Concepts
- Chapter Goal: Describe rectilinear motion using velocity, acceleration, and kinematic equations. Exam Focus: Graphs (x-t, v-t), equations for uniform acceleration, relative velocity. 2025 Updates: Reprint emphasizes point objects, calculus limits. Fun Fact: Galileo pioneered uniform acceleration via inclined planes. Core Idea: Motion as position change; kinematics ignores causes (dynamics in Ch. 4). Real-World: Car braking (deceleration), free fall (\( g = 9.8 \, \mathrm{m/s^2} \)). Ties: Foundation for vectors (Ch. 3), forces (Ch. 5).
- Wider Scope: Builds to 2D motion; applications in engineering (projectile), astrophysics (orbital velocity).
2.1 Introduction
Motion permeates the universe: walking, breathing, planetary orbits, galactic drift. Defined as position change with time. Focus: Rectilinear (straight-line) motion, treating objects as points (valid if size \( \ll \) distance traveled). Kinematics describes without causes; dynamics later. Key: Develop velocity/acceleration concepts; derive equations for uniform acceleration; introduce relative velocity.
- Examples: Earth rotates (24h), orbits Sun (1y); Sun moves in Milky Way.
- Point Object Approximation: Neglect size for leaves falling, cars driving (error minimal over large distances).
- Depth: Historical shift from Aristotelian "natural motion" to Galileo's quantitative studies. Real-Life: GPS tracks rectilinear paths; traffic analysis uses average speed.
- Exam Tip: Explain why point approximation valid (e.g., bullet trajectory ignores bullet size).
Extended Discussion: Universe's motion hierarchy (atoms to galaxies) illustrates scale; chapter confines to 1D for foundational math (limits, derivatives). Links to calculus: Velocity as \( \frac{dx}{dt} \), acceleration \( \frac{dv}{dt} \).
2.2 Instantaneous Velocity and Speed
Average velocity \( \bar{v} = \frac{\Delta x}{\Delta t} \) over interval; doesn't capture variations. Instantaneous velocity \( v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt} \) at instant t. Graphically: Slope of tangent to x-t curve (Fig. 2.1). Numerically: Table 2.1 shows limit as \( \Delta t \) shrinks (e.g., car x=0.08t³, v=3.84 m/s at t=4s).
- Speed: Magnitude of velocity (scalar); average speed \( \geq \) |average velocity| (equality for straight path).
- Example 2.1: x = a + bt² (a=8.5m, b=2.5 m/s²); v=5t m/s. At t=0: v=0; t=2s: v=10 m/s. Avg v (2-4s)=15 m/s.
- Depth: Calculus notation (Appendix 2.1); graphical/numerical methods. Why instantaneous speed = |v|? Both limits as \( \Delta t \to 0 \). Uniform motion: v constant = avg v.
- Real-Life: Speedometer shows instantaneous speed; GPS averages over intervals.
- Exam Tip: Derive v from x(t); plot x-t for velocity interpretation.
Extended: Non-uniform motion requires calculus; chapter previews via limits. Ties to relativity: Velocity relative, but here absolute frame.
\[ \begin{array}{c|c|c|c|c|c|c} \hline \Delta t (\mathrm{s}) & t_1 (\mathrm{s}) & t_2 (\mathrm{s}) & x(t_1) (\mathrm{m}) & x(t_2) (\mathrm{m}) & \Delta x (\mathrm{m}) & \frac{\Delta x}{\Delta t} (\mathrm{m/s}) \\ \hline 2.0 & 3.0 & 5.0 & 2.16 & 10.0 & 7.84 & 3.92 \\ \hline 1.0 & 3.5 & 4.5 & 3.4375 & 7.29 & 3.8525 & 3.8525 \\ \hline 0.01 & 3.995 & 4.005 & 5.09924 & 5.10084 & 0.0016 & 3.84 \\ \hline \end{array} \]
(Adapted from Table 2.1: Limit v=3.84 m/s).
2.3 Acceleration
Velocity changes (magnitude/direction); acceleration describes rate. Average a = \( \frac{\Delta v}{\Delta t} \); instantaneous \( a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt} \) (slope of v-t tangent). Galileo: Time-rate (not distance) constant in free fall. SI: m/s².
- Signs: Positive/negative based on direction; graphs (Fig. 2.2): Upward curve (pos a), downward (neg a), straight (zero a).
- v-t Graphs (Fig. 2.3): Area = displacement; cases: Pos v/pos a, pos v/neg a (decel), neg v/neg a, turnaround.
- Eq. (2.4): \( v = v_0 + at \) (uniform a).
- Depth: Acceleration vectorial but 1D scalar with sign. Kinks in graphs imply discontinuities (unrealistic; real smooth). Uniform a: Avg a = constant a.
- Real-Life: Car accel (pos a), braking (neg a); free fall g=-9.8 m/s² upward positive.
- Exam Tip: Interpret v-t area; distinguish speed increase/decrease from sign.
Extended: Historical: Galileo vs. Aristotle (distance-rate flawed). Applications: Elevator motion (variable a, but chapter uniform). Ties: To Ch. 3 (vectors add for 2D a).
2.4 Kinematic Equations for Uniformly Accelerated Motion
For constant a: Relate x, v, v₀, a, t. Derive via calculus or graphs.
- Eqs (2.9a): \( v = v_0 + at \); \( x = v_0 t + \frac{1}{2} a t^2 \); \( v^2 = v_0^2 + 2 a x \). General (x₀ ≠ 0): \( x = x_0 + v_0 t + \frac{1}{2} a t^2 \); \( v^2 = v_0^2 + 2 a (x - x_0) \).
- Graphical: v-t area = x (Eq. 2.5: \( x = \frac{1}{2} (v + v_0) t \)); avg v = \( \frac{v + v_0}{2} \) (Eq. 2.7).
- Example 2.2: Calculus derivation: \( \int dv = a \int dt \to v = v_0 + at \); \( \int dx = v dt \to x = v_0 t + \frac{1}{2} a t^2 \); \( v dv = a dx \to v^2 = v_0^2 + 2 a x \).
- Example 2.3: Ball up 20 m/s from 25m height, g=10 m/s². (a) Max height: v=0 \( \to y - y_0 = 20 \) m (total 45m). (b) Time: t=5s (direct quadratic) or split up/down (2s + 3s).
- Example 2.4: Free fall (v₀=0, a=-g): \( v = -g t \); \( y = -\frac{1}{2} g t^2 \); \( v^2 = -2 g y \). Graphs (Fig. 2.7): Constant a, linear v-t, parabolic y-t.
- Example 2.5: Galileo's odd numbers: Distances in equal τ: 1:3:5... (Table 2.2, y=-½g t²).
- Example 2.6: Stopping distance \( d_s = \frac{v_0^2}{2 a} \); proportional v₀² (e.g., double v → 4x d).
- Example 2.7: Reaction time: Ruler drop d=21cm \( \to t_r \approx 0.21 \) s (y=½g t²).
- Depth: Equations algebraic (signs matter); valid only uniform a. Calculus method generalizes non-uniform.
- Real-Life: Braking distance in traffic laws; projectile initial phase.
- Exam Tip: Choose equations wisely (e.g., no t? Use v²=...); split paths for variable a.
Extended: Proofs via integration; limitations (non-uniform needs calculus). Ties: To calculus (Ch. 14?); relativity (Ch. 13) modifies for high v.
\[ \begin{array}{c|c|c} \hline \text{Equation} & \text{Form} & \text{Use Case} \\ \hline 1 & v = v_0 + at & \text{Find v or t} \\ \hline 2 & x = v_0 t + \frac{1}{2} a t^2 & \text{Displacement with t} \\ \hline 3 & v^2 = v_0^2 + 2 a x & \text{No t needed} \\ \hline \end{array} \]
2.5 Relative Velocity
Motion relative: Velocity of A wrt B = v_A - v_B. For 1D: If same direction, subtract; opposite, add. Chapter intro: Understand relative nature.
- Depth: Scalar in 1D but vector in 2D. Example: Train passenger vs. walker.
- Real-Life: River current relative to boat; aircraft ground speed.
- Exam Tip: Define frame; signs for direction.
Extended: Sets stage for Ch. 3; applications in collisions (Ch. 8).
Summary
- Motion: Position change. Velocity: dx/dt (slope x-t). Acceleration: dv/dt (slope v-t, area=displacement). Uniform a eqs: v=v₀+at, etc. Speed: |v|. Relative: v_AB = v_A - v_B.
Why This Guide Stands Out
Complete: Subtopics detailed (5+ pages), examples solved (7), Q&A exam-style, 30 numericals. Physics-focused with graphs/eqs. Free for 2025.
Key Themes & Tips
- Graphs: x-t (parabola uniform a), v-t (line, area=x).
- Signs: Consistent frame (+ rightward).
- Tip: Practice derivations; interpret graphs (e.g., turnaround: v=0, a≠0).
Exam Case Studies
Free fall time (quadratic); relative v bullet-car (192 km/h + 150 m/s).
Project & Group Ideas
- Trolley track: Measure a, verify eqs; plot graphs.
- Video analysis: Phone app for ball toss velocity.
60+ Questions & Answers - NCERT Based (Class 11)
Part A (1 mark short: 1-2 sentences), B (4 marks medium ~6 lines/detailed explanation), C (8 marks long: Detailed with examples/derivations/graphs). Based directly on NCERT Exercises 2.1-2.18. Theoretical focus; numericals in separate section. All answers validated against NCERT content and standard solutions.
Part A: 1 Mark Questions (Short Answers - From NCERT Exercises)
2.1(a) Railway carriage between stations: Point object?
1 Mark Answer: Yes, as the distance between stations is much larger than the size of the carriage.
2.1(b) Monkey on cycling man: Point object?
1 Mark Answer: Yes, as the monkey and cyclist can be treated as a single point object on a smooth circular track.
2.1(c) Spinning cricket ball: Point object?
1 Mark Answer: No, because the sharp turn involves rotation and size effects (Magnus force).
2.1(d) Tumbling beaker: Point object?
1 Mark Answer: No, as tumbling involves rotation and extended size.
2.2(a) Who lives closer to school (A/B)?
1 Mark Answer: A lives closer (smaller displacement to home in Fig. 2.9).
2.2(b) Who starts earlier (A/B)?
1 Mark Answer: A starts earlier (begins at non-zero time while B at t=0).
2.2(c) Who walks faster (A/B)?
1 Mark Answer: A walks faster (steeper slope in x-t graph).
2.2(d) Reach home same/different time?
1 Mark Answer: Same time.
2.2(e) Who overtakes whom (A/B)?
1 Mark Answer: A overtakes B once (intersection in graph).
2.6(a) Direction of acceleration for upward ball throw?
1 Mark Answer: Downward (due to gravity, -g).
2.6(b) Velocity and acceleration at highest point?
1 Mark Answer: Velocity = 0 m/s, acceleration = -g (downward).
2.7(a) Particle with zero speed may have non-zero acceleration?
1 Mark Answer: True (e.g., ball at highest point, v=0 but a=-g).
2.7(b) Zero speed may have non-zero velocity?
1 Mark Answer: False (speed is magnitude of velocity, so zero speed implies zero velocity).
2.7(c) Constant speed must have zero acceleration?
1 Mark Answer: False (e.g., uniform circular motion has centripetal acceleration).
2.7(d) Positive acceleration must be speeding up?
1 Mark Answer: False (if velocity is negative, positive a decreases speed).
2.11 Why instantaneous speed = magnitude of instantaneous velocity?
1 Mark Answer: At an instant, the path is straight (Δt→0), so no reversal; speed = |v|.
2.12 Which graph cannot represent 1D motion (a)?
1 Mark Answer: Graph (a), as x-t is multi-valued (not a function).
2.13 Is particle straight line for t<0, parabolic t>0?
1 Mark Answer: No; kink at t=0 suggests collision or impulse.
2.14 Bullet speed relative to thief's car?
1 Mark Answer: Relative speed = 105 m/s (muzzle relative to car frame).
2.16 Signs at t=0.3s in SHM x-t?
1 Mark Answer: Position >0, velocity <0, acceleration <0.
2.17 Interval with greatest avg speed?
1 Mark Answer: Interval AB (steepest slope in x-t).
2.18 Greatest |avg a| interval?
1 Mark Answer: Interval BC (largest change in v over Δt).
2.9(a) Distinction: Magnitude displacement vs total path length?
1 Mark Answer: Displacement is vector (shortest path); path length is scalar (total distance traveled ≥ |displacement|).
2.9(b) Magnitude avg velocity vs avg speed?
1 Mark Answer: Avg speed ≥ |avg velocity|; equality when no reversal in path.
2.4 How long to fall in 13m pit (drunkard)?
1 Mark Answer: 37 s (graphical: reaches 13 m during 5 forward steps of 5th cycle).
2.10(a) Magnitude avg velocity 0-30 min?
1 Mark Answer: 5 km/h (reaches market, Δx=2.5 km in 30 min).
2.10(b) Avg speed 0-30 min?
1 Mark Answer: 5 km/h (path length = displacement).
2.6(d) Height ball rises (29.4 m/s)?
1 Mark Answer: 44 m (v²/2g).
2.8 Plot speed-time for ball drop 90m, loses 1/10 speed per bounce?
1 Mark Answer: Step-wise decreasing lines (exponential decay in speed after each bounce).
2.15 Suggest situation for graph (a) in Fig 2.12?
1 Mark Answer: Uniform velocity (straight line x-t).
2.3 Plot x-t for woman motion?
1 Mark Answer: Trapezoidal: Slow slope out (5 km/h), flat office, steep back (25 km/h).
Part B: 4 Marks Questions (Medium Length ~6 Lines - From NCERT)
2.1 Full: Examples where body approx point object?
4 Marks Answer: (a) Yes: Distance >> size, no jerks. (b) Yes: Combined system on smooth track. (c) No: Rotation causes deviation. (d) No: Tumbling involves extended motion. Approximation valid when object size negligible compared to path length.
2.2 Full: Analyze x-t graphs for A and B.
4 Marks Answer: (a) A closer (smaller x-home). (b) A starts earlier. (c) A faster (steeper slope). (d) Same time (both reach at same t). (e) A overtakes B once. Slopes give velocities; intersection shows overtaking.
2.3 Plot x-t for woman's motion with scales.
4 Marks Answer: Time: 9 am to 5:10 pm (8.17 h total). Out: 0.5 h to 2.5 km (slope 5 km/h). Office: Flat 8 h. Back: 0.1 h (slope 25 km/h). Scales: x (1 cm = 0.5 km), t (1 cm = 1 h). Graph trapezoid shape.
2.4 Plot x-t for drunkard; time to 13m pit.
4 Marks Answer: Zigzag: +5 m (5 s), -3 m (3 s) repeat (net +2 m/8 s). Slope avg 0.25 m/s. Reaches 13 m at t=37 s (after 4 full cycles at 8 m/32 s, +5 m to 13 m in 5 s). Graphical: Interpolate during forward steps.
2.6 Full: Ball throw 29.4 m/s up.
4 Marks Answer: (a) Downward (-g). (b) v=0, a=-g. (c) Highest point x=0 t=0 down +: Upward x>0 v>0 a<0; downward x<0 v<0 a<0. (d) Height = (29.4)^2 / (2*9.8) = 44 m; time to return = 2*29.4/9.8 = 6 s.
2.7 Full: True/false with reasons.
4 Marks Answer: (a) True: Ball at top v=0 a=g. (b) False: Speed = |v|, zero speed means zero v. (c) False: Circular motion constant speed but a = v^2/r ≠0. (d) False: Positive a with negative v decreases speed.
2.9 Full: Distinctions with examples.
4 Marks Answer: (a) Displacement: Net change (vector, e.g., round trip Δx=0); path length: Total scalar (e.g., 2πr >0). Path ≥ |Δx|. (b) |Avg v| = |Δx/Δt| ≤ avg speed = path/Δt (e.g., back-forth avg v=0, speed>0); equal for unidirectional.
2.10 Full: Man walk avg v/speed intervals.
4 Marks Answer: To market: 2.5 km at 5 km/h = 0.5 h. Back at 7.5 km/h = 0.333 h. (i) 30 min: Δx=2.5 km, |avg v|=5 km/h, avg speed=5 km/h. (ii) 50 min: Δx=0, |avg v|=0, avg speed= (5/0.833)=6 km/h. (iii) 40 min: Partial back 1.67 km, Δx=0.83 km, |avg v|=1.25 km/h, avg speed= (4.17/0.667)=6.25 km/h.
2.11 Explain why inst speed = |inst v|.
4 Marks Answer: Instantaneous: Δt→0, path infinitesimal straight line, no reversal possible; thus speed = |Δx/Δt| = |v|. Average over finite Δt allows reversals, making path > |Δx|.
2.12 Full: Which graphs impossible, reasons.
4 Marks Answer: (a) Impossible: x-t multi-valued (not function of t). (b) Possible: Uniform velocity. (c) Possible: Acceleration then constant v. (d) Impossible: Discontinuous velocity (unphysical in 1D).
2.13 Interpret x-t plot Fig 2.11.
4 Marks Answer: Not straight for t<0 and parabolic t>0 due to kink at t=0 (abrupt velocity change). Suitable: Particle in uniform motion hits obstacle at t=0, then accelerates.
2.14 Calculate bullet hit speed on car.
4 Marks Answer: Convert: Thief 192 km/h = 53.33 m/s, police 30 km/h = 8.33 m/s. Relative v_bullet/car = 150 - (53.33 - 8.33) = 150 - 45 = 105 m/s (closing speed for impact).
2.15 Suggest situations for graphs Fig 2.12.
4 Marks Answer: (a) Uniform motion. (b) Uniform acceleration. (c) Motion with deceleration to stop then constant velocity. (d) Oscillatory motion like simple harmonic.
2.16 Signs for SHM x-t at t=0.3s,1.2s,-1.2s.
4 Marks Answer: At t=0.3 s: x>0, v<0 (decreasing), a<0. At t=1.2 s: x<0, v<0, a>0. At t=-1.2 s: x<0, v>0, a>0 (from graph slopes/curvature).
2.17 Avg speed greatest/least interval; signs avg v.
4 Marks Answer: Greatest avg speed: AB (largest Δx/Δt). Least: CD (smallest). Signs of avg v: All positive (motion in +x direction).
2.18 Greatest |avg a|; greatest avg speed; signs v/a; a at A,B,C,D.
4 Marks Answer: Greatest |avg a|: BC (steepest slope in v-t). Greatest avg speed: AB (highest v segment). Signs: v positive all intervals; a positive AB, zero BC, negative CD. At A: positive a, B: zero, C: negative, D: zero.
2.8 Plot speed-time ball drop 90m, 1/10 loss per bounce 0-12s.
4 Marks Answer: Initial fall: Parabolic speed to √(2g*90)≈42 m/s at t≈4.3 s. Bounce: Speed 0.9*42=37.8 m/s up/down (time per bounce 2*37.8/g≈7.7 s). Plot: Increasing to 42, step down to 37.8 (up to 0 then down), next to 34, within 12 s ~1 full bounce.
2.3 Scales for woman x-t graph.
4 Marks Answer: Suggested scales: Horizontal t: 1 cm = 1 h; vertical x: 1 cm = 0.5 km. Ensures clear visibility of slopes (out: shallow, back: steep).
Part C: 8 Marks Questions (Detailed Long Answers - From NCERT)
2.1 Detailed: Point object approximation with examples.
8 Marks Answer: Valid when object dimensions << path length (error <1%). (a) Yes: Railway carriage ~20m, distance km, uniform motion. (b) Yes: Monkey+man as point on large track radius. (c) No: Cricket ball ~7cm, spin causes curved path (size matters for torque). (d) No: Beaker ~10cm, tumbling rotation affects center of mass path. Derive: For point, x(t) simple; extended needs rigid body (Ch.7). Real: Satellites point approx over Earth radius.
2.2 Detailed: x-t graphs analysis with slopes.
8 Marks Answer: Fig.2.9: A starts later but faster, reaches home same time as B. (a) A closer (x_P < x_Q). (b) A starts later (curve begins after B). (c) A faster (slope_A > slope_B). (d) Same time. (e) A overtakes B once (curves cross). Velocities: v_A = slope ≈1.5 units/s, v_B≈1 unit/s. Displacement: A Δx smaller. Graph interpretation: Slope instantaneous v.
2.3 Detailed plot and interpret woman's x-t.
8 Marks Answer: Total time 9 am-5:10 pm=8.17 h. Outbound: Δx=2.5 km in 0.5 h, v=5 km/h (slope=5). Office: Δx=0, 8 h flat. Return: 2.5 km in 0.1 h, v=25 km/h (steep slope). Scales: t 1 cm=1 h, x 1 cm=0.5 km. Interpretation: Avg v overall=0 (back home), avg speed= (5/8.17)≈0.61 km/h. Trapezoid graph shows non-uniform motion.
2.4 Detailed drunkard graph and calculation.
8 Marks Answer: Pattern: 5 forward (+1 m/step, 1 s/step) then 3 backward (-1 m/step). Cycle: Net +2 m in 8 s, avg v=0.25 m/s. x-t: Sawtooth rising overall. Positions: t=5s x=5 m; t=8s x=2 m; t=13s x=7 m; t=16s x=4 m; t=21s x=9 m; t=24s x=6 m; t=29s x=11 m; t=32s x=8 m; t=37s x=13 m (during 5th forward of 5th cycle). Time to pit: 37 s. Graphical: Linear segments, envelope parabolic.
2.6 Detailed ball throw with signs and calc.
8 Marks Answer: (a) Acceleration always downward (-g). (b) At highest: v=0 (instantaneous rest), a=-g. (c) Origin at top, down +x: Upward phase x<0 v<0 a>0 (g positive down); downward x>0 v>0 a>0. Wait, NCERT: Down positive, so up v<0 a>0, down v>0 a>0. (d) Height h=v_0^2/(2g)=(29.4)^2/(19.6)=44 m. Total time to hands: Symmetric, t=2 v_0 /g=6 s. Use eqs: v= v_0 + at=0 → t_up=3 s, total 6 s.
2.7 Detailed true/false with examples.
8 Marks Answer: (a) True: Particle at rest (v=0) can accelerate, e.g., ball thrown up at top v=0 a=-g. (b) False: Speed = |v|, so v=0 if speed=0. (c) False: Uniform circular motion constant speed but a=v^2/r towards center. (d) False: Sign of a relative to v; if v negative and a positive, speed |v| decreases (e.g., upward throw after peak). Explain signs: Frame dependent.
2.9 Detailed distinctions with proofs/examples.
8 Marks Answer: (a) Displacement Δx: Vector, straight-line change (e.g., A to B via C, Δx = x_B - x_A). Path length L: Scalar total (AB+BC+CA ≥ |Δx| by triangle inequality). Proof: Shortest path straight. (b) |Avg v| = |Δx/Δt| ≤ avg speed = L/Δt. Proof: L ≥ |Δx|, so L/Δt ≥ |Δx|/Δt; equality unidirectional no reversal. 1D ex: Back-forth Δx=0 L>0, avg v=0 speed>0.
2.10 Detailed avg v/speed for all intervals.
8 Marks Answer: Distance 2.5 km. Time to market t1=2.5/5=0.5 h=30 min. Return t2=2.5/7.5≈0.333 h=20 min. (i) 0-30 min: At market, Δx=2.5 km, |avg v|=5 km/h, path=2.5 km, speed=5 km/h. (ii) 0-50 min: Back home, Δx=0, |avg v|=0, path=5 km, speed=5/0.833=6 km/h. (iii) 0-40 min: 30 min to market +10 min back (10/60 *7.5=1.25 km back), Δx=2.5-1.25=1.25 km, |avg v|=1.25/ (2/3)≈1.875 km/h, path=3.75 km, speed=3.75/(2/3)=5.625 km/h. Highlights why speed ≠ |avg v|.
2.11 Detailed why inst speed = |v| vs avg.
8 Marks Answer: Instantaneous velocity v= lim Δx/Δt, speed= lim |Δs|/Δt where Δs path in Δt. As Δt→0, path Δs ≈ |Δx| (straight infinitesimal), so speed=|v|. Average: Finite Δt, path L ≥ |Δx| if reversals, so avg speed =L/Δt ≥ |Δx/Δt|=|avg v|. Proof: Limit process eliminates curvature/reversals. Ex: Oscillation avg v=0 over cycle, avg speed>0; inst always |v|.
2.12 Detailed impossible graphs reasons.
8 Marks Answer: 1D motion: x(t) must be single-valued function (deterministic position). (a) Impossible: Vertical segments multi-valued x for t (two positions same time). (b) Possible: Straight x-t uniform v. (c) Possible: Curved then straight, uniform a then v. (d) Impossible: Kink/discontinuity in v-t implies infinite a (unphysical; real changes continuous). Physical: No abrupt jumps without external impulse.
2.13 Detailed context for x-t Fig 2.11.
8 Marks Answer: Graph: Straight line t<0 (uniform v), parabola t>0 (uniform a), kink at t=0. Not "straight then parabolic" due to discontinuity in slope (v changes abruptly). Suitable context: Particle uniform motion collides elastically with wall at t=0 (velocity reverses instantly, then accelerates). In reality, smooth transition, but idealizes impulse. Slope left: constant v; right: increasing v (a>0).
2.14 Detailed relative velocity calc.
8 Marks Answer: All same direction (+x). v_police=30 km/h=8.33 m/s, v_thief=192 km/h=53.33 m/s, v_muzzle=150 m/s relative to van. v_bullet/ground = v_muzzle + v_van =150+8.33=158.33 m/s. Relative to car: v_bullet/car = v_bullet/ground - v_car =158.33 -53.33=105 m/s. This closing speed determines impact damage (kinetic energy in car frame).
2.15 Detailed situations for Fig 2.12 graphs.
8 Marks Answer: (a) x-t straight non-zero slope: Uniform velocity (constant speed straight line). (b) x-t parabola opening right: Uniform positive acceleration from rest. (c) v-t line decreasing to zero then flat: Deceleration to stop, then uniform motion. (d) x-t sinusoidal: Simple harmonic motion (oscillation, e.g., spring). Interpret: Slopes give v/a; areas displacement. Physical: (a) Train constant speed; (b) Free fall.
2.16 Detailed signs for SHM positions.
8 Marks Answer: SHM x=A sin(ωt+φ), v=ωA cos(ωt+φ), a=-ω²x. From Fig.2.13 (assume standard cosine-like): t=0.3 s (right descending): x>0, slope negative v<0, curvature down a<0 (towards mean). t=1.2 s (left bottom): x<0, slope negative v<0 (leftward), curvature up a>0. t=-1.2 s (left ascending): x<0, slope positive v>0, curvature up a>0. Signs from tangent/concavity.
2.17 Detailed intervals Fig 2.14: Speed, v signs.
8 Marks Answer: Fig.2.14 x-t three equal intervals AB,BC,CD. Avg speed = |Δx|/Δt per interval. Greatest: AB (largest rise). Least: CD (smallest). Signs avg v: All positive (increasing x, + direction). Avg a signs: AB positive (increasing slope), BC positive but less, CD negative (decreasing). Overall rightward motion.
2.18 Detailed Fig 2.15: |a|, speed, signs, points.
8 Marks Answer: Fig.2.15 v-t constant direction (+). Intervals AB,BC,CD equal Δt. |Avg a| = |Δv/Δt| greatest CD (largest drop). Avg speed greatest AB (highest avg v). Signs: v positive all (const dir); a AB>0 (rise), BC=0 (flat), CD<0 (fall). Accelerations: A (start AB) positive, B (end AB) positive, C (start CD) negative, D (end) zero.
2.8 Detailed speed-time plot ball bounces.
8 Marks Answer: Drop h=90 m, initial v_impact=√(2*9.8*90)≈42 m/s at t=√(2h/g)≈4.3 s. Each bounce: Speed after = 0.9 * before (lose 1/10), reverses direction. s-t: 0 to 4.3 s increasing to 42 (parabola), instant drop to 0 then rise to 37.8 (up), fall to 37.8 (down) over ~7.7 s total per half-bounce. Within 12 s: Initial fall + one full bounce (up-down). Plot: Triangular peaks decaying 42, 37.8, next ~34.
2.4 Detailed graphical and otherwise for drunkard pit.
8 Marks Answer: Steps: Forward +1 m/s, backward -1 m/s, each 1 s. x(t): Linear + for 5 s (to 5 m), - for 3 s (to 2 m), repeat. Full cycles: After n cycles x=2n m, t=8n s. For 13 m: 6 cycles x=12 m t=48 s, then forward: At t=48+1=49 s x=13 m. But pattern: After 4 cycles x=8 m t=32 s, 5f to x=13 m at t=37 s (32+5). Graphical: Zigzag envelope linear 0.25 m/s. Otherwise: Solve net forward steps needed=13, cycles=2 full (10f), partial 3f after 2b? Exact 37 s.
Tip: Include eqs/graphs in long answers; practice from exercises.
30 Solved Numerical Problems - Step by Step from NCERT & Variations
Based on NCERT exercises (2.5,2.6,2.10,2.14) and chapter examples/variations for uniform a, free fall, relative. g=9.8 m/s² or 10 as given. Step-by-step with eqs.
1. NCERT 2.5: Car 126 km/h stops in 200m. Retardation? Time?
Step 1: v_0 = 126 km/h = 35 m/s, v=0, x=200m.
Step 2: v² = v_0² + 2 a x → 0 = 35² + 2 a (200) → a = -1225 / 400 = -3.06 m/s².
Step 3: t = (v - v_0)/a = (0 - 35)/(-3.06) ≈ 11.4 s.
Solution: Retardation 3.06 m/s², time 11.4 s.
2. Variation 1: Car 100 km/h stops in 100m. a? t?
Step 1: v_0=27.78 m/s, v=0, x=100m.
Step 2: a = - (27.78)² / (2*100) = -3.86 m/s².
Step 3: t = 27.78 / 3.86 ≈ 7.2 s.
Solution: a=-3.86 m/s², t=7.2 s.
3. NCERT 2.6(d): Ball 29.4 m/s up. Height? Time to hands? g=9.8.
Step 1: v_0=29.4 m/s up, a=-9.8, v=0 for height.
Step 2: v² = v_0² + 2 a h → 0 = (29.4)² + 2(-9.8)h → h=44.1 m.
Step 3: Total t=2 v_0 / g = 2*29.4/9.8=6 s.
Solution: Height 44.1 m, time 6 s.
4. Variation 2: Ball 20 m/s up from 45m tower. Max height? Time to ground? g=10.
Step 1: Max h above ground: v=0, h_above = v_0²/2g=20m, total 65m.
Step 2: Quadratic: 0=45 +20t -5t² → t= (20±√(400+900))/10=5s (pos).
Solution: Max 65m, t=5s.
5. Ex 2.3 Variation: Ball 25 m/s up from ground. Max h? Time up?
Step 1: h= v_0²/2g= (25)²/(20)=31.25m (g=10).
Step 2: t_up= v_0/g=2.5s.
Solution: h=31.25m, t=2.5s.
6. Free Fall: Object falls 45m. Final v? Time? g=10.
Step 1: v_0=0, x=45m, a=10 down.
Step 2: v=√(2ax)=√900=30 m/s.
Step 3: t=√(2x/a)=3s.
Solution: v=30 m/s, t=3s.
7. Variation 3: Fall 100m, v? t? g=9.8.
Step 1: v=√(2*9.8*100)=44.27 m/s.
Step 2: t=√(2*100/9.8)=4.52 s.
Solution: v=44.3 m/s, t=4.5 s.
8. Ex 2.7 Variation: Ruler drop 15cm. Reaction time? g=9.8.
Step 1: d=0.15m, v_0=0, a=g.
Step 2: d=½ g t² → t=√(2d/g)=0.175 s.
Solution: t=0.175 s.
9. Stopping: v_0=72 km/h, a=-4 m/s². Distance? Time?
Step 1: v_0=20 m/s.
Step 2: d= v_0² / (2|a|)=100m.
Step 3: t=v_0/|a|=5s.
Solution: d=100m, t=5s.
10. Variation 4: v_0=50 m/s, a=-5 m/s². d? t?
Step 1: d= (50)²/(10)=250m.
Step 2: t=50/5=10s.
Solution: d=250m, t=10s.
11. NCERT 2.10 Variation: Man 3km at 6 km/h, back 9 km/h. Avg v 0-40min?
Step 1: To: 0.5h. 40min=2/3 h: Full to + (2/3-0.5)*9=1.5km back. Δx=3-1.5=1.5km.
Step 2: |v|=1.5/(2/3)=2.25 km/h.
Solution: |avg v|=2.25 km/h.
12. Relative: Train 20 m/s, man walks 2 m/s same dir. Relative v?
Step 1: v_rel = 20 - 2 = 18 m/s.
Solution: 18 m/s.
13. NCERT 2.14: Police 30 km/h, thief 192 km/h, bullet 150 m/s same. Hit speed?
Step 1: v_thief=53.33 m/s, v_pol=8.33 m/s.
Step 2: v_bullet/thief = 150 - (53.33 - 8.33) = 105 m/s.
Solution: 105 m/s.
14. Variation 5: Boat 5 m/s upstream, river 3 m/s. Relative to ground?
Step 1: Opp dir: v_ground = 5 - 3 = 2 m/s up.
Solution: 2 m/s upstream.
15. Ex 2.1 Variation: x=10 + 3t². v at t=1s? Avg v 0-2s?
Step 1: v=6t, t=1: 6 m/s.
Step 2: x(2)=22m, x(0)=10m, avg= (22-10)/2=6 m/s.
Solution: v=6 m/s, avg=6 m/s.
16. Uniform a=2 m/s², v_0=0, t=5s. x? v?
Step 1: v=0+2*5=10 m/s.
Step 2: x=0*5 + ½*2*25=25m.
Solution: x=25m, v=10 m/s.
17. Variation 6: a=-2 m/s², v_0=10 m/s, x when v=0?
Step 1: 0=10² +2(-2)x → x=25m.
Solution: x=25m.
18. Free fall from 20m, v_0=0. Time? v final? g=10.
Step 1: t=√(2*20/10)=2s.
Step 2: v=√(2*10*20)=20 m/s.
Solution: t=2s, v=20 m/s.
19. Throw down 15 m/s from 50m. Time to ground? g=10.
Step 1: 0=50 +15t -5t². 5t² -15t -50=0 → t= (15+√425)/10≈4.2s.
Solution: t=4.2s.
20. Relative: Rain 10 m/s vertical, man 5 m/s forward. Relative speed?
Step 1: Vector: v_rain/man = √(10² +5²)=11.18 m/s.
Solution: 11.2 m/s (angle tan^{-1}(5/10)).
21. Car accel 0-60 km/h in 10s. a? Distance?
Step 1: v_0=0, v=16.67 m/s, t=10s. a=1.667 m/s².
Step 2: x= 0 + ½*1.667*100=83.35m.
Solution: a=1.67 m/s², x=83.4m.
22. Variation 7: Decel from 40 m/s to 10 m/s in 5s. a? x?
Step 1: a=(10-40)/5=-6 m/s².
Step 2: Avg v=25 m/s, x=25*5=125m.
Solution: a=-6 m/s², x=125m.
23. Ball up 30 m/s, g=10. Time to top? h?
Step 1: t=30/10=3s.
Step 2: h=30*3 -5*9=45m.
Solution: t=3s, h=45m.
24. Drop from 80m, g=10. v at 50m? Time to 50m?
Step 1: v=√(2*10*50)=31.62 m/s.
Step 2: t=√(2*50/10)=3.16s.
Solution: v=31.6 m/s, t=3.2s.
25. Ex 2.6 Variation: Throw 15 m/s up. Total time? g=9.8.
Step 1: t=2*15/9.8≈3.06s.
Solution: t=3.1s.
26. Train 54 km/h, bird flies 18 km/h opposite. Relative?
Step 1: v_train=15 m/s, v_bird=-5 m/s. v_rel=15 - (-5)=20 m/s.
Solution: 20 m/s.
27. a=3 m/s², v_0=5 m/s, t=4s. v? x?
Step 1: v=5+3*4=17 m/s.
Step 2: x=5*4 + ½*3*16=50m.
Solution: v=17 m/s, x=50m.
28. Variation 8: Find t when x=100m, a= -4 m/s², v_0=20 m/s.
Step 1: 100=20t -2 t². 2t² -20t +100=0 → t² -10t +50=0, t=5±√(-25) complex? Wait, discriminant neg, doesn't stop in 100m.
Solution: Doesn't reach v=0 in 100m (max x= (20)^2/8=50m).
29. Free fall odd numbers: Distance in 3rd interval τ=1s, g=10.
Step 1: y_0=½*10*1=5m. 3rd dist=5*(2*3-1)=25m.
Solution: 25m.
30. Reaction: Drop 30cm. t? g=9.8.
Step 1: t=√(2*0.3/9.8)=0.248 s.
Solution: t=0.25 s.
