NCERT Plus One (Class 11) Physics • Chapter 2

Motion in a Straight Line

Q: Can a body have zero velocity and still have non-zero acceleration? Give one example.

Yes! When a stone is thrown vertically upward, at its highest point instantaneous velocity v = 0 , while acceleration remains a = g = 9.8 m/s² downward.

Q: What physical quantities do (a) the slope and (b) the area under a velocity–time (v–t) graph represent?

(a) Slope of v–t graph: Instantaneous acceleration ( a = dv/dt ). (b) Area under v–t graph: Displacement (signed area) / Distance (sum of magnitudes).

Q: Distinguish between distance (path length) and displacement with two key points.

1. Path length is a scalar (always positive for motion) equal to total route length; Displacement is a vector ( Δx = x₂ − x₁ ) that can be positive, negative, or zero. 2. Always Path Length ≥ |Displacement| (equal only for unidirectional straight motion).

Q: Draw the position–time (x–t) graphs for (i) a stationary object and (ii) a uniformly accelerated object starting from rest.

(i) Stationary object: Straight line parallel to the time axis ( x = constant , slope = 0). (ii) Uniform acceleration from rest: Upward-opening parabola ( x = ½at² ) passing through origin with increasing slope.

Q: Using a velocity–time (v–t) graph, derive the kinematic equations v = u + at and s = ut + ½at² for uniformly accelerated motion.

Consider velocity increasing uniformly from u at t = 0 to v at time t : 1. Slope = a: a = (v − u)/t ⇒ v = u + at . 2. Displacement s = Area under v–t trapezium: s = Area(rectangle) + Area(triangle) = u·t + ½·t·(v − u) . Substituting v − u = at gives s = ut + ½at² .

Q: Define stopping distance of a moving vehicle. Show that doubling speed makes stopping distance four times larger. A car moving at 54 km/h stops in 15 m; find its deceleration.

Stopping distance d_s is the distance travelled after brakes are applied until rest ( v = 0 ). From v² − u² = 2(−a)d_s : 0 − u² = −2ad_s ⇒ d_s = u²/(2a) ∝ u² . Thus doubling u quadruples d_s . Numerical: u = 54 × (5/18) = 15 m/s . a = u²/(2d_s) = 225/(2×15) = 7.5 m/s² .

Q: A ball is thrown vertically upwards with a velocity of 20 m/s from the top of a 25 m tall building. How high does it rise from the ground, and how long before it hits the ground? (Take g = 10 m/s²)

Take launch point as y₀ = 0 , upward + , u = +20 m/s , a = −10 m/s² . 1. Rise above roof: 0 − 20² = 2(−10)h ⇒ h = 20 m . Height from ground = 25 + 20 = 45 m . 2. Total time to ground ( y = −25 m ): −25 = 20t − 5t² ⇒ t² − 4t − 5 = 0 ⇒ (t − 5)(t + 1) = 0 ⇒ t = 5 s .

Q: Two parallel rail tracks run north–south. Train A moves north at 54 km/h and Train B moves south at 90 km/h. Find the velocity of B with respect to A, and the velocity of ground with respect to B.

Take North as + : v_A = +54 × 5/18 = +15 m/s , v_B = −90 × 5/18 = −25 m/s . 1. v_BA = v_B − v_A = −25 − 15 = −40 m/s ( 40 m/s South). 2. v_GB = 0 − v_B = +25 m/s ( 25 m/s North).

Position, displacement, x-t & v-t graphs, kinematic equations, free fall, stopping distance & 1D relative velocity.

📄 Official NCERT Textbook PDF

Topic 2.1: Position, Path Length & Displacement

Imagine cycling straight from your house gate (0 m) along the Kollam beach road to a bakery +60 m east, grabbing a puff, and turning back 20 m to meet your friend at +40 m. Your bicycle wheel rolled 80 m in total — that is path length. Yet if your friend looks only at where you started and ended, you are just +40 m east of the gate — that straight shortcut with direction is displacement!

PATH LENGTH (DISTANCE)Actual route covered. Scalar, always ≥ 0, never shrinks during motion.

DISPLACEMENT (Δx)Δx = x₂ − x₁ (final minus initial). Vector (+, −, or 0).

GOLDEN INEQUALITYPath Length ≥ |Displacement| (equal only when moving one way without turning!).

🧠 Memory Trick: Distance = Odometer (tracks every step, never resets); Displacement = Google Maps straight arrow from Start to Finish!

💡 Simple Kid-Check: To say something is moving, you first pick a zero-point called the Origin (plus a watch). Moving right of origin is +, moving left is −. If you run around a 400 m circular school track and stop back at the starting chalk line, your path length is 400 m, but your displacement is ZERO!

1
Frame of Reference: A coordinate axis (say x-axis) fixed to a reference object (like the ground or a bus stop) plus a clock to record position x(t) at each instant.
2
Path Length (Scalar): Total length of the actual path traversed. It depends on the route taken and always increases (or stays constant) with time.
3
Displacement (Vector in 1D): Change in position Δx = x₂ − x₁. In straight-line motion, its sign tells direction: + means right/forward, − means left/backward.

Core Relations: Δx = x_final − x_initial | Path Length ≥ |Δx|

🧒 Super-Easy Example: A KSRTC bus moves from O (0 m) to P (+120 m) and reverses to Q (+30 m).
• Path length = 120 + (120 − 30) = 210 m.
• Displacement = x_Q − x_O = +30 − 0 = +30 m.

Watch out when an object turns around! Magnitude of displacement is NOT always equal to path length — they match only if the object moves in one constant direction without reversing. Also, displacement can be negative (for instance, x₁ = +50 m to x₂ = +10 m gives Δx = −40 m)!

Topic 2.2: Instantaneous Velocity & Speed (x–t Graphs)

On an Ernakulam–Thrissur highway trip, your car covers 60 km in 1 hour, so your average speed is 60 km/h. Does that mean your speedometer needle stayed glued at 60? Of course not — you slowed to 20 km/h at traffic signals and zoomed to 80 km/h on open roads! That live needle reading at one exact second is instantaneous velocity.

AVG VELOCITY vs SPEEDv_avg = Δx / Δt (can be 0 or −) | Avg Speed = Path / Δt ≥ |v_avg|

x–t GRAPH SECRETSlope of tangent on x–t curve = Instantaneous Velocity v = dx/dt!

TWO-HALF SHORTCUTEqual distances at v₁, v₂ → v_avg = 2v₁v₂/(v₁+v₂); Equal times → (v₁+v₂)/2

🧠 Memory Trick: x–t Slope = Velocity! Flat line = Sleeping (rest); Steep upward line = Fast forward; Downward slant = Reversing!

💡 Simple Kid-Check: Average velocity looks at your whole journey using a straight chord between start and end on the x–t graph. Shrink the time gap Δt to an eyeblink (Δt → 0), and that chord snaps right onto the curve as a tangent line — telling you your exact speedometer reading!

1
Average Velocity & Average Speed: v_avg = (x₂ − x₁)/(t₂ − t₁) depends only on endpoints, whereas Average Speed = Total path length / Total time.
2
Instantaneous Velocity: v = lim(Δt→0) Δx/Δt = dx/dt. Geometrically, it is the slope of the tangent drawn to the position-time (x–t) graph at instant t.
3
Instantaneous Speed: Simply the magnitude |v|. Even when average speed differs from |v_avg|, at any single instant, instantaneous speed ALWAYS equals |instantaneous velocity|!

v = dxdt = slope of x–t graph | Equal-distance avg speed: v_avg = 2 v₁ v₂v₁ + v₂

🧒 Super-Easy Example: A scooter rides to a shop at 20 km/h and returns the exact same distance at 30 km/h.
Average speed = 2(20)(30) / (20 + 30) = 1200 / 50 = 24 km/h (NOT 25 km/h!).
What is average velocity for the round trip? 0 km/h because net displacement is zero!

Classic Kerala HSE trap: Never take (v₁ + v₂)/2 when a vehicle covers equal distances (like going and returning)! Use the harmonic mean 2v₁v₂/(v₁+v₂). Also remember: an x–t graph can NEVER be vertical (that would require infinite speed!) or bend backward in time.

Topic 2.3: Acceleration & v–t Graph Slope / Area

When a Kochi Metro train pulls out from Aluva station, your back presses gently into the seat as speed climbs from 0 to 72 km/h in 10 s; when it reaches the next station, your body leans forward as brakes smoothly wipe out that speed. Changing velocity every second — whether speeding up OR slowing down — is acceleration!

ACCELERATION FORMULAa = dvdt = d²xdt² | SI unit: m/s²

v–t GRAPH SLOPESlope of v–t graph = Instantaneous Acceleration a!

v–t GRAPH AREASigned area = Displacement | Sum of |Area| = Total Distance!

🧠 Memory Trick: Think x → v → a! Moving right (x → v → a) = take SLOPE; moving left (a → v → x) = take AREA!

💡 Simple Kid-Check: Look at a v–t graph like a treasure map with two secrets built in: how steep the hill climbs tells you the acceleration, and how much coloured paint fits under the curve tells you how far you travelled!

1
Average & Instantaneous Acceleration: a_avg = (v₂ − v₁)/(t₂ − t₁) and a = dv/dt. Positive, negative, or zero depending on how velocity changes.
2
Reading a v–t Graph: Horizontal line = constant velocity (a = 0); upward straight slant = constant positive a; downward straight slant = constant negative a.
3
Area Above vs Below Axis: Area above the t-axis (v > 0) is forward displacement +A₁; area below (v < 0) is backward displacement −A₂. Net displacement = A₁ − A₂, whereas total distance = A₁ + A₂!

Speeding Up vs Slowing Down: Same sign (v & a both + or both −) ⇒ Speeding UP! | Opposite signs ⇒ Slowing DOWN (Retardation)!

🧒 Quick Graph Read: A bike moves at +20 m/s for 4 s (Area = +80 m), then reverses at −10 m/s for 2 s (Area = −20 m).
• Net Displacement = 80 − 20 = +60 m.
• Total Distance = 80 + 20 = 100 m!

Two giant 1-mark traps! First: Negative acceleration does NOT always mean slowing down! If a particle moves left (v = −10 m/s) with a = −2 m/s², after 1 second v = −12 m/s — it is actually speeding up! Second: At the peak of a vertical throw, v = 0 while a = −g ≠ 0!

Topic 2.4: Kinematic Equations for Uniform Acceleration

Imagine a drag-racing car at the Cochin airfield accelerating steadily from rest at 4 m/s². Do you need a stopwatch and measuring tape at every metre to predict its speed after 5 seconds? Nope — Galileo and Newton condensed all constant-acceleration motion into three golden kinematic equations!

1. VELOCITY–TIMEv = u + at (no s!)

2. POSITION–TIMEs = ut + ½at² (no v!)

3. VELOCITY–POSITIONv² − u² = 2as (no t!)

🧠 Memory Trick: Pick the equation missing the variable NOT in your question! Bonus formula for n-th second: sₙ = u + a2(2n − 1)!

💡 Simple Kid-Check: Think of the v–t graph of uniform acceleration as a simple right trapezoid starting at height u and sloping up to v over width t. Its slope immediately gives v = u + at, and its trapezium area ½(u + v)t gives BOTH s = ut + ½at² and v² − u² = 2as!

v = u + at | s = ut + ½at² | v² = u² + 2as | s = u + v2 t | sₙ = u + a(n − ½)

1
Convert Units First! Exams love km/h: multiply by 518 to convert km/h → m/s (e.g. 36 km/h = 10 m/s, 72 km/h = 20 m/s).
2
Pick Signs Carefully: Assign one direction as + (usually initial motion or upward) and plug u, a, s with proper + or − signs!
3
Average Velocity Shortcut: ONLY when acceleration a is constant, v_avg = u + v2!

🧒 Super-Easy Example: A train starts from rest (u = 0) and reaches 72 km/h (20 m/s) in 50 m.
No time given → pick equation #3: 20² − 0² = 2 a (50) ⇒ 400 = 100a ⇒ a = 4 m/s²!

GOLDEN LIMITATION: These three SUVAT equations work ONLY for constant (uniform) acceleration! If a problem gives a = 3t or v = 4t², acceleration is changing — you MUST use calculus (dv/dt and integration), NEVER s = ut + ½at²!

Topic 2.4: Free Fall & Stopping Distance

Drop a coconut from a tall Thrissur palm tree (u = 0), and Earth pulls it down faster every single second by 9.8 m/s! Meanwhile, on a rainy highway, a car moving twice as fast doesn't need double the braking distance — it needs FOUR TIMES the road to stop safely!

FREE FALL (FROM REST)a = g ≈ 9.8 m/s² downward | h = ½gt², v = √(2gh)

GALILEO ODD LAWDistances in equal 1 s drops from rest follow 1 : 3 : 5 : 7 : ...!

STOPPING DISTANCEd_s = u²2a ⇒ d_s ∝ u²! (2× speed → 4× distance!)

🧠 Memory Trick: Free fall up & down is a mirror! Throw up at u: t_up = u/g, T_total = 2u/g, H_max = u²/(2g), and it hits your hand back at speed u!

💡 Simple Kid-Check: Ignore air drag, and Earth pulls a marble and a bowling ball with the exact same acceleration g = 9.8 m/s²! Because distance grows with t² (1², 2², 3²... = 1, 4, 9...), subtracting previous heights gives Galileo's magical odd sequence 1, 3, 5, 7...!

1
Dropping from Height H: Take downward as +, u = 0, a = +g ⇒ t = √(2H/g) and impact speed v = √(2gH) — completely independent of object mass!
2
Vertical Upward Throw: Take upward +, so a = −g throughout! At highest point, v = 0, giving H_max = u²/(2g) and time to top t_up = u/g. Total flight time = 2u/g.
3
Stopping Distance of Vehicles: Using 0² − u² = 2(−a) d_s gives d_s = u²/(2a). (Add reaction-time distance u · t_react if reaction time is specified!)

🧒 Super-Easy Example: A car moving at 10 m/s stops in 5 m under maximum braking. What is its stopping distance at 30 m/s (3× speed)?
Since d_s ∝ u², new stopping distance = 3² × 5 m = 9 × 5 = 45 m!

At the very top of a vertical throw, students think "everything is zero!" — FALSE! Velocity v = 0 for an instant, while acceleration is STILL a = −9.8 m/s² downward! (If acceleration became zero too, the ball would float in mid-air forever!)

Topic 2.5: Relative Velocity in One Dimension

Sitting inside a KSRTC bus cruising at 50 km/h, if a car overtakes you at 65 km/h, it seems to creep past gently at just 15 km/h. Yet if that same car comes from the opposite lane, it flashes past your window at a terrifying 115 km/h! Motion depends completely on who is watching.

MASTER DEFINITIONv_AB = v_A − v_B (Velocity of A as seen from B!)

SAME vs OPPOSITESame way → Subtract speeds | Opposite ways → Add speeds!

x–t GRAPH CLUESv_A = v_B → Parallel lines | v_A ≠ v_B → Intersecting lines (meeting!)

🧠 Memory Trick: Read v_AB backwards as "A minus B" — always keep signs (+ right, − left) BEFORE subtracting!

💡 Simple Kid-Check: To see the world through Person B's eyes, mathematically freeze Person B by subtracting v_B from EVERY object! That is why velocity of A relative to B is simply v_AB = v_A − v_B, and relative displacement is x_A(t) − x_B(t)!

1
Signed Equation: Write v_AB = v_A − v_B and v_BA = v_B − v_A = −v_AB. Never drop signs!
2
Train Crossing Formula: Two trains of lengths L₁ and L₂ must clear a combined relative distance L₁ + L₂, so crossing time t = L₁ + L₂|v_rel|!
3
x–t Graph Interpretation: Equal velocities give parallel lines (never catch up!). Unequal slopes cross at the exact meeting time and position!

v_AB = v_A − v_B | x_B(t) − x_A(t) = (x_B0 − x_A0) + (v_B − v_A)t

🧒 Super-Easy Example: Train A (120 m long) goes East at +20 m/s. Train B (130 m long) approaches West at −30 m/s.
• Relative speed: v_AB = (+20) − (−30) = +50 m/s.
• Time to completely cross each other = (120 + 130) / 50 = 250 / 50 = 5 s!

Classic 2-mark mistake: When two trains move in opposite directions, students forget the minus sign on the second velocity and accidentally subtract (20 − 30) instead of subtracting a negative (20 − (−30) = 50 m/s)! Also remember crossing distance is sum of both train lengths L₁ + L₂.

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 1 Marks): Can a body have zero velocity and still have non-zero acceleration? Give one example.

Yes! When a stone is thrown vertically upward, at its highest point instantaneous velocity v = 0, while acceleration remains a = g = 9.8 m/s² downward.

Q2 (Kerala HSE Model • 1 Marks): What physical quantities do (a) the slope and (b) the area under a velocity–time (v–t) graph represent?

(a) Slope of v–t graph: Instantaneous acceleration (a = dv/dt).
(b) Area under v–t graph: Displacement (signed area) / Distance (sum of magnitudes).

Q3 (Kerala HSE March • 2 Marks): Distinguish between distance (path length) and displacement with two key points.

1. Path length is a scalar (always positive for motion) equal to total route length; Displacement is a vector (Δx = x₂ − x₁) that can be positive, negative, or zero.
2. Always Path Length ≥ |Displacement| (equal only for unidirectional straight motion).

Q4 (Kerala HSE Say • 2 Marks): Draw the position–time (x–t) graphs for (i) a stationary object and (ii) a uniformly accelerated object starting from rest.

(i) Stationary object: Straight line parallel to the time axis (x = constant, slope = 0).
(ii) Uniform acceleration from rest: Upward-opening parabola (x = ½at²) passing through origin with increasing slope.

Q5 (Kerala HSE March • 3 Marks): Using a velocity–time (v–t) graph, derive the kinematic equations v = u + at and s = ut + ½at² for uniformly accelerated motion.

Consider velocity increasing uniformly from u at t = 0 to v at time t:
1. Slope = a: a = (v − u)/t ⇒ v = u + at.
2. Displacement s = Area under v–t trapezium:
s = Area(rectangle) + Area(triangle) = u·t + ½·t·(v − u).
Substituting v − u = at gives s = ut + ½at².

Q6 (NCERT / DHSE Model • 3 Marks): Define stopping distance of a moving vehicle. Show that doubling speed makes stopping distance four times larger. A car moving at 54 km/h stops in 15 m; find its deceleration.

Stopping distance d_s is the distance travelled after brakes are applied until rest (v = 0).
From v² − u² = 2(−a)d_s: 0 − u² = −2ad_s ⇒ d_s = u²/(2a) ∝ u². Thus doubling u quadruples d_s.
Numerical: u = 54 × (5/18) = 15 m/s. a = u²/(2d_s) = 225/(2×15) = 7.5 m/s².

Q7 (Kerala HSE March • 3 Marks): A ball is thrown vertically upwards with a velocity of 20 m/s from the top of a 25 m tall building. How high does it rise from the ground, and how long before it hits the ground? (Take g = 10 m/s²)

Take launch point as y₀ = 0, upward +, u = +20 m/s, a = −10 m/s².
1. Rise above roof: 0 − 20² = 2(−10)h ⇒ h = 20 m. Height from ground = 25 + 20 = 45 m.
2. Total time to ground (y = −25 m): −25 = 20t − 5t² ⇒ t² − 4t − 5 = 0 ⇒ (t − 5)(t + 1) = 0 ⇒ t = 5 s.

Q8 (Kerala HSE Say • 2 Marks): Two parallel rail tracks run north–south. Train A moves north at 54 km/h and Train B moves south at 90 km/h. Find the velocity of B with respect to A, and the velocity of ground with respect to B.

Take North as +: v_A = +54 × 5/18 = +15 m/s, v_B = −90 × 5/18 = −25 m/s.
1. v_BA = v_B − v_A = −25 − 15 = −40 m/s (40 m/s South).
2. v_GB = 0 − v_B = +25 m/s (25 m/s North).