Motion is the most numerically-heavy chapter in the AMU 11 Physics syllabus. Every year 3-4 questions come from this chapter — mostly on equations of motion, graphs, and average speed. This note covers distance vs displacement, speed vs velocity, acceleration, retardation, scalars vs vectors, equations of motion with graphical derivation, projectile motion, relative velocity, and every type of graph you'll face. 15+ PYQs and practice exercises included. Practice the numericals — that's where marks live.
1) Rest, Motion, and Types of Motion
Rest and Motion are Relative
A body is said to be at rest if it does not change its position with respect to a reference point. A body is in motion if it changes its position with respect to a reference point. Important: Rest and motion are relative — a person sitting in a moving train is at rest with respect to the train but in motion with respect to the platform.
Types of Motion
| Type of Motion | Description | Example |
|---|---|---|
| Translatory motion | Body moves from one point to another in a straight or curved line | A car moving on a road, a bullet fired from a gun |
| Rotatory motion | Body rotates about a fixed axis | Earth rotating on its axis, a spinning top |
| Oscillatory / Vibratory motion | Body moves to and fro about a mean position | Pendulum of a clock, a vibrating tuning fork |
| Periodic motion | Motion that repeats after a fixed time interval | Motion of Earth around Sun, pendulum |
| Random motion | Motion with no fixed path or direction | Movement of dust particles in air, Brownian motion |
🔥 AMU PYQ: Which of the following is an example of periodic motion?
(A) A car moving on a straight road (B) Motion of a pendulum (C) A ball thrown upward (D) A falling apple
Ans: B (pendulum motion repeats after a fixed time interval)
(A) A car moving on a straight road (B) Motion of a pendulum (C) A ball thrown upward (D) A falling apple
Ans: B (pendulum motion repeats after a fixed time interval)
2) Scalars and Vectors (NEW)
Scalars
Quantities that have only magnitude (size) and no direction are called scalars. They obey ordinary algebra rules.
Examples: Mass, length, time, speed, distance, temperature, energy, work, power, density, volume, area, pressure, electric charge.
Vectors
Quantities that have both magnitude and direction are called vectors. They obey vector algebra (triangle law, parallelogram law).
Examples: Displacement, velocity, acceleration, force, momentum, weight, impulse, electric field, magnetic field.
| Feature | Scalar | Vector |
|---|---|---|
| Magnitude | Yes | Yes |
| Direction | No | Yes |
| Addition | Ordinary algebra | Triangle / Parallelogram law |
| Multiplication | Ordinary multiplication | Dot product / Cross product |
| Representation | Just a number with unit | Arrow (length = magnitude, arrowhead = direction) |
| Examples | Speed, distance, mass, time | Velocity, displacement, force, momentum |
Vector Addition — Triangle Law
If two vectors are represented by two sides of a triangle in magnitude and direction (taken in order), then the third side represents their resultant in the opposite order.
R = √(A² + B² + 2AB cos θ)
tan α = (B sin θ) / (A + B cos θ)
θ = angle between A and B, α = angle between R and A
Vector Addition — Parallelogram Law
If two vectors are represented by adjacent sides of a parallelogram, the diagonal through the common point gives the resultant.
Same formula: R = √(A² + B² + 2AB cos θ)Special cases:
θ = 0° → R = A + B (maximum)
θ = 90° → R = √(A² + B²)
θ = 180° → R = |A - B| (minimum)
🔥 AMU PYQ: Which of the following is a vector quantity?
(A) Speed (B) Distance (C) Velocity (D) Time
Ans: C (Velocity has both magnitude and direction)
🔥 AMU PYQ: Two forces of 3 N and 4 N act at a point at right angles. The magnitude of the resultant force is:
(A) 5 N (B) 7 N (C) 1 N (D) 12 N
Ans: A (R = √(3² + 4²) = √25 = 5 N)
(A) Speed (B) Distance (C) Velocity (D) Time
Ans: C (Velocity has both magnitude and direction)
🔥 AMU PYQ: Two forces of 3 N and 4 N act at a point at right angles. The magnitude of the resultant force is:
(A) 5 N (B) 7 N (C) 1 N (D) 12 N
Ans: A (R = √(3² + 4²) = √25 = 5 N)
3) Distance vs Displacement
Distance
Distance is the total path length travelled by an object. It is a scalar quantity (only magnitude). It is always positive and depends on the actual path taken.
Displacement
Displacement is the shortest straight-line distance between the initial and final positions. It is a vector quantity (magnitude + direction). It can be positive, negative, or zero.
| Feature | Distance | Displacement |
|---|---|---|
| Type | Scalar | Vector |
| Path dependent? | Yes | No (only initial & final) |
| Can be zero? | Only if no motion | Yes (if return to start) |
| Can be negative? | No (always ≥ 0) | Yes (depends on direction) |
| Symbol | s or d | s (with arrow) or Δx |
| SI unit | m (metre) | m (metre) |
Numerical Example:
A person walks 3 m east, then 4 m north.
Distance = 3 + 4 = 7 m
Displacement = √(3² + 4²) = 5 m (north-east direction)
A person walks 3 m east, then 4 m north.
Distance = 3 + 4 = 7 m
Displacement = √(3² + 4²) = 5 m (north-east direction)
🟡 Senior's Trick: Distance ≥ Displacement always! They're equal only when motion is in a straight line without changing direction.
🔥 AMU PYQ: The numerical ratio of displacement to distance for a moving object is:
(A) always less than 1 (B) always equal to 1 (C) always more than 1 (D) equal to or less than 1
Ans: (D) equal to or less than 1
Reason: Displacement ≤ Distance, so ratio ≤ 1.
(A) always less than 1 (B) always equal to 1 (C) always more than 1 (D) equal to or less than 1
Ans: (D) equal to or less than 1
Reason: Displacement ≤ Distance, so ratio ≤ 1.
4) Speed vs Velocity
Speed
Speed = distance travelled per unit time. It is a scalar quantity. It is always positive.
Speed = Distance / Time
SI unit: m/s | CGS unit: cm/s
Velocity
Velocity = displacement per unit time. It is a vector quantity. It can be positive, negative, or zero.
Velocity = Displacement / Time
SI unit: m/s
Types of Speed / Velocity
| Type | Definition | Formula |
|---|---|---|
| Average speed | Total distance / total time | vavg = Total distance / Total time |
| Average velocity | Total displacement / total time | vavg = Δx / Δt |
| Instantaneous speed | Speed at a particular instant | v = ds/dt |
| Instantaneous velocity | Velocity at a particular instant | v = dx/dt |
| Uniform speed | Equal distances in equal intervals | v = constant |
| Non-uniform speed | Speed changes with time | v ≠ constant |
🟡 Trick — Average speed formulas for AMU:
Case 1: Body covers half distance at v₁ and other half at v₂:
Case 2: Body moves at v₁ for half time and v₂ for other half:
Case 3: Body covers three equal distances at v₁, v₂, v₃:
Case 1: Body covers half distance at v₁ and other half at v₂:
Average speed = (2 v₁ v₂) / (v₁ + v₂)Case 2: Body moves at v₁ for half time and v₂ for other half:
Average speed = (v₁ + v₂) / 2Case 3: Body covers three equal distances at v₁, v₂, v₃:
Average speed = 3 / (1/v₁ + 1/v₂ + 1/v₃)
🔥 AMU PYQ: A car covers first half of distance at 40 km/h and second half at 60 km/h. Average speed?
(A) 40 km/h (B) 48 km/h (C) 50 km/h (D) 60 km/h
Ans: (B) 48 km/h
🔥 AMU PYQ: A car travels at 20 m/s for 10 s and then at 30 m/s for 5 s. Average speed is:
(A) 22 m/s (B) 23.3 m/s (C) 25 m/s (D) 26.7 m/s
Ans: B (Total distance = 20×10 + 30×5 = 200 + 150 = 350 m, Total time = 15 s, v_avg = 350/15 = 23.3 m/s)
(A) 40 km/h (B) 48 km/h (C) 50 km/h (D) 60 km/h
Ans: (B) 48 km/h
v_avg = (2 × 40 × 60) / (40 + 60) = 4800/100 = 48 km/h🔥 AMU PYQ: A car travels at 20 m/s for 10 s and then at 30 m/s for 5 s. Average speed is:
(A) 22 m/s (B) 23.3 m/s (C) 25 m/s (D) 26.7 m/s
Ans: B (Total distance = 20×10 + 30×5 = 200 + 150 = 350 m, Total time = 15 s, v_avg = 350/15 = 23.3 m/s)
5) Acceleration & Retardation
Acceleration
Acceleration = rate of change of velocity. It is a vector quantity. When velocity increases with time, we call it acceleration (positive).
Acceleration (a) = (v - u) / t
SI unit: m/s²
CGS unit: cm/s²
Retardation (Deceleration)
When velocity decreases with time, it's called retardation or deceleration. It is simply negative acceleration.
| Quantity | When velocity increases | When velocity decreases |
|---|---|---|
| Acceleration | Positive (a > 0) | Negative (a < 0) → Retardation |
| Example | Car speeding up | Car applying brakes |
Uniform vs Non-uniform Acceleration
Uniform acceleration: Velocity changes by equal amounts in equal time intervals (a = constant). Example: free fall under gravity (g = 9.8 m/s²).
Non-uniform acceleration: Velocity changes by unequal amounts in equal time intervals (a varies). Example: a car on a crowded road.
Numerical Example:
A car increases speed from 18 km/h to 36 km/h in 5 seconds. Find acceleration.
u = 18 × 5/18 = 5 m/s
v = 36 × 5/18 = 10 m/s
a = (10 - 5)/5 = 1 m/s²
A car increases speed from 18 km/h to 36 km/h in 5 seconds. Find acceleration.
u = 18 × 5/18 = 5 m/s
v = 36 × 5/18 = 10 m/s
a = (10 - 5)/5 = 1 m/s²
🔥 AMU PYQ: A vehicle moving at 10 m/s is brought to rest in 5 seconds. Its retardation is:
(A) 1 m/s² (B) 2 m/s² (C) 3 m/s² (D) 4 m/s²
Ans: B (a = (0-10)/5 = -2 m/s², retardation = 2 m/s²)
🔥 AMU PYQ: A body starts from rest with acceleration 2 m/s². Its velocity after 5 s is:
(A) 5 m/s (B) 10 m/s (C) 15 m/s (D) 20 m/s
Ans: B (v = u + at = 0 + 2×5 = 10 m/s)
(A) 1 m/s² (B) 2 m/s² (C) 3 m/s² (D) 4 m/s²
Ans: B (a = (0-10)/5 = -2 m/s², retardation = 2 m/s²)
🔥 AMU PYQ: A body starts from rest with acceleration 2 m/s². Its velocity after 5 s is:
(A) 5 m/s (B) 10 m/s (C) 15 m/s (D) 20 m/s
Ans: B (v = u + at = 0 + 2×5 = 10 m/s)
6) Relative Velocity (NEW)
Relative velocity of body A with respect to body B is the velocity of A as measured from the frame of reference of B.
vAB = vA - vB
vBA = vB - vA = -vAB
Cases for 1D Motion:
| Situation | Relative velocity |
|---|---|
| Both moving in same direction | vAB = vA - vB (if vA > vB, A sees B moving backward) |
| Both moving in opposite directions | vAB = vA - (-vB) = vA + vB |
| One body at rest | vAB = vA (B is stationary, so A sees B at relative velocity equal to its own speed) |
Numerical Example:
Two trains A and B are moving on parallel tracks. A moves at 60 km/h east, B moves at 40 km/h east. Find relative velocity of A w.r.t B.
vAB = 60 - 40 = 20 km/h east (A sees B moving backward at 20 km/h)
If B moves west (opposite direction): vAB = 60 - (-40) = 100 km/h
Two trains A and B are moving on parallel tracks. A moves at 60 km/h east, B moves at 40 km/h east. Find relative velocity of A w.r.t B.
vAB = 60 - 40 = 20 km/h east (A sees B moving backward at 20 km/h)
If B moves west (opposite direction): vAB = 60 - (-40) = 100 km/h
🔥 AMU PYQ: Two cars A and B move at 30 m/s and 20 m/s in the same direction. The relative velocity of A w.r.t B is:
(A) 50 m/s (B) 10 m/s (C) -10 m/s (D) 0 m/s
Ans: B (30 - 20 = 10 m/s)
(A) 50 m/s (B) 10 m/s (C) -10 m/s (D) 0 m/s
Ans: B (30 - 20 = 10 m/s)
7) Equations of Motion (For Constant Acceleration)
These 3 equations are the most important tools for solving AMU motion numericals. Memorize them like your name.
① v = u + at
② s = ut + ½ at²
③ v² = u² + 2as
| Symbol | Meaning | SI unit |
|---|---|---|
| u | Initial velocity | m/s |
| v | Final velocity | m/s |
| a | Constant acceleration | m/s² |
| t | Time taken | s |
| s | Displacement | m |
7.1 Derivation of v = u + at
Acceleration = Rate of change of velocity
a = (v - u) / t
Cross multiply: at = v - u
→ v = u + at
a = (v - u) / t
Cross multiply: at = v - u
→ v = u + at
7.2 Derivation of s = ut + ½ at²
Method 1: Graphical (area under v-t graph)
On a v-t graph with initial velocity u and constant acceleration a:
Area under v-t = displacement
Area = Area of rectangle (u × t) + Area of triangle (½ × t × at)
s = ut + ½ × t × at
→ s = ut + ½ at²
Method 2: Using average velocity
Average velocity = (u + v)/2 = (u + u + at)/2 = u + at/2
s = avg velocity × time = (u + at/2) × t
→ s = ut + ½ at²
On a v-t graph with initial velocity u and constant acceleration a:
Area under v-t = displacement
Area = Area of rectangle (u × t) + Area of triangle (½ × t × at)
s = ut + ½ × t × at
→ s = ut + ½ at²
Method 2: Using average velocity
Average velocity = (u + v)/2 = (u + u + at)/2 = u + at/2
s = avg velocity × time = (u + at/2) × t
→ s = ut + ½ at²
7.3 Derivation of v² = u² + 2as
From eq 1: t = (v - u)/a
Substitute in eq 2: s = u[(v-u)/a] + ½ a[(v-u)/a]²
s = (uv - u²)/a + ½(v² - 2uv + u²)/a
Multiply both sides by 2a: 2as = 2uv - 2u² + v² - 2uv + u²
2as = v² - u²
→ v² = u² + 2as
Substitute in eq 2: s = u[(v-u)/a] + ½ a[(v-u)/a]²
s = (uv - u²)/a + ½(v² - 2uv + u²)/a
Multiply both sides by 2a: 2as = 2uv - 2u² + v² - 2uv + u²
2as = v² - u²
→ v² = u² + 2as
7.4 Distance Covered in nth Second
sn = u + a(2n - 1)/2
Distance travelled in the nth second (not total distance!)
Numerical Example:
A body starts from rest with a = 2 m/s². Find distance in 4th second.
u = 0, a = 2 m/s², n = 4
s₄ = 0 + 2(2×4 - 1)/2 = 2(8-1)/2 = 2×7/2 = 7 m
A body starts from rest with a = 2 m/s². Find distance in 4th second.
u = 0, a = 2 m/s², n = 4
s₄ = 0 + 2(2×4 - 1)/2 = 2(8-1)/2 = 2×7/2 = 7 m
8) Motion Under Gravity (Free Fall)
When a body falls only under the influence of gravity (no air resistance), it's called free fall. The acceleration due to gravity, g = 9.8 m/s² ≈ 10 m/s² (for quick AMU calculations).
Equations for free fall (taking downward as positive):
v = u + g t
s = u t + ½ g t²
v² = u² + 2 g s
Sign Convention:
| Case | u | a (or g) | s |
|---|---|---|---|
| Body dropped from rest | u = 0 | +g (downward) | + (downward displacement) |
| Body thrown upward | u = + (upward) | -g (downward) | + (upward displacement) |
| Body thrown downward | u = - (downward) | +g (downward) | + (downward) |
Key formulas for vertical motion:
Time to reach highest point: t = u/g
Maximum height: H = u²/2g
Time of flight (total): T = 2u/g
Velocity at impact (when thrown upward): v = u (same speed, opposite direction)
Velocity after falling height h: v = √(2gh)
Numerical Example:
A ball is thrown upward with velocity 30 m/s. (g = 10 m/s²)
(a) Max height: H = u²/2g = 900/20 = 45 m
(b) Time to reach top: t = u/g = 30/10 = 3 s
(c) Total time of flight: T = 2u/g = 2×30/10 = 6 s
(d) Velocity after 2 s: v = u - gt = 30 - 20 = 10 m/s upward
A ball is thrown upward with velocity 30 m/s. (g = 10 m/s²)
(a) Max height: H = u²/2g = 900/20 = 45 m
(b) Time to reach top: t = u/g = 30/10 = 3 s
(c) Total time of flight: T = 2u/g = 2×30/10 = 6 s
(d) Velocity after 2 s: v = u - gt = 30 - 20 = 10 m/s upward
🔥 AMU PYQ: A body is thrown vertically upward with velocity 20 m/s. The maximum height reached is (g = 10 m/s²):
(A) 10 m (B) 20 m (C) 30 m (D) 40 m
Ans: B (H = u²/2g = 400/20 = 20 m)
🔥 AMU PYQ: A stone dropped from a height reaches the ground in 4 seconds. Height of the tower is (g = 10 m/s²):
(A) 40 m (B) 60 m (C) 80 m (D) 160 m
Ans: C (s = 0 + ½ × 10 × 4² = 5 × 16 = 80 m)
(A) 10 m (B) 20 m (C) 30 m (D) 40 m
Ans: B (H = u²/2g = 400/20 = 20 m)
🔥 AMU PYQ: A stone dropped from a height reaches the ground in 4 seconds. Height of the tower is (g = 10 m/s²):
(A) 40 m (B) 60 m (C) 80 m (D) 160 m
Ans: C (s = 0 + ½ × 10 × 4² = 5 × 16 = 80 m)
9) Graphical Representation of Motion
AMU loves graph-based questions. At least one question every year comes from interpreting s-t, v-t, or a-t graphs.
9.1 Distance-Time (s-t) Graph
| Nature of s-t graph | Interpretation |
|---|---|
| Straight line sloping upward | Uniform speed (constant velocity) |
| Straight line parallel to time axis | Body at rest (v = 0) |
| Curve (parabola) opening upward | Uniform acceleration |
| Curve (parabola) opening downward | Uniform retardation |
Key fact: Slope of s-t graph = velocity
9.2 Velocity-Time (v-t) Graph
| Nature of v-t graph | Interpretation |
|---|---|
| Straight line parallel to time axis | Uniform velocity (a = 0) |
| Straight line sloping upward | Uniform acceleration |
| Straight line sloping downward | Uniform retardation |
| Curve | Non-uniform acceleration |
Two key facts:
• Slope of v-t graph = acceleration
• Area under v-t graph = displacement
• Slope of v-t graph = acceleration
• Area under v-t graph = displacement
9.3 Acceleration-Time (a-t) Graph
| Nature of a-t graph | Interpretation |
|---|---|
| Straight line parallel to time axis | Uniform acceleration (a = constant) |
| Straight line sloping | Non-uniform acceleration |
Key fact: Area under a-t graph = change in velocity (Δv)
9.4 Quick Comparison Table
| Graph type | Slope gives | Area gives |
|---|---|---|
| s-t | Velocity (v) | — |
| v-t | Acceleration (a) | Displacement (s) |
| a-t | — | Change in velocity (Δv) |
🟡 Senior's Trick for graphs:
• s-t slope → velocity
• v-t slope → acceleration
• v-t area → displacement
• If v-t graph is a straight line, a = constant and s-t is a parabola
• Area of trapezium on v-t = ½(u+v)t = displacement!
• s-t slope → velocity
• v-t slope → acceleration
• v-t area → displacement
• If v-t graph is a straight line, a = constant and s-t is a parabola
• Area of trapezium on v-t = ½(u+v)t = displacement!
🔥 AMU PYQ: Area under velocity-time graph gives:
(A) Acceleration (B) Displacement (C) Speed (D) Force
Ans: (B) Displacement
🔥 AMU PYQ: Slope of displacement-time graph gives:
(A) Acceleration (B) Velocity (C) Speed (D) Force
Ans: (B) Velocity
(A) Acceleration (B) Displacement (C) Speed (D) Force
Ans: (B) Displacement
🔥 AMU PYQ: Slope of displacement-time graph gives:
(A) Acceleration (B) Velocity (C) Speed (D) Force
Ans: (B) Velocity
10) Uniform Circular Motion
When a body moves in a circular path with constant speed, its direction changes continuously, so its velocity changes even though speed is constant. This is uniform circular motion.
Angular Displacement (θ)
The angle swept by the radius vector. Measured in radians.
Angular Velocity (ω)
Rate of change of angular displacement.
ω = θ / t
SI unit: rad/s
Relation Between Linear and Angular Velocity
v = r ω
where r = radius of circular path
Time Period (T) and Frequency (f)
T = time for one complete revolution = 2πr / v = 2π / ω
f = number of revolutions per second = 1 / T
ω = 2π / T = 2π f
v = 2π r / T = 2π r f
Centripetal Acceleration (ac)
The acceleration directed towards the centre of the circle that keeps the body in circular motion.
ac = v² / r = r ω²
Direction: Towards the centre of the circle
🔥 AMU PYQ: In uniform circular motion, which of the following remains constant?
(A) Velocity (B) Speed (C) Acceleration (D) Displacement
Ans: (B) Speed (velocity changes due to changing direction)
🔥 AMU PYQ: The SI unit of angular velocity is:
(A) rad/s (B) m/s (C) m/s² (D) Hz
Ans: A
(A) Velocity (B) Speed (C) Acceleration (D) Displacement
Ans: (B) Speed (velocity changes due to changing direction)
🔥 AMU PYQ: The SI unit of angular velocity is:
(A) rad/s (B) m/s (C) m/s² (D) Hz
Ans: A
11) Projectile Motion (Basic Concept for Class 10 Level)
When a body is thrown at an angle to the horizontal, it moves along a parabolic path. This is called projectile motion. The motion can be analyzed as two independent motions: horizontal (uniform) and vertical (accelerated).
Key quantities (for projection at angle θ):
Horizontal component: ux = u cos θ (constant)
Vertical component: uy = u sin θ (changes due to g)
Time of flight: T = 2u sin θ / g
Maximum height: H = u² sin²θ / 2g
Horizontal range: R = u² sin 2θ / g
Trajectory equation: y = x tan θ - (g x²) / (2 u² cos²θ)
🟡 Senior's Tips for Projectile Motion:
• Maximum range is achieved at θ = 45° (Rmax = u²/g)
• Range is same for θ and (90° - θ) — complementary angles give same range
• At highest point: vy = 0, v = u cos θ (only horizontal component remains)
• Time of flight depends only on vertical component: T = 2uy/g
• Maximum range is achieved at θ = 45° (Rmax = u²/g)
• Range is same for θ and (90° - θ) — complementary angles give same range
• At highest point: vy = 0, v = u cos θ (only horizontal component remains)
• Time of flight depends only on vertical component: T = 2uy/g
🔥 AMU PYQ: The path of a projectile is:
(A) Circular (B) Parabolic (C) Elliptical (D) Straight line
Ans: B (Parabolic)
(A) Circular (B) Parabolic (C) Elliptical (D) Straight line
Ans: B (Parabolic)
12) Important Formulae at a Glance
| Situation | Formula |
|---|---|
| Speed | v = s / t |
| Velocity | v = Δx / t |
| Acceleration | a = (v - u) / t |
| 1st eq of motion | v = u + at |
| 2nd eq of motion | s = ut + ½ at² |
| 3rd eq of motion | v² = u² + 2as |
| Distance in nth second | sn = u + a/2 (2n - 1) |
| Angular velocity | ω = θ / t = 2π / T = 2πf |
| Linear-circular relation | v = r ω |
| Centripetal acceleration | ac = v² / r = r ω² |
| Relative velocity | vAB = vA - vB |
| Max height (projectile) | H = u² sin²θ / 2g |
| Range (projectile) | R = u² sin 2θ / g |
| Time of flight (projectile) | T = 2u sin θ / g |
| Free fall max height | H = u²/2g |
| Free fall time of flight | T = 2u/g |
| km/h → m/s | Multiply by 5/18 |
| m/s → km/h | Multiply by 18/5 |
| Resultant of two vectors | R = √(A² + B² + 2AB cos θ) |
13) Common Mistakes Students Make in Motion
⚠️ Mistake 1: Forgetting to convert km/h to m/s.
Fix: Always multiply km/h by 5/18 before using equations of motion.
⚠️ Mistake 2: Sign convention confusion in free fall.
Fix: Pick a direction as positive and stick to it. If upward is positive, g = -9.8 m/s².
⚠️ Mistake 3: Using distance instead of displacement in equations of motion.
Fix: s in equations means displacement, not distance!
⚠️ Mistake 4: Forgetting that slope of v-t = a and area = s.
Fix: Write this on your rough sheet: "slope = a, area = s"
⚠️ Mistake 5: Confusing average speed with average velocity.
Fix: Avg speed = total distance/total time. Avg velocity = net displacement/total time.
⚠️ Mistake 6: Thinking "starts from rest" means u = anything other than 0.
Fix: u = 0 when body starts from rest!
⚠️ Mistake 7: Getting the formula for distance in nth second wrong (using n instead of 2n-1).
Fix: sn = u + a(2n-1)/2 (not u + a n)
Fix: Always multiply km/h by 5/18 before using equations of motion.
⚠️ Mistake 2: Sign convention confusion in free fall.
Fix: Pick a direction as positive and stick to it. If upward is positive, g = -9.8 m/s².
⚠️ Mistake 3: Using distance instead of displacement in equations of motion.
Fix: s in equations means displacement, not distance!
⚠️ Mistake 4: Forgetting that slope of v-t = a and area = s.
Fix: Write this on your rough sheet: "slope = a, area = s"
⚠️ Mistake 5: Confusing average speed with average velocity.
Fix: Avg speed = total distance/total time. Avg velocity = net displacement/total time.
⚠️ Mistake 6: Thinking "starts from rest" means u = anything other than 0.
Fix: u = 0 when body starts from rest!
⚠️ Mistake 7: Getting the formula for distance in nth second wrong (using n instead of 2n-1).
Fix: sn = u + a(2n-1)/2 (not u + a n)
14) PYQ Battle — 17 AMU Past Questions on Motion
Q1. A body starts from rest and moves with uniform acceleration of 2 m/s². Its velocity after 5 seconds is:
(A) 5 m/s (B) 10 m/s (C) 15 m/s (D) 20 m/s
Ans: B (v = u + at = 0 + 2×5 = 10 m/s)
(A) 5 m/s (B) 10 m/s (C) 15 m/s (D) 20 m/s
Ans: B (v = u + at = 0 + 2×5 = 10 m/s)
Q2. A body starts from rest. If it travels with an acceleration of 2 m/s², its displacement at the end of 3 seconds is:
(A) 6 m (B) 9 m (C) 12 m (D) 18 m
Ans: B (s = 0 + ½ × 2 × 3² = 9 m)
(A) 6 m (B) 9 m (C) 12 m (D) 18 m
Ans: B (s = 0 + ½ × 2 × 3² = 9 m)
Q3. A body starts from rest and travels with uniform acceleration a to make a displacement of 6 m. If its velocity after that displacement is 6 m/s, then a is:
(A) 1 m/s² (B) 2 m/s² (C) 3 m/s² (D) 4 m/s²
Ans: C (v² = u² + 2as → 36 = 0 + 2×a×6 → a = 3 m/s²)
(A) 1 m/s² (B) 2 m/s² (C) 3 m/s² (D) 4 m/s²
Ans: C (v² = u² + 2as → 36 = 0 + 2×a×6 → a = 3 m/s²)
Q4. A body moving with a velocity of 10 m/s comes to rest in 5 seconds. Its retardation is:
(A) 1 m/s² (B) 2 m/s² (C) 3 m/s² (D) 4 m/s²
Ans: B (a = (0-10)/5 = -2 m/s² → retardation = 2 m/s²)
(A) 1 m/s² (B) 2 m/s² (C) 3 m/s² (D) 4 m/s²
Ans: B (a = (0-10)/5 = -2 m/s² → retardation = 2 m/s²)
Q5. The ratio of average velocity to average speed of a body is:
(A) always 1 (B) always < 1 (C) ≤ 1 (D) ≥ 1
Ans: C (≤ 1, equal only for straight line motion without reversal)
(A) always 1 (B) always < 1 (C) ≤ 1 (D) ≥ 1
Ans: C (≤ 1, equal only for straight line motion without reversal)
Q6. A body is thrown vertically upward with velocity 20 m/s. The maximum height reached is (g = 10 m/s²):
(A) 10 m (B) 20 m (C) 30 m (D) 40 m
Ans: B (H = u²/2g = 400/20 = 20 m)
(A) 10 m (B) 20 m (C) 30 m (D) 40 m
Ans: B (H = u²/2g = 400/20 = 20 m)
Q7. A car initially at rest accelerates uniformly to 20 m/s in 10 seconds. The distance covered is:
(A) 50 m (B) 100 m (C) 150 m (D) 200 m
Ans: B (s = ½ × (0+20) × 10 = 100 m)
(A) 50 m (B) 100 m (C) 150 m (D) 200 m
Ans: B (s = ½ × (0+20) × 10 = 100 m)
Q8. The SI unit of angular velocity is:
(A) rad/s (B) m/s (C) m/s² (D) Hz
Ans: A
(A) rad/s (B) m/s (C) m/s² (D) Hz
Ans: A
Q9. Which of the following is a vector quantity?
(A) Speed (B) Distance (C) Velocity (D) Time
Ans: C
(A) Speed (B) Distance (C) Velocity (D) Time
Ans: C
Q10. A car travels at 20 m/s for 10 s and then at 30 m/s for 5 s. Average speed is:
(A) 22 m/s (B) 23.3 m/s (C) 25 m/s (D) 26.7 m/s
Ans: B (Total distance = 20×10 + 30×5 = 350 m, Total time = 15 s, v_avg = 350/15 = 23.3 m/s)
(A) 22 m/s (B) 23.3 m/s (C) 25 m/s (D) 26.7 m/s
Ans: B (Total distance = 20×10 + 30×5 = 350 m, Total time = 15 s, v_avg = 350/15 = 23.3 m/s)
Q11. A car covers first half of distance at 40 km/h and second half at 60 km/h. Average speed:
(A) 40 (B) 48 (C) 50 (D) 60 km/h
Ans: B (2×40×60)/(40+60) = 4800/100 = 48 km/h
(A) 40 (B) 48 (C) 50 (D) 60 km/h
Ans: B (2×40×60)/(40+60) = 4800/100 = 48 km/h
Q12. The numerical value of displacement is always ______ distance.
(A) equal to (B) less than (C) less than or equal to (D) greater than
Ans: C
(A) equal to (B) less than (C) less than or equal to (D) greater than
Ans: C
Q13. A stone is dropped from a cliff. Its velocity after 3 seconds is (g = 10 m/s²):
(A) 10 m/s (B) 20 m/s (C) 30 m/s (D) 40 m/s
Ans: C (v = u + gt = 0 + 10×3 = 30 m/s)
(A) 10 m/s (B) 20 m/s (C) 30 m/s (D) 40 m/s
Ans: C (v = u + gt = 0 + 10×3 = 30 m/s)
Q14. The distance covered by a body in the nth second is given by:
(A) u + a(2n-1)/2 (B) u + an (C) u + a(n-1) (D) u + a(2n+1)/2
Ans: A
(A) u + a(2n-1)/2 (B) u + an (C) u + a(n-1) (D) u + a(2n+1)/2
Ans: A
Q15. If a body moves in a circular path with uniform speed, its acceleration is:
(A) zero (B) directed towards centre (C) directed outward (D) along tangent
Ans: B (centripetal acceleration is always towards the centre)
(A) zero (B) directed towards centre (C) directed outward (D) along tangent
Ans: B (centripetal acceleration is always towards the centre)
Q16. A car moving at 30 m/s applies brakes with retardation of 5 m/s². Stopping distance:
(A) 60 m (B) 90 m (C) 120 m (D) 180 m
Ans: B (v² = u² + 2as → 0 = 900 + 2(-5)s → s = 900/10 = 90 m)
(A) 60 m (B) 90 m (C) 120 m (D) 180 m
Ans: B (v² = u² + 2as → 0 = 900 + 2(-5)s → s = 900/10 = 90 m)
Q17. Two vectors of magnitude 3 and 4 act at right angles. Resultant magnitude:
(A) 5 (B) 7 (C) 1 (D) 12
Ans: A (R = √(3² + 4²) = 5)
(A) 5 (B) 7 (C) 1 (D) 12
Ans: A (R = √(3² + 4²) = 5)
15) Practice Exercise — 15 Problems
Solve these. Cover the answers and try yourself first.
1. A car accelerates from 5 m/s to 25 m/s in 10 seconds. Find acceleration.
Answer
a = (25-5)/10 = 2 m/s²
2. Find the distance covered by the above car in those 10 seconds.
Answer
s = ut + ½at² = 5×10 + ½×2×100 = 50 + 100 = 150 m
3. A ball is thrown upward with velocity 15 m/s. Find max height. (g = 10 m/s²)
Answer
H = u²/2g = 225/20 = 11.25 m
4. Find time of flight for the ball in Q3.
Answer
T = 2u/g = 30/10 = 3 s
5. A body starts from rest. Its velocity after 6 seconds is 18 m/s. Find acceleration and distance.
Answer
a = 18/6 = 3 m/s²; s = ½×3×36 = 54 m
6. Convert 90 km/h to m/s.
Answer
90 × 5/18 = 25 m/s
7. A particle moves in a circle of radius 0.5 m with speed 5 m/s. Find centripetal acceleration.
Answer
ac = v²/r = 25/0.5 = 50 m/s²
8. A train moving at 72 km/h applies brakes and stops after 200 m. Find retardation.
Answer
u = 72×5/18 = 20 m/s; v² = u² + 2as → 0 = 400 + 2×a×200 → a = -1 m/s² (retardation = 1 m/s²)
9. Find the distance covered in the 5th second by a body starting from rest with a = 4 m/s².
Answer
s5 = 0 + 4(2×5-1)/2 = 4×9/2 = 18 m
10. A stone dropped from a tower hits ground in 5 seconds. Find tower height. (g = 9.8 m/s²)
Answer
h = ½gt² = ½×9.8×25 = 122.5 m
11. Two cars A (30 m/s) and B (20 m/s) move towards each other. Find relative velocity.
Answer
vAB = 30 - (-20) = 50 m/s
12. Find angular velocity of a particle moving in a circle of radius 2 m with linear speed 4 m/s.
Answer
ω = v/r = 4/2 = 2 rad/s
13. A body is thrown upward with velocity 40 m/s. Find velocity after 3 seconds. (g = 10 m/s²)
Answer
v = 40 - 10×3 = 10 m/s (upward)
14. If displacement-time graph is a horizontal line, what is the body doing?
Answer
At rest (v = 0)
15. A projectile is fired at 30° with speed 20 m/s. Find time of flight. (g = 10 m/s²)
Answer
T = 2u sin θ / g = 2×20×0.5/10 = 2 s16) Quick Revision — One-liners
- Distance = scalar, Displacement = vector, Distance ≥ Displacement
- Speed = scalar, Velocity = vector
- Average speed = total distance / total time
- Half-distance avg speed = (2v₁v₂)/(v₁+v₂)
- Acceleration = (v - u)/t, SI unit: m/s²
- Retardation = negative acceleration
- 3 equations: v = u + at, s = ut + ½at², v² = u² + 2as
- Distance in nth second: sn = u + a(2n-1)/2
- For free fall: a = g = 9.8 m/s² (use 10 in AMU)
- Max height = u²/2g, Time of flight = 2u/g
- Slope of s-t = velocity, Slope of v-t = acceleration
- Area under v-t = displacement, Area under a-t = Δv
- Angular velocity ω = θ/t = 2π/T = 2πf
- Linear velocity in circular motion: v = rω
- Centripetal acceleration: ac = v²/r
- Relative velocity: vAB = vA - vB
- Vector resultant: R = √(A² + B² + 2AB cos θ)
- Projectile path = parabola; Max range at 45°
- km/h to m/s: multiply by 5/18
- Scalars: speed, distance, mass, time. Vectors: velocity, displacement, force, momentum
✅ Motion done. This chapter alone can get you 4-5 marks in AMU Physics if you practice the numericals. Next up: Force and Newton's Laws — where motion meets the cause behind it.