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State Board · Class 9 · Science and Technology

Lesson 1: Laws of Motion

5 concepts · 40 practice questions with answers · Free to read

This lesson is also in: मराठी

Distance, displacement, speed and velocity

From daily life

You walk to school by a winding lane that is 900 m long, but the school gate is only 600 m from your house in a straight line. Your path and your straight-line gap are two different things.

Simple definition

Distance (अंतर) is the length of the actual path travelled. Displacement (विस्थापन) is the shortest straight-line distance from the start to the finish.

Walking 3 m east then 4 m north gives a distance of 7 m but a displacement of only 5 mACB3 m east4 m northdisplacement = 5 mDistance = 3 + 4 = 7 mDisplacement = 5 m

In the figure : In the figure distance follows the path walked, while displacement is the straight line from start to finish.

Rule 1: Distance and displacement — Distance has only size (a scalar, अदिश). Displacement has size and direction (a vector, सदिश). Displacement can be zero even when the distance is not.

  • Easy : Walk 3 km east and then 4 km north: distance = 3 + 4 = 7 km, displacement = 5 km (3, 4, 5 right triangle).
  • Medium : A runner completes one full round of a 400 m track: distance = 400 m, but she is back at the start, so displacement = 0.
  • Exam-type : A cyclist goes 6 km north and returns 2 km south. Distance = 6 + 2 = 8 km; displacement = 6 − 2 = 4 km north.

Rule 2: Speed and velocity — Speed (चाल) = total distance ÷ total time (scalar). Velocity (वेग) = displacement ÷ time (vector). The SI unit of both is m/s. Velocity changes when speed changes, direction changes, or both.

  • Easy : A bus covers 120 m in 20 s: speed = 120 ÷ 20 = 6 m/s.
  • Medium : Convert: 36 km/h = 36 × 1000 ÷ 3600 = 10 m/s.
  • Exam-type : The 400 m round took 80 s: average speed = 400 ÷ 80 = 5 m/s, but average velocity = 0 ÷ 80 = 0, as displacement is zero.
Live figure

Distance, displacement, speed and velocity

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Remember

Distance and speed are scalars; displacement and velocity are vectors. Speed = distance ÷ time.

Common mistake

Thinking displacement is always equal to distance. It is equal only for straight-line motion in one direction.

Shortcut

Memory clue: "Distance = Door-to-door path; Displacement = Direct line".

Spot the difference

PointDistanceDisplacement
MeaningPath lengthStraight line start to end
TypeScalarVector
Can it be zero?No, if movedYes, if back at start

Practice questions

Easy
  1. 1What is distance?
  2. 2What is displacement?
  3. 3Which of speed and velocity is a vector?
Medium
  1. 4What is the SI unit of speed?
  2. 5A car covers 150 m in 15 s. Find its speed.
  3. 6Convert 54 km/h into m/s.
Challenging
  1. 7A boy walks 8 m east and then 6 m north. Find the distance and the displacement.
  2. 8A runner finishes a 200 m circular track in 40 s. Find the average speed and average velocity.
Show answers (try them yourself first)
  1. 1The length of the actual path travelled.
  2. 2The shortest straight-line distance from start to finish with direction.
  3. 3Velocity
  4. 4metre per second (m/s)
  5. 5150 ÷ 15 = 10 m/s
  6. 654 × 1000 ÷ 3600 = 15 m/s
  7. 7Distance = 14 m; displacement = √(8² + 6²) = 10 m.
  8. 8Speed = 200 ÷ 40 = 5 m/s; velocity = 0 because displacement is zero.

Acceleration and motion graphs

From daily life

When a cricket ball is rolling and slows down on the grass, or a train leaves a station and picks up speed, the speed is changing. We need a way to say how quickly it changes.

Simple definition

Acceleration (त्वरण) is the rate of change of velocity: a = (v − u) ÷ t.

A straight distance time line of slope 5 metres per second246810203040xyDistance–time graph (uniform motion)slope = speed= (30 − 10)/(6 − 2) = 5 m/s

In the figure : In the figure the slope of a distance–time line is the speed.

Rule 1: Acceleration — a = (v − u) ÷ t, where u is the initial velocity, v the final velocity and t the time. It is positive when velocity increases, negative (deceleration, मंदन) when it decreases, and zero when constant. The unit is m/s². Uniform motion means equal distances in equal times.

  • Easy : A scooter goes from rest to 20 m/s in 5 s: a = (20 − 0) ÷ 5 = 4 m/s².
  • Medium : A car slows from 30 m/s to 10 m/s in 4 s: a = (10 − 30) ÷ 4 = −5 m/s², a deceleration.
  • Exam-type : A distance–time graph shows 60 m at 10 s along a straight line from the origin: speed = slope = 60 ÷ 10 = 6 m/s.
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Acceleration and motion graphs

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A velocity time graph rising to 10 metres per second in 5 seconds then flat; the areas give 25 and 50 metres24681024681012xyVelocity–time graph: area = distance25 m50 ma = 2 m/s²uniform speed

In the figure : In the figure the area under a velocity–time graph is the distance travelled.

Rule 2: Graphs of motion — In a distance–time graph a straight line means uniform motion and its slope is the speed. In a velocity–time graph a straight line means uniform acceleration, and the area under the graph is the distance travelled.

  • Easy : A flat v–t line at 5 m/s for 8 s: area = 5 × 8 = 40 m.
  • Medium : A v–t line rising from 0 to 8 m/s in 4 s: area of the triangle = ½ × 4 × 8 = 16 m.
  • Exam-type : The speed rises from 0 to 8 m/s in 4 s and then stays at 8 m/s for 6 s: distance = 16 + 8 × 6 = 16 + 48 = 64 m.
Live figure

Acceleration and motion graphs

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Remember

a = (v − u) ÷ t. Slope of d–t graph = speed. Area under v–t graph = distance.

Common mistake

Treating a negative acceleration as an error. It only means the velocity is falling.

Shortcut

Memory clue: "Slope of d–t = Speed; Area of v–t = Distance" (S-S, A-D).

Spot the difference

GraphStraight line meansSlope givesArea gives
Distance–timeUniform speedSpeedNothing useful
Velocity–timeUniform accelerationAccelerationDistance

Practice questions

Easy
  1. 1What is acceleration?
  2. 2Write the formula and unit of acceleration.
  3. 3What does a negative acceleration mean?
Medium
  1. 4What does the slope of a distance–time graph give?
  2. 5A bike goes from 0 to 15 m/s in 5 s. Find a.
  3. 6A train slows from 25 m/s to 5 m/s in 10 s. Find a.
Challenging
  1. 7A v–t graph is a flat line at 6 m/s for 10 s. Find the distance.
  2. 8A car speeds up from 0 to 10 m/s in 5 s and then moves at 10 m/s for 5 s. Find the total distance.
Show answers (try them yourself first)
  1. 1The rate of change of velocity.
  2. 2a = (v − u) ÷ t; m/s².
  3. 3The velocity is decreasing (deceleration).
  4. 4The speed.
  5. 5(15 − 0) ÷ 5 = 3 m/s²
  6. 6(5 − 25) ÷ 10 = −2 m/s²
  7. 76 × 10 = 60 m
  8. 8½ × 5 × 10 + 10 × 5 = 25 + 50 = 75 m

Equations of motion

From daily life

A cyclist rolls down a slope without pedalling. Every second she goes faster by the same amount. If we know how fast she started, how fast she speeds up and for how long, we can tell her final speed and how far she rolled, without measuring on the road.

Simple definition

The equations of motion (गतीची समीकरणे) connect u, v, a, t and s for uniform acceleration.

A velocity time trapezoid split into a rectangle ut and a triangle half a t squared, total 32 metres123452468101214xyWhy s = ut + ½at²ut = 4 × 4 = 16½at² = 16u = 4v = 12a = 2s = 32 m

In the figure : In the figure the equations of motion come from the area of the v–t graph.

Rule 1: The three equations — v = u + at; s = ut + ½at²; v² = u² + 2as. Here u is the initial velocity, v the final velocity, a the acceleration, t the time and s the displacement. They come from the v–t graph.

  • Easy : A cyclist with u = 5 m/s and a = 2 m/s² for t = 6 s: v = 5 + 2 × 6 = 17 m/s.
  • Medium : From rest with a = 2 m/s² for 5 s: s = 0 + ½ × 2 × 25 = 25 m.
  • Exam-type : A ball starts from rest and rolls 8 m with a = 4 m/s²: v² = 0 + 2 × 4 × 8 = 64, so v = 8 m/s.

Rule 2: Solving problems — List the given values (u, v, a, t, s), pick the equation with the missing one, and use a minus sign for deceleration.

  • Easy : Stopping: v = 0. A bike at 12 m/s with a = −3 m/s²: t = (0 − 12) ÷ (−3) = 4 s.
  • Medium : The stopping distance of that bike: s = (0 − 144) ÷ (2 × −3) = 24 m.
  • Exam-type : A train at 20 m/s brakes with a = −2.5 m/s². Stops in t = 20 ÷ 2.5 = 8 s and covers s = 400 ÷ 5 = 80 m; check: 20 × 8 − ½ × 2.5 × 64 = 160 − 80 = 80 m.
Live figure

Equations of motion

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Remember

v = u + at; s = ut + ½at²; v² = u² + 2as. Pick the one with the unknown.

Common mistake

Forgetting that u = 0 when starting from rest, and that v = 0 when stopping.

Shortcut

Memory clue: "Each equation hides one letter": no s in the first, no v in the second, no t in the third.

Spot the difference

EquationMissing quantityUse when
v = u + atsDistance not asked
s = ut + ½at²vFinal speed not given
v² = u² + 2astTime not given

Practice questions

Easy
  1. 1Write the three equations of motion.
  2. 2What is u for a body that starts from rest?
  3. 3What is v for a body that comes to a stop?
Medium
  1. 4From which graph are the equations obtained?
  2. 5A car with u = 4 m/s, a = 3 m/s² runs for 5 s. Find v.
  3. 6A body starts from rest with a = 4 m/s². Find s in 6 s.
Challenging
  1. 7A scooter at 10 m/s decelerates at 2 m/s². Find the time to stop and the stopping distance.
  2. 8A ball dropped from rest reaches 20 m/s after 2 s. Find its acceleration and the distance fallen.
Show answers (try them yourself first)
  1. 1v = u + at; s = ut + ½at²; v² = u² + 2as.
  2. 20
  3. 30
  4. 4The velocity–time graph.
  5. 54 + 3 × 5 = 19 m/s
  6. 6½ × 4 × 36 = 72 m
  7. 7t = 10 ÷ 2 = 5 s; s = 100 ÷ 4 = 25 m
  8. 8a = 20 ÷ 2 = 10 m/s²; s = ½ × 10 × 4 = 20 m

Circular motion and Newton's first law

From daily life

A stone tied to a string and whirled in a circle keeps changing its direction. When the string snaps, the stone does not go on circling; it shoots off straight. Also, when a bus starts suddenly, standing passengers lurch backwards. Both are about how objects like to keep their motion.

Simple definition

Inertia (जडत्व) is the tendency of an object to keep its state of rest or of uniform straight-line motion. A tangent (स्पर्शिका) is the straight line touching a circle at one point.

A ball moving on a circle; if released it flies off along the tangentreleased:goes along the tangentspeed = 2πr / tconstant speed,direction keeps changing

In the figure : In the figure the speed in circular motion is constant but the direction changes, and a released body goes along the tangent.

Rule 1: Uniform circular motion — In uniform circular motion the speed is constant but the direction keeps changing, so the velocity changes and the body is accelerated. Speed = 2πr ÷ t. A body released from the circle moves along the tangent.

  • Easy : A stone on a string released at the top of its circle flies off along the tangent.
  • Medium : With π = 22/7, a toy car goes round a circle of radius 7 m in 22 s: speed = 2 × 22/7 × 7 ÷ 22 = 44 ÷ 22 = 2 m/s.
  • Exam-type : A bangle of radius 3.5 cm: its circumference = 2 × 22/7 × 3.5 = 22 cm, so an ant walking round it in 11 s has speed 2 cm/s.

Rule 2: Newton's first law — An object stays at rest or in uniform straight-line motion unless an external unbalanced force acts on it. This is the law of inertia. Greater mass means greater inertia.

  • Easy : A passenger lurches forward when a bus brakes, as the body wants to keep moving.
  • Medium : A coin on a card over a glass drops into the glass when the card is flicked away: the coin tends to stay at rest.
  • Exam-type : An empty cart and a loaded cart: the loaded one is harder to start and harder to stop, as it has more mass and so more inertia.
Remember

Circular motion: constant speed, changing direction, so accelerated. First law: inertia.

Common mistake

Saying a body in uniform circular motion has no acceleration. Its direction changes, so it does.

Shortcut

Memory clue: "Released = Tangent" and "Lazy objects = Inertia".

Spot the difference

PointUniform linearUniform circular
SpeedConstantConstant
DirectionConstantChanges
AccelerationZeroNot zero

Practice questions

Easy
  1. 1State Newton's first law.
  2. 2What is inertia?
  3. 3Along which line does a released body leave a circle?
Medium
  1. 4Which has more inertia, a cricket ball or a tennis ball?
  2. 5Write the formula for speed in circular motion.
  3. 6Find the speed of a toy on a circle of radius 14 m that takes 44 s per round (π = 22/7).
Challenging
  1. 7Why does a rider on a turning bike lean outward in the moment of the turn? Link it to inertia.
  2. 8A car moves round a circular track at a constant 20 m/s. Is its velocity constant? Is it accelerated?
Show answers (try them yourself first)
  1. 1An object stays at rest or in uniform straight-line motion unless an external unbalanced force acts on it.
  2. 2The tendency of an object to keep its state of rest or uniform motion.
  3. 3The tangent.
  4. 4The cricket ball, as it has more mass.
  5. 5Speed = 2πr ÷ t
  6. 62 × 22/7 × 14 ÷ 44 = 88 ÷ 44 = 2 m/s
  7. 7The body tends to continue in a straight line, while the bike turns.
  8. 8No; the direction changes, so velocity changes and the car is accelerated.

Second and third laws, momentum

From daily life

A small boat tied to the shore moves backwards when you jump from it onto the land. A tennis ball hits the racket and flies away. A full truck is hard to stop but an empty cycle is easy. These things are explained by force and momentum.

Simple definition

Momentum (संवेग) p = mv. Force (बल) F = ma.

A gun recoils backwards at 1 metre per second when a 0.02 kg bullet leaves at 200 metres per secondbullet 0.02 kg200 m/sgun 4 kg? m/sTotal momentum before = 0, after = 04 × v + 0.02 × 200 = 0 ⇒ v = −1 m/s (backwards)

In the figure : In the figure conservation of momentum: a gun recoils when the bullet is fired.

Rule 1: Momentum and the second law — Momentum p = mv (a vector, unit kg m/s). Newton's second law: the rate of change of momentum is proportional to the force, which gives F = ma. 1 N = 1 kg m/s² and 1 dyne = 1 g cm/s², so 1 N = 10⁵ dyne.

  • Easy : A 2 kg ball moving at 3 m/s: p = 2 × 3 = 6 kg m/s.
  • Medium : A 5 kg trolley with a = 2 m/s²: F = 5 × 2 = 10 N.
  • Exam-type : A force of 20 N acts on a 4 kg cart: a = 20 ÷ 4 = 5 m/s². In dyne, 20 N = 20 × 10⁵ = 2 × 10⁶ dyne.

Rule 2: Third law and conservation — Third law: every action has an equal and opposite reaction, and they act on different objects. Conservation of momentum: if no external force acts, total momentum before = total momentum after. Recoil of a gun: m₁v₁ + m₂v₂ = 0.

  • Easy : A swimmer pushes water backwards, and the water pushes her forwards.
  • Medium : A 2 kg gun fires a 0.02 kg bullet at 300 m/s: 0.02 × 300 + 2 × v = 0, so v = −3 m/s (backwards recoil).
  • Exam-type : A 2 kg trolley at 4 m/s hits an equal trolley at rest and they stick: 2 × 4 = 4 × v, so v = 2 m/s.
Live figure

Second and third laws, momentum

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Remember

p = mv; F = ma; 1 N = 10⁵ dyne. Action and reaction act on different bodies. Momentum is conserved.

Common mistake

Thinking action and reaction cancel each other. They act on different objects, so they do not cancel.

Shortcut

Memory clue: "Force = Mass × Acceleration (FMA)" and "Momentum = Mass × Velocity (MV)".

Spot the difference

LawStatementExample
FirstInertiaPassenger lurches
SecondF = maPushing a cart
ThirdAction = reactionSwimming

Practice questions

Easy
  1. 1Define momentum.
  2. 2State Newton's third law.
  3. 3What is the SI unit of force?
Medium
  1. 4How many dynes make 1 newton?
  2. 5Find the momentum of a 3 kg body moving at 4 m/s.
  3. 6What force gives a 6 kg body an acceleration of 3 m/s²?
Challenging
  1. 7A 55 kg boy standing at rest on a skateboard throws a 5 kg bag forward at 6 m/s. Find his recoil speed.
  2. 8A horse pulls a cart and the cart pulls the horse back with equal force. Why does the cart still move?
Show answers (try them yourself first)
  1. 1The product of mass and velocity, p = mv.
  2. 2Every action has an equal and opposite reaction.
  3. 3newton (N), 1 N = 1 kg m/s².
  4. 410⁵ dyne
  5. 53 × 4 = 12 kg m/s
  6. 66 × 3 = 18 N
  7. 75 × 6 = 55 × v, so v = 30 ÷ 55 ≈ 0.55 m/s backwards.
  8. 8The two forces act on different bodies; the cart moves because the ground pushes the horse forwards more than the cart pulls it back.

Quick revision

  • Distance and speed are scalars; displacement and velocity are vectors.
  • v = u + at; s = ut + (1/2)at2; v2 = u2 + 2as.
  • F = ma; p = mv; momentum is conserved.