Chapter 4: Describing Motion Around Us — Part 2 | Class 9 Science

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Class 9 Science · Chapter 4 · Part 2 of 2

Describing Motion Around Us

Graphical Representation · Kinematic Equations · Motion in a Plane · Uniform Circular Motion

📊 Position-Time Graph
📈 Velocity-Time Graph
🧮 Kinematic Equations
🔄 Circular Motion

⬅️ Part 1 covered: Basics of Motion

Position, Distance & Displacement, Average Speed & Velocity, Average Acceleration

PART 2
This part covers: Graphical Representation of Motion → Kinematic Equations → Motion in a Plane → Uniform Circular Motion → Summary → Final Quiz.
📌 Jump to Section
Graphs Intro
Position-Time Graph
Velocity-Time Graph
Kinematic Equations
Motion in a Plane
Circular Motion
Summary
Definitions
Exam FAQs
Final Quiz

📊 4.2 Graphical Representation of Motion

Graphs provide a visual representation of how position, velocity, and acceleration change with time. They help in comparing motion of two objects, calculating physical quantities, and identifying uniform vs non-uniform motion.

💡 Important Note

All graphs in this chapter are for motion in a straight line in one direction only. In this case: distance = magnitude of displacement; speed = magnitude of velocity. So position-time graph = distance-time graph (if position is 0 at t=0), and velocity-time graph = speed-time graph.

Steps to Plot a Graph (Activity 4.3)

Take graph paper (pre-divided into small squares)
Draw two perpendicular lines — origin O, horizontal = X-axis, vertical = Y-axis
Decide which quantity goes on which axis (e.g., time on X, position on Y)
Choose a suitable scale (e.g., 5 divisions = 1 s)
Mark values along each axis using the chosen scale
Plot each data point
Connect points to form the graph (straight line or curve)

⭐ Note — Graph is NOT a Route Map

A graph does not show the actual route/path — it shows how position of the object changes with time with respect to the origin.

📈 4.2.2 Position-Time Graphs

📖 What does the SHAPE tell us?

Straight line position-time graph → object moving with constant velocity (uniform motion)
Curved line position-time graph → velocity is NOT constant → object is in accelerated motion
Straight horizontal line (parallel to time axis) → object is at rest (stationary)

Fig 4.13: Straight line = constant velocity | Curve = accelerated (changing) velocity

Calculating Velocity from Position-Time Graph (Activity 4.4)

The slope of a position-time graph gives velocity. Slope = steepness of the line = (change in position) ÷ (change in time).

Average velocity from graph
v = (s₂ − s₁) ÷ (t₂ − t₁) = Slope of the line

Example calculation: v = (80m − 40m) ÷ (4s − 2s) = 40m ÷ 2s = 20 m s⁻¹

📝 Example 4.6 & 4.7 — Interpreting Graphs

Ex 4.6: A horizontal line parallel to time-axis (e.g., constant 40m position) → vehicle is at rest at 40m from origin.

Ex 4.7: Two objects A and B — whichever line has a steeper slope has higher velocity. If B’s line is steeper than A’s, then velocity of B > velocity of A.

✅ Check Your Understanding — Position-Time Graphs
Q1. A straight line position-time graph indicates the object is moving with:
AZero velocity
BConstant velocity
CIncreasing velocity
DDecreasing velocity
Q2. A curved position-time graph indicates:
AObject at rest
BAccelerated motion (velocity changing)
CObject moving backward only
DZero displacement
Q3. The slope of a position-time graph gives:
AAcceleration
BVelocity
CDistance
DTime
📋 Answer Key
Q1B — Constant velocity. A straight (non-horizontal) line on a position-time graph means equal displacement in equal time = constant velocity.
Q2B — Accelerated motion. A curve means the velocity (slope) is changing at different points, indicating acceleration.
Q3B — Velocity. Slope = (change in position) ÷ (change in time) = velocity.

📉 4.2.3 Velocity-Time Graphs

📖 What does the SHAPE tell us?

Horizontal straight line (parallel to X-axis) → constant velocity, acceleration = zero
Straight line going UP → velocity increasing → constant positive acceleration (in direction of velocity)
Straight line going DOWN → velocity decreasing → constant negative acceleration (opposite to direction of velocity)

Fig 4.17(a)/(b): Horizontal line = zero acceleration | Rising line = positive acceleration

Two Things You Can Calculate from a Velocity-Time Graph

What to find	How	Result
Acceleration	Slope of the velocity-time line = (v−u)÷(t₂−t₁)	Rate of change of velocity
Displacement	Area enclosed by the line and the time axis	Net change in position

📝 Example — Area Under Velocity-Time Graph = Displacement

Case 1 (Constant velocity, rectangle): velocity = 20 m/s, time = 6s

Displacement = Area of rectangle = 20 m/s × 6 s = 120 m

Case 2 (Changing velocity, rectangle + triangle): Between 10s-20s, velocity goes from 5 m/s to 10 m/s

Displacement = (5 m/s × 10s) + (½ × 10s × 5 m/s) = 50m + 25m = 75 m
🎯 Exam Point — Key Formula Summary

Slope of position-time graph = Velocity
Slope of velocity-time graph = Acceleration
Area under velocity-time graph (with time axis) = Displacement

✅ Check Your Understanding — Velocity-Time Graphs
Q4. The slope of a velocity-time graph gives:
ADisplacement
BDistance
CAcceleration
DPosition
Q5. The area under a velocity-time graph (with time axis) gives:
ADisplacement
BAcceleration
CVelocity
DTime
Q6. A horizontal line on a velocity-time graph means:
AObject at rest
BConstant velocity, zero acceleration
CIncreasing acceleration
DObject reversing direction
📋 Answer Key
Q4C — Acceleration. Slope of velocity-time graph = (change in velocity)÷(change in time) = acceleration.
Q5A — Displacement. The enclosed area between the velocity-time line and the time axis equals displacement.
Q6B — Constant velocity, zero acceleration. No change in height (velocity) over time means zero slope = zero acceleration.

🧮 4.3 Kinematic Equations for Motion with Constant Acceleration

📖 The Three Kinematic Equations

For motion in a straight line with constant acceleration, five quantities relate: displacement (s), time (t), initial velocity (u), final velocity (v), acceleration (a).

Eq 4.4a — First Equation
v = u + at
Eq 4.4b — Second Equation
s = ut + ½at²
Eq 4.4c — Third Equation
v² = u² + 2as
🎯 Exam Point — Memory Trick

v = u + at (no s) | s = ut + ½at² (no v) | v² = u² + 2as (no t)

Each equation is missing exactly one of the 5 quantities — useful for choosing which equation to use!

💡 Important Note

These kinematic equations are valid only when acceleration is constant. In straight-line motion (both directions), the sign of u, v, a, and s tells us the direction of that quantity.

📝 Example 4.8 — Braking Distance

Problem: Brakes cause acceleration of −4 m/s². Find stopping distance if initial velocity is (i) 54 km/h, (ii) 108 km/h.

Given: a = −4 m/s², v = 0 m/s

Using v² = u² + 2as → 0 = u² − 8s → s = u²÷8

(i) u = 15 m/s → s = 225÷8 = 28.1 m
(ii) u = 30 m/s → s = 900÷8 = 112.5 m

Key insight: Doubling the speed increases stopping distance by 4 times (not 2 times)! This is why high speed driving is so dangerous.

🌏 Bridging Science and Society — Safe Driving Distance

Stopping distance depends on: velocity, road surface (wet/dry), braking capacity, and driver’s reaction time. This is why maintaining a safe distance from the vehicle ahead is critical — and why this distance must increase with speed. Vehicle-to-Vehicle (V2V) communication technology is being developed (including in India) to warn drivers of possible collisions.

✅ Check Your Understanding — Kinematic Equations
Q7. Which kinematic equation does NOT involve displacement (s)?
Av = u + at
Bs = ut + ½at²
Cv² = u² + 2as
DNone of these
Q8. Kinematic equations are valid only when:
AVelocity is zero
BAcceleration is constant
CTime is zero
DObject is at rest
📋 Answer Key
Q7A — v = u + at. This equation relates v, u, a, t — no ‘s’ (displacement) involved.
Q8B — Acceleration is constant. These equations are derived assuming constant acceleration throughout the motion.

🌐 4.4 Motion in a Plane

📖 Definition

Motion in a plane (2D) — Motion that doesn’t follow a straight line, e.g., a vehicle overtaking, path of a kicked ball, satellite moving in a circular path.

💡 Ready to Go Beyond — Motion in 3D

Motion in space (3D) — e.g., car climbing a mountain road, bird flying, aircraft moving through air.

🔄 4.4.1 Uniform Circular Motion

Consider a child on a merry-go-round moving from A→B→C (a curved path).

Distance travelled = the curve ABC (along the circular path)
Displacement = the straight line AC (these are NOT equal)
Distance in one revolution = circumference = 2πR
Displacement after one full revolution = zero (back to start)

Eq 4.5 — Average Speed in Circular Motion
v_av = 2πR ÷ T (R = radius, T = time for one revolution)
📖 Definition — Uniform Circular Motion

When an object moves in a circular path with constant (uniform) speed, its motion is called uniform circular motion.

Why is Uniform Circular Motion Accelerated?

An athlete running along closed paths (rectangle → hexagon → circle): as the number of sides increases, the athlete changes direction more frequently. A circle is the limiting case where direction changes continuously.

🎯 Exam Point — Crucial Concept!

In uniform circular motion, speed is constant but the direction of velocity changes continuously (velocity is always along the tangent to the circle). Since velocity changes (in direction), the motion is accelerated — even though speed doesn’t change!

💡 Note — Everyday Misconception

In everyday life, we say a vehicle is “accelerating” only when speed changes. But acceleration can also occur when only the direction of velocity changes — like in circular motion!

🧪 Activity 4.5 — Marble in a Ring
Setup

Place a ring (tape ring) flat on a smooth surface. Throw a marble inside so it rotates along the inner boundary.

Action

After 1-2 revolutions, lift the ring without disturbing the marble’s motion.

Observation

The marble moves in a straight line (not circular) once released!

Conclusion

Once released, the marble continues in the direction it was moving at that instant (tangent direction).

🔩 Ready to Go Beyond — Tangent to a Circle

The velocity at a point on a circular path is along the tangent to the circle at that point, in the direction of motion. A tangent is a straight line that touches the circle at one and only one point.

💡 Real-World Application

Uniform circular motion is an idealised model — real conditions (constant speed + perfect circle) are rarely fully met. Still, it’s foundational for understanding: planets orbiting the Sun, vehicles making circular turns, satellites, etc.

✅ Check Your Understanding — Circular Motion
Q9. After one complete revolution in a circular path, the displacement is:
AEqual to 2πR
BZero
CEqual to the radius R
DEqual to diameter
Q10. In uniform circular motion, the object is accelerated because:
ASpeed is changing
BDirection of velocity is changing
CIt is moving in a straight line
DIt is at rest
Q11. Velocity at any point on a circular path is directed:
ATowards the centre
BAway from the centre
CAlong the tangent at that point
DOpposite to motion
📋 Answer Key
Q9B — Zero. After one revolution, the object returns to its starting position, so net displacement = 0.
Q10B — Direction of velocity is changing. Even though speed is constant, the continuously changing direction makes the motion accelerated.
Q11C — Along the tangent. Velocity at any instant on a circular path points along the tangent to the circle at that point, in the direction of motion.

📋 Chapter Summary — Full Chapter

📍Position & MotionPosition = distance + direction from reference point. Object in motion if position changes with time.
📏Distance & DisplacementDistance = scalar, total path. Displacement = vector, net position change. Equal only if no turning back.
⚡Speed & VelocitySpeed = distance÷time (scalar). Velocity = displacement÷time (vector).
🚀AccelerationChange in velocity ÷ time. Can be zero even at high speed if velocity is constant.
📈GraphsSlope of position-time = velocity. Slope of velocity-time = acceleration. Area under v-t graph = displacement.
🧮Kinematic Equationsv=u+at | s=ut+½at² | v²=u²+2as — valid only for constant acceleration.
🔄Circular MotionUniform circular motion = constant speed, but direction changes continuously → always accelerated.

🔑 Keywords — Part 2

Position-Time GraphVelocity-Time GraphSlopeArea Under GraphKinematic EquationsMotion in a PlaneTwo DimensionsCircular MotionUniform Circular MotionRevolutionCircumferenceTangentThree DimensionsV2V CommunicationStopping Distance

📖 Important Definitions — Part 2

SlopeThe steepness of a line on a graph; gives the rate of change of the Y-axis quantity with respect to the X-axis quantity.
Kinematic EquationsThree equations (v=u+at, s=ut+½at², v²=u²+2as) relating displacement, time, initial velocity, final velocity, and acceleration — valid only for constant acceleration.
Motion in a Plane (2D)Motion not confined to a single straight line, e.g., a kicked ball’s path, a vehicle overtaking another.
Uniform Circular MotionMotion of an object in a circular path with constant (uniform) speed. The direction of velocity changes continuously, making it accelerated motion.
TangentA straight line that touches a circle at exactly one point. The velocity of an object in circular motion is directed along the tangent at that point.

❓ Frequently Asked Exam Concepts — Part 2

Why is uniform circular motion considered accelerated motion even though speed is constant? ▶
Acceleration occurs whenever velocity changes — and velocity is a vector quantity with both magnitude (speed) AND direction. In uniform circular motion, the speed remains constant, but the direction of velocity changes continuously as the object moves around the circle (velocity is always along the tangent, which keeps rotating). Since the direction component of velocity is constantly changing, the motion is accelerated, even without any change in speed.
How do you determine velocity and acceleration from graphs? ▶
From a position-time graph, the SLOPE of the line gives velocity (since velocity = change in position ÷ change in time). From a velocity-time graph, the SLOPE gives acceleration (since acceleration = change in velocity ÷ change in time), and the AREA enclosed between the line and the time axis gives displacement (since displacement = velocity × time, which is exactly what area represents geometrically).
Why does doubling speed quadruple the stopping distance? ▶
Using v² = u² + 2as, with v=0 (final velocity at stop): s = u²/(2|a|). Since distance is proportional to the SQUARE of initial velocity (u²), doubling u means s increases by a factor of 2² = 4. This is why driving at high speeds dramatically increases stopping distance and the risk of accidents — not just proportionally, but quadratically.
What happens to a circularly moving object if the centripetal force/constraint is suddenly removed? ▶
As shown in Activity 4.5 (marble in a ring), when the constraint (the ring) is removed, the object continues moving in a STRAIGHT LINE — in the direction it was moving at that exact instant (i.e., along the tangent to the circle at that point). It does NOT continue in a circular path. This demonstrates that circular motion requires a continuous force/constraint to keep changing the direction of motion.

🏆 Final Practice Quiz
15 Mixed MCQs · Full Chapter Coverage (Part 1 + Part 2)
Q1 — The SI unit of displacement is:
AMetre
BSecond
Cm/s
Dm/s²
Q2 — Which quantity is a vector (requires direction)?
ADistance
BDisplacement
CSpeed
DTime
Q3 — Total distance and magnitude of displacement are equal only when an object:
AMoves in a circle
BMoves in one direction (no turning back)
CIs at rest
DAccelerates
Q4 — Average speed is calculated using:
ATotal distance ÷ time interval
BDisplacement ÷ time interval
CVelocity ÷ acceleration
DTime ÷ distance
Q5 — A car moving at constant velocity on a highway has acceleration equal to:
AZero
BEqual to its velocity
C9.8 m/s²
DMaximum possible
Q6 — A curved position-time graph indicates the object’s motion is:
AAt rest
BUniform (constant velocity)
CAccelerated (velocity changing)
DCircular
Q7 — The area under a velocity-time graph (with time axis) gives:
ADisplacement
BAcceleration
CSpeed
DTime
Q8 — Which equation relates v, u, a and s (no time, t)?
Av = u + at
Bs = ut + ½at²
Cv² = u² + 2as
DNone
Q9 — In uniform circular motion, after one revolution, the displacement is:
AZero
B2πR
CπR
DR
Q10 — Why is uniform circular motion considered accelerated motion?
ASpeed is increasing
BDirection of velocity is continuously changing
CThe object stops periodically
DSpeed is decreasing
Q11 — A bus moving at 36 km/h slows down (negative acceleration). The acceleration is acting:
AOpposite to the direction of velocity
BIn the direction of velocity
CPerpendicular to velocity
DZero
Q12 — The velocity at any point in circular motion is directed:
ATowards the centre
BAlong the tangent
CAway from centre, radially
DVertically upward always
Q13 — If initial speed doubles (with same negative acceleration), the stopping distance becomes:
ADouble
BFour times
CSame
DHalf
Q14 — Average speed has no direction because it is calculated from:
ADistance travelled (which has no direction)
BDisplacement
CAcceleration
DTime interval
Q15 — Slope of a position-time graph represents:
AVelocity
BAcceleration
CDisplacement
DTime
Your Final Quiz Score
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📋 Final Quiz — Answer Key with Explanations
Q1A — Metre. Displacement’s SI unit is metre, same as distance.
Q2B — Displacement. Displacement requires both magnitude and direction, making it a vector.
Q3B — Moves in one direction. Without reversing direction, total distance equals displacement magnitude.
Q4A — Total distance ÷ time interval. This is Eq 4.1, the formula for average speed.
Q5A — Zero. Constant velocity means no change, hence zero acceleration regardless of speed value.
Q6C — Accelerated. A curve indicates the slope (velocity) is changing — i.e., accelerated motion.
Q7A — Displacement. Area under v-t graph = velocity × time = displacement.
Q8C — v² = u² + 2as. This equation has no ‘t’ (time) term.
Q9A — Zero. The object returns to its starting position after one revolution, so net displacement = 0.
Q10B — Direction continuously changing. Speed stays constant but velocity direction changes constantly = acceleration.
Q11A — Opposite to velocity. When velocity decreases, acceleration acts opposite to the direction of motion.
Q12B — Along the tangent. Velocity in circular motion always points along the tangent at that instant’s position.
Q13B — Four times. Since s ∝ u² (from v²=u²+2as with v=0), doubling u quadruples s.
Q14A — Distance travelled. Distance has no direction (scalar), so speed calculated from it also has no direction.
Q15A — Velocity. Slope = (change in position)÷(change in time) = velocity, by definition.