MOTION

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Chapter 7 – Motion (Class 9 Science)
This chapter introduces the fundamental concepts of motion in physics. It helps students understand how objects move, how their motion is described and measured, and how to interpret different types of motion using graphs and equations.

Contents hide

1 🔹 Key Concepts in Motion

2 Rest and Motion

2.1 Types of Motion

2.2 Distance and Displacement

2.3 Speed and Velocity

2.4 Acceleration

3 🔹 Graphical Representation

3.1 a) Distance-Time Graph

3.2 b) Velocity-Time Graph

4 🔹 Equations of Uniformly Accelerated Motion

5 🔹 Uniform Circular Motion

6 🔹 Important Points

7 ✅ Conclusion

8 FAQs

🔹 Key Concepts in Motion

Rest and Motion

Rest: An object is said to be at rest if it does not change its position with respect to its surroundings.
Motion: An object is said to be in motion if it changes its position with respect to its surroundings over time.

Example: A tree is at rest; a moving car is in motion.

Types of Motion

Rectilinear motion: Movement in a straight line (e.g., a car on a straight road).
Circular motion: Movement in a circular path (e.g., a merry-go-round).
Periodic motion: Motion repeated after equal intervals of time (e.g., pendulum).

Distance and Displacement

Distance: Total path length covered by an object; scalar quantity.
Displacement: Shortest distance between initial and final position; vector quantity.

Example: If a car moves in a circle and returns to the starting point, distance ≠ 0, but displacement = 0.

1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

Yes, an object can have zero displacement even if it has moved through a distance. This happens when the initial and final positions of the object are the same.

Example: If a person walks around a circular track and returns to the starting point, the distance covered is the total path length, but the displacement is zero.

2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Side of square = 10 m, so perimeter = 4 × 10 = 40 m
Time for 1 round = 40 seconds
Total time = 2 minutes 20 seconds = 140 seconds
Number of rounds = 140 ÷ 40 = 3.5 rounds
→ After 3 full rounds, the farmer is back at the starting point (displacement = 0).
→ In the next half round (0.5 round), the farmer reaches the opposite corner of the square.
Displacement = Diagonal of the square = √(10² + 10²) = √200 ≈ 14.14 m

Final Answer: 14.14 meters

3. Which of the following is true for displacement?

(a) It cannot be zero.
(b) Its magnitude is greater than the distance travelled by the object.

Answer: None of the above options are correct. The correct statement is:
Displacement can be zero, and its magnitude is always less than or equal to the distance travelled.

Speed and Velocity

Velocity includes direction, speed does not.

Speed: Distance travelled per unit time; scalar quantity.
- Speed = Total distance / Total time
Velocity: Displacement per unit time; vector quantity.
- Velocity = Displacement / Time

Uniform speed/velocity: Constant over time
Non-uniform speed/velocity: Changes with time

1. Distinguish between speed and velocity.

Speed: It is the distance travelled by an object per unit time. It is a scalar quantity.
Velocity: It is the displacement of an object per unit time in a specific direction. It is a vector quantity.

Example: A car moving 60 km/h is its speed, but 60 km/h towards north is its velocity.

2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

The magnitude of average velocity is equal to average speed when the total displacement is equal to total distance. This happens when the object moves in a straight line without changing direction.

3. What does the odometer of an automobile measure?

The odometer in a vehicle measures the total distance travelled by the automobile.

4. What does the path of an object look like when it is in uniform motion?

When an object is in uniform motion, it moves in a straight line, covering equal distances in equal intervals of time. So the path appears as a straight line.

5. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 10⁸ m/s.

Given:
Time = 5 minutes = 5 × 60 = 300 seconds
Speed of light = 3 × 10⁸ m/s

Distance = Speed × Time
= (3 × 10⁸) × 300
= 9 × 10¹⁰ meters

Final Answer: 9 × 10¹⁰ m

Acceleration

Acceleration: Rate of change of velocity.
- Acceleration $a=\dfrac{v-u}{t}$
- Where:
- $u$ = initial velocity
- $v$ = final velocity
- $t$ = time taken

Positive acceleration: Speeding up
Negative acceleration (retardation): Slowing down

1. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?

(i) Uniform acceleration: A body is said to be in uniform acceleration when it changes its velocity by equal amounts in equal intervals of time.

(ii) Non-uniform acceleration: A body is said to be in non-uniform acceleration when the change in velocity is not equal in equal intervals of time.

2. A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find the acceleration of the bus.

Given:
Initial speed, u = 80 km/h = 80 × (1000/3600) = 22.22 m/s
Final speed, v = 60 km/h = 60 × (1000/3600) = 16.67 m/s
Time, t = 5 s

Using formula: a = (v - u) / t
= (16.67 – 22.22) / 5 = -5.55 / 5 = -1.11 m/s²

Final Answer: -1.11 m/s² (negative sign shows retardation)

3. A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/h in 10 minutes. Find its acceleration.

Given:
Initial speed, u = 0 (starts from rest)
Final speed, v = 40 km/h = 40 × (1000/3600) = 11.11 m/s
Time, t = 10 minutes = 600 s

Using formula: a = (v - u) / t
= (11.11 – 0) / 600 = 11.11 / 600 = 0.0185 m/s²

Final Answer: 0.0185 m/s²

🔹 Graphical Representation

a) Distance-Time Graph

Straight line = Uniform motion
Curved line = Non-uniform motion

Slope = Speed

b) Velocity-Time Graph

Straight horizontal line = Constant velocity
Sloped line = Uniform acceleration

Area under the graph = Distance

1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?

– For uniform motion, the distance-time graph is a straight line. This shows that the object covers equal distances in equal intervals of time.
– For non-uniform motion, the distance-time graph is a curved line, indicating unequal distances in equal intervals of time.

2. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?

If the distance-time graph is a straight line parallel to the time axis, it means the distance is not changing with time. Hence, the object is at rest (no motion).

3. What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?

A speed-time graph parallel to the time axis shows that speed is constant over time. This means the object is moving with uniform speed.

4. What is the quantity which is measured by the area occupied below the velocity-time graph?

The area under the velocity-time graph represents the displacement (or distance, if velocity is not negative) travelled by the object during the given time interval.

🔹 Equations of Uniformly Accelerated Motion

Three equations that describe motion mathematically.

$v=u+at$
$s=ut+\frac{1}{2}at^2$
$v^2=u^2+2as$

Where:

$s$ = displacement
$u$ = initial velocity
$v$ = final velocity
$a$ = acceleration
$t$ = time

1. A bus starting from rest moves with a uniform acceleration of 0.1 m/s² for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.

Given:
u = 0 m/s, a = 0.1 m/s², t = 2 min = 120 s

(a) Speed acquired:
v = u + at = 0 + (0.1 × 120) = 12 m/s

(b) Distance travelled:
s = ut + ½ at² = 0 + ½ × 0.1 × (120)² = 0.05 × 14400 = 720 m

Answers:
Speed = 12 m/s
Distance = 720 m

2. A train is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of −0.5 m/s². Find how far the train will go before it is brought to rest.

Given:
u = 90 km/h = 90 × 1000 / 3600 = 25 m/s
v = 0 (final speed when train stops)
a = -0.5 m/s²

Using: v² = u² + 2as
0 = (25)² + 2 × (-0.5) × s
0 = 625 – s
s = 625 / 1 = 625 m

Answer: 625 meters

3. A trolley, while going down an inclined plane, has an acceleration of 2 cm/s². What will be its velocity 3 s after the start?

Given:
u = 0 (starts from rest), a = 2 cm/s² = 0.02 m/s², t = 3 s

Using: v = u + at = 0 + 0.02 × 3 = 0.06 m/s

Answer: 0.06 m/s

4. A racing car has a uniform acceleration of 4 m/s². What distance will it cover in 10 s after start?

Given:
u = 0, a = 4 m/s², t = 10 s
Using: s = ut + ½ at²
s = 0 + ½ × 4 × 100 = 2 × 100 = 200 m

Answer: 200 meters

5. A stone is thrown in a vertically upward direction with a velocity of 5 m/s. If the acceleration of the stone during its motion is 10 m/s² in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Given:
u = 5 m/s, v = 0 (at highest point), a = -10 m/s²

To find height (s):
Using v² = u² + 2as
0 = 25 + 2 × (-10) × s ⇒ -25 = -20s ⇒ s = 25 / 20 = 1.25 m

To find time (t):
v = u + at ⇒ 0 = 5 + (-10)t ⇒ t = 5 / 10 = 0.5 s

Answers:
Height = 1.25 meters
Time = 0.5 seconds

🔹 Uniform Circular Motion

Even constant speed can have acceleration due to changing direction.

Motion in a circle with constant speed.
Velocity is changing due to change in direction.
It is an accelerated motion.

Example: Revolution of Earth around the Sun

🔹 Important Points

Scalar quantities: Only magnitude (e.g., distance, speed)
Vector quantities: Magnitude + direction (e.g., displacement, velocity)
SI Unit of speed/velocity: m/s
SI Unit of acceleration: m/s²

\dfrac{\text{Displacement}}{\text{Time}}

Quantity	Formula	Unit
Speed	$\dfrac{\text{Distance}}{\text{Time}}$	m/s
Velocity	$\dfrac{\text{Displacement}}{\text{Time}}$	m/s
Acceleration	$\dfrac{v - u}{t}$	m/s²
Equation 1	$v = u + at$
Equation 2	$s = ut + \dfrac{1}{2}at^2$
Equation 3	$v^2 = u^2 + 2as$

✅ Conclusion

This chapter builds the foundation of mechanics by explaining motion, its types, and the mathematical tools (graphs and equations) used to describe it. Understanding these basic concepts is essential for grasping advanced topics in physics.

FAQs

1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?

Given:
Diameter = 200 m ⇒ Circumference = π × d = 3.14 × 200 = 628 m
Time for 1 round = 40 s
Total time = 2 min 20 s = 140 s
Number of rounds = 140 ÷ 40 = 3.5 rounds

Distance covered:
= 3.5 × 628 = 2198 m
Displacement:
After 3 rounds = back to start (displacement = 0)
After 0.5 round = at opposite end of circle ⇒ displacement = diameter = 200 m

Answers:
Distance = 2198 m
Displacement = 200 m

2. Joseph jogs from point A to B (300 m) in 2 min 30 s and then turns and jogs 100 m back to point C in 1 minute. Find average speed and velocity for (a) A to B, (b) A to C.

(a) From A to B:
Distance = 300 m, Time = 150 s
Average speed = 300 / 150 = 2 m/s
Displacement = 300 m (straight line) ⇒ Velocity = 2 m/s

(b) From A to C:
Distance = 300 + 100 = 400 m
Time = 150 + 60 = 210 s
Average speed = 400 / 210 ≈ 1.90 m/s
Displacement = 300 – 100 = 200 m ⇒ Velocity = 200 / 210 ≈ 0.95 m/s

Answers:
(a) Speed = Velocity = 2 m/s
(b) Avg Speed ≈ 1.90 m/s, Avg Velocity ≈ 0.95 m/s

3. Abdul drives to school with avg speed 20 km/h and returns at 30 km/h. What is the average speed for the entire trip?

Let distance = d km (same both ways)
Time (to school) = d / 20, Time (return) = d / 30
Total distance = 2d, Total time = d/20 + d/30 = (3d + 2d)/60 = 5d/60

Average speed = Total distance / Total time
= 2d / (5d / 60) = (2 × 60) / 5 = 24 km/h

Answer: 24 km/h

4. A motorboat starting from rest accelerates at 3.0 m/s² for 8 s. How far does it travel?

Given: u = 0, a = 3 m/s², t = 8 s
s = ut + ½ at² = 0 + ½ × 3 × 8² = 1.5 × 64 = 96 m

Answer: 96 meters

5. A car is travelling at 52 km/h and applies brakes. (a) Shade area on graph that shows distance during braking. (b) Which part represents uniform motion?

(a) The area under the speed-time graph during deceleration (i.e., where the line slopes down) represents the distance covered while braking. It is the area of the triangle beneath the curve.

(b) The horizontal portion of the speed-time graph (a flat line) represents uniform motion, where speed remains constant.

6. Study the distance-time graph of three objects A, B, and C and answer the following:

(a) Which of the three is travelling the fastest?
→ Object B is travelling fastest because it has the steepest slope (greater distance in lesser time).

(b) Are all three ever at the same point on the road?
→ No, the three lines never intersect at the same point, so they are never at the same position simultaneously.

(c) How far has C travelled when B passes A?
→ B passes A at around 1.2 h. From the graph, at 1.2 h, C is at 8 km.

(d) How far has B travelled by the time it passes C?
→ From the graph, B passes C at 1.4 h. At that time, B has travelled approximately 12 km.

7. A ball is dropped from a height of 20 m. If velocity increases at 10 m/s², what velocity will it strike the ground with? After what time?

Given: u = 0, s = 20 m, a = 10 m/s²
Using: v² = u² + 2as = 0 + 2 × 10 × 20 = 400 ⇒ v = √400 = 20 m/s
Time: v = u + at ⇒ 20 = 0 + 10 × t ⇒ t = 2 s

Answer: Velocity = 20 m/s, Time = 2 s

8. Speed-time graph for a car is given. (a) Find distance in first 4 seconds. (b) Which part represents uniform motion?

(a) Area under speed-time graph gives distance.
Approximate as area of a trapezium or by counting boxes (graph dependent). If exact curve is known:
Use: Area ≈ (1/2) × (sum of parallel sides) × height
For linear rise from 0 to 6 m/s in 4 s:
Area = ½ × (0 + 6) × 4 = 12 m

(b) The flat part of the graph (from 6 to 10 seconds) shows speed is constant ⇒ uniform motion.

9. State which of the following situations are possible. Give example for each.

(a) An object with constant acceleration but zero velocity –
✔️ Possible at turning point (e.g., stone thrown vertically up, velocity = 0 at top but acceleration = g).

(b) Object with acceleration but uniform speed –
✔️ Possible in circular motion, where speed is constant but direction (hence velocity) is changing.

(c) Object moving in one direction with acceleration in perpendicular direction –
✔️ Possible in projectile motion (horizontal velocity + vertical acceleration).

10. An artificial satellite is moving in circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve.

Given: r = 42250 km = 4.225 × 10⁷ m, T = 24 h = 86400 s
Circumference = 2πr = 2 × 3.14 × 4.225 × 10⁷ ≈ 2.65 × 10⁸ m
Speed = Distance / Time
= 2.65 × 10⁸ / 86400 ≈ 3.07 × 10³ m/s

Answer: 3.07 km/s (approx.)