Motion

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CBSE Learning Objectives – Key Concepts & Skills You Must Know

Learning Objectives

Understand the concept of motion and its various forms.
Describe motion in terms of distance and displacement.
Differentiate between uniform and non-uniform motion.
Calculate speed and velocity, including average speed.
Apply kinematic equations to solve problems involving uniformly accelerated motion.
Analyze motion using graphs, including distance-time and speed-time graphs.
Recognize the significance of reference points in describing motion.
Explore real-life examples of circular motion and its implications.

CBSE Revision Notes & Quick Summary for Last-Minute Study

Chapter 7: Motion

7.1 Describing Motion

Definition of Motion: Change of position over time.
Reference Point: The position of an object is described relative to a reference point (e.g., a school is 2 km north of a railway station).

Types of Motion

Straight Line Motion: Objects may move in a straight line, circular path, rotate, or vibrate.
Uniform Motion: Equal distances in equal intervals of time (e.g., a car moving at a constant speed).
Non-Uniform Motion: Unequal distances in equal intervals of time (e.g., a car in traffic).

7.2 Measuring the Rate of Motion

Speed: Distance covered per unit time.
Velocity: Displacement per unit time.
Acceleration: Change in velocity per unit time.

Key Equations

Final Velocity:

$v = u + at$
- Where:
  - $v$ = final velocity
  - $u$ = initial velocity
  - $a$ = acceleration
  - $t$ = time
Displacement:

$s = ut + \frac{1}{2} at^2$
- Where:
  - $s$ = displacement
Acceleration Relation:

$2as = v^2 - u^2$

7.3 Graphical Representation of Motion

Speed-Time Graph: Illustrates speed against time, showing acceleration and uniform motion.
Distance-Time Graph: Shows distance against time, indicating speed and motion type.

Example Data

Time (s)	Distance travelled by Feroz (km)	Distance travelled by Sania (km)
8:00 am	0	0
8:05 am	1.0	0.8
8:10 am	1.9	1.6
8:15 am	2.8	2.3
8:20 am	3.6	3.0
8:25 am	-	3.6

7.4 Circular Motion

Uniform Circular Motion: Motion in a circular path at constant speed.
Example: The motion of the moon around the earth or a cyclist on a circular track.

7.5 Applications of Motion

Real-Life Examples: Motion of vehicles, athletes in sports, and natural phenomena like the movement of planets.

7.6 Exercises

Calculate distance and displacement for an athlete completing a circular track.
Average speed and velocity calculations for a jogger.
Average speed for a round trip with different speeds.
Distance covered by a motorboat under constant acceleration.
Analyze a speed-time graph for distance traveled.

Conclusion

Understanding motion is crucial for analyzing various physical phenomena and applications in everyday life.

CBSE Exam Tips, Important Questions & Common Mistakes to Avoid

Common Mistakes and Exam Tips

Common Pitfalls

Confusing Distance and Displacement: Students often think distance and displacement are the same. Remember, distance is the total path length covered, while displacement is the shortest distance from the initial to the final position.
Misunderstanding Uniform and Non-Uniform Motion: Many confuse uniform motion (equal distances in equal intervals) with non-uniform motion (unequal distances in equal intervals). Ensure to identify examples correctly.
Ignoring Reference Points: When describing motion, failing to specify a reference point can lead to incorrect interpretations of an object's position.
Assuming All Motion is Perceived: Students might think that all motion can be directly observed. Some motions, like the Earth's rotation, are inferred through indirect evidence.

Tips for Exam Preparation

Practice with Graphs: Familiarize yourself with distance-time and speed-time graphs. Understand how to interpret them and identify uniform motion.
Use Real-Life Examples: Relate concepts to everyday experiences, such as driving or walking, to better understand motion.
Review Kinematic Equations: Make sure to memorize and understand the application of key equations like:
- v = u + at
- s = ut + ½ at²
- 2as = v² - u²
Conduct Experiments: Engage in hands-on activities to visualize concepts of motion, such as measuring distances and times in various scenarios.

CBSE Quiz & Practice Test – MCQs, True/False Questions with Solutions

Multiple Choice Questions

3.07 km/s

1.77 km/s

2.97 km/s

4.07 km/s

Correct Answer: B

Solution:

The circumference of the orbit is

2\pi \times 42250\, km

. The speed

v = \frac{2\pi \times 42250}{24 \times 3600}\, km/s = 1.77\, km/s

70 km

30 km

40 km

10 km

Correct Answer: A

Solution:

The total distance covered by the car is the sum of the distances from X to Y and Y to Z, which is 40 km + 30 km = 70 km.

3 km

5 km

8 km

13 km

Correct Answer: A

Solution:

The cyclist ends up 3 km north of the starting point, so the displacement is 3 km.

Motion is always observed directly.

Motion can be inferred through indirect evidence.

Motion is only possible in a straight line.

Motion does not require a reference point.

Correct Answer: B

Solution:

Motion can be inferred through indirect evidence, such as observing the movement of dust or leaves.

50 m

100 m

150 m

200 m

Correct Answer: B

Solution:

Using the equation

s = ut + \frac{1}{2}at^2

, where

u = 0

a = 2 \text{ m/s}^2

, and

t = 10 \text{ s}

, we get

s = 0 + \frac{1}{2} \times 2 \times 10^2 = 100 \text{ m}

The stone moves in a circular path.

The stone moves in a straight line tangential to the circle.

The stone falls straight down.

The stone stops moving.

Correct Answer: B

Solution:

When the string is released, the stone moves in a straight line tangential to the circular path due to inertia.

5 km

7 km

12 km

1 km

Correct Answer: B

Solution:

The total distance traveled is the sum of all distances covered, which is 3 km + 4 km = 7 km.

0 m

10 m

20 m

5 m

Correct Answer: A

Solution:

Displacement is the shortest distance from the initial to the final position. Since the object returns to its starting point, the displacement is 0 m.

10 m/s

12 m/s

14 m/s

16 m/s

Correct Answer: C

Solution:

Using the equation

v = u + at

, where

u = 2 \text{ m/s}

a = 3 \text{ m/s}^2

, and

t = 4 \text{ s}

, we get

v = 2 + 3 \times 4 = 14 \text{ m/s}

2.62 m/s

3.14 m/s

4.18 m/s

5.24 m/s

Correct Answer: B

Solution:

The circumference of the track is

\pi \times 100 \text{ m} = 314 \text{ m}

. Completing two rounds, the total distance is

2 \times 314 \text{ m} = 628 \text{ m}

. The time taken is 4 minutes = 240 seconds. Therefore, the average speed is

\frac{628 \text{ m}}{240 \text{ s}} = 2.62 \text{ m/s}

96 m

48 m

192 m

144 m

Correct Answer: A

Solution:

Using the equation of motion

s = ut + \frac{1}{2}at^2

where initial velocity

u = 0

, acceleration

a = 3 \text{ m/s}^2

, and time

t = 8 \text{ s}

, the distance

s = 0 \times 8 + \frac{1}{2} \times 3 \times 8^2 = 96 \text{ m}

10 m/s

20 m/s

15 m/s

25 m/s

Correct Answer: B

Solution:

Using the equation

v^2 = u^2 + 2as

, where

u = 0

a = 10\, m/s^2

s = 20\, m

. Solving for

v

v = \sqrt{0 + 2 \times 10 \times 20} = 20\, m/s

60 km

120 km

30 km

90 km

Correct Answer: A

Solution:

The total distance covered is the sum of the distance from A to B and back from B to A, which is 120 km. Since the displacement is 0 km, points A and B are the same, and the distance between them is half of the total distance, i.e., 60 km.

20 m/s

25 m/s

15 m/s

30 m/s

Correct Answer: B

Solution:

Using the equation

v = u + at

, where

u = 5 \text{ m/s}

a = 4 \text{ m/s}^2

, and

t = 5 \text{ s}

, we find

v = 5 + 4 \times 5 = 25 \text{ m/s}

Distance: 300 m, Displacement: 200 m

Distance: 300 m, Displacement: 100 m

Distance: 400 m, Displacement: 200 m

Distance: 400 m, Displacement: 100 m

Correct Answer: C

Solution:

The circumference of the track is

2\pi \times 100 = 200\pi

m. For one and a half rounds, the distance is

1.5 \times 200\pi = 300\pi

m. The displacement is the diameter of the circle, which is 200 m.

60 km

25 km

85 km

35 km

Correct Answer: D

Solution:

The displacement is the shortest distance from the initial to the final position. Here, the displacement is 35 km from O to B.

3,070 m/s

4,420 m/s

7,560 m/s

1,770 m/s

Correct Answer: A

Solution:

The speed of the satellite can be calculated using the formula for the circumference of a circle and the time period:

\text{Speed} = \frac{2\pi r}{T}

. Here,

r = 42,250 \text{ km} = 42,250,000 \text{ m}

and

T = 24 \times 3600 \text{ s}

. Thus,

\text{Speed} = \frac{2 \times 3.1416 \times 42,250,000}{86,400} \approx 3,070 \text{ m/s}

100 m

200 m

150 m

250 m

Correct Answer: A

Solution:

Using the equation of motion

s = ut + \frac{1}{2}at^2

, where

u = 0

v = 20

m/s,

t = 10

s, and

a = \frac{v-u}{t} = 2

m/s². The distance

s = 0 \times 10 + \frac{1}{2} \times 2 \times 10^2 = 100

200 m

400 m

0 m

100 m

Correct Answer: C

Solution:

Displacement is the shortest distance between the initial and final positions. After one full round, the displacement is 0 m as the athlete returns to the starting point.

2 m/s

1.67 m/s

1.5 m/s

1.25 m/s

Correct Answer: B

Solution:

The total distance covered by Joseph from A to C is 300 m + 100 m = 400 m. The total time taken is 2.5 minutes + 1 minute = 3.5 minutes = 210 seconds. The average speed is given by

\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{400}{210} \approx 1.67 \text{ m/s}

400 m

200 m

0 m

800 m

Correct Answer: C

Solution:

After completing one full round on a circular track, the runner returns to the starting point, making the displacement 0 m.

It is always equal to the distance travelled.

It can be zero even if distance is not zero.

It is always greater than the distance travelled.

It is a scalar quantity.

Correct Answer: B

Solution:

Displacement can be zero if the initial and final positions are the same, even if the distance travelled is not zero.

40 km

100 km

70 km

0 km

Correct Answer: B

Solution:

The total distance is the sum of all segments: 30 km + 40 km + 30 km = 100 km.

0 meters

10 meters

20 meters

5 meters

Correct Answer: A

Solution:

The displacement is the shortest distance from the initial to the final position. Since the person returns to the starting point, the displacement is 0 meters.

100 km

20 km

60 km

40 km

Correct Answer: A

Solution:

The total distance covered is the sum of the distances traveled in each direction: 60 km + 40 km = 100 km.

Tangentially upward

Tangentially downward

Radially inward

Radially outward

Correct Answer: A

Solution:

When an object in circular motion is released, it continues to move in the direction it was moving at the moment of release. At the topmost point, the stone moves tangentially upward.

4 m/s²

5 m/s²

10 m/s²

20 m/s²

Correct Answer: A

Solution:

Using the equation

v = u + at

, where

v = 20

m/s,

u = 0

m/s, and

t = 5

s, we find

a = \frac{v - u}{t} = \frac{20}{5} = 4

m/s².

6 km

14 km

10 km

4 km

Correct Answer: A

Solution:

The displacement is the straight-line distance from the initial to the final position. It is 10 km - 4 km = 6 km.

Equal to the circumference of the circle

Zero

Equal to the diameter of the circle

Equal to the radius of the circle

Correct Answer: B

Solution:

Displacement is the shortest distance from the initial to the final position. Since the cyclist returns to the starting point, the displacement is zero.

48 m

96 m

72 m

144 m

Correct Answer: B

Solution:

Using the equation of motion:

s = ut + \frac{1}{2}at^2

, where

u = 0

a = 3.0\, m/s^2

t = 8.0\, s

. Thus,

s = 0 + \frac{1}{2} \times 3.0 \times 8.0^2 = 96\, m

75 km

150 km

300 km

0 km

Correct Answer: A

Solution:

The displacement being 0 km indicates that the car returned to its starting point. Therefore, the distance between X and Y is half of the total distance covered: 150 km / 2 = 75 km.

15 m/s

10 m/s

5 m/s

20 m/s

Correct Answer: A

Solution:

Using the equation

v = u + at

, where initial velocity

u = 0

a = 3 \text{ m/s}^2

, and

t = 5 \text{ s}

, we get

v = 0 + 3 \times 5 = 15 \text{ m/s}

96 m

48 m

24 m

12 m

Correct Answer: A

Solution:

Using the equation

s = ut + \frac{1}{2} at^2

, where

u = 0

a = 3.0 \text{ m/s}^2

, and

t = 8 \text{ s}

, we find

s = 0 + \frac{1}{2} \times 3.0 \times 8^2 = 96 \text{ m}

The object covers equal distances in unequal intervals of time.

The object covers unequal distances in equal intervals of time.

The object covers equal distances in equal intervals of time.

The object does not move at all.

Correct Answer: C

Solution:

In uniform motion, an object covers equal distances in equal intervals of time.

Along the circular path

Towards the center of the circle

Tangential to the circular path

Opposite to the direction of motion

Correct Answer: C

Solution:

When the string is released, the stone will continue to move in the direction it was moving at the instant of release, which is tangential to the circular path.

5.24 m/s

3.14 m/s

6.28 m/s

10.47 m/s

Correct Answer: A

Solution:

The circumference of the circular track is given by

2\pi r = 2\times3.14\times100 = 628

m. The time taken to complete one revolution is 2 minutes or 120 seconds. Therefore, the average speed is

\frac{628}{120} \approx 5.24

m/s.

60 km

120 km

0 km

30 km

Correct Answer: B

Solution:

The car travels 60 km to point B and then 60 km back to point A, making the total distance covered 120 km.

200 km

100 km

0 km

50 km

Correct Answer: C

Solution:

The train returns to its starting point, so the displacement, which is the shortest distance from the initial to the final position, is zero.

0 m

100 m

141.42 m

200 m

Correct Answer: B

Solution:

The athlete's final position forms a right triangle with the starting point. The displacement is the hypotenuse, calculated as

\sqrt{(100)^2 + (100)^2} = 141.42

30 km

60 km

120 km

180 km

Correct Answer: C

Solution:

The total distance covered is speed multiplied by time: 60 km/h × 2 h = 120 km.

216 m

432 m

288 m

144 m

Correct Answer: A

Solution:

Using the equation

s = ut + \frac{1}{2}at^2

, where

u = 0

a = 3 \text{ m/s}^2

, and

t = 12 \text{ s}

, we find

s = 0 + \frac{1}{2} \times 3 \times (12)^2 = 216 \text{ m}

50 km/h

66.67 km/h

33.33 km/h

100 km/h

Correct Answer: B

Solution:

The total distance covered by the car is 100 km (to Y) + 100 km (back to X) = 200 km. The total time taken is 3 hours. Therefore, the average speed is

\frac{200 \text{ km}}{3 \text{ h}} = 66.67 \text{ km/h}

30 km/h

60 km/h

90 km/h

120 km/h

Correct Answer: B

Solution:

The total distance covered by the car is 60 km (A to B) + 60 km (B to A) = 120 km. The total time taken is 2 hours. Therefore, the average speed is

\frac{120 \text{ km}}{2 \text{ h}} = 60 \text{ km/h}

50 m

100 m

25 m

75 m

Correct Answer: B

Solution:

Using the equation

s = ut + \frac{1}{2}at^2

, where

u = 0

a = \frac{20}{5} = 4

m/s², and

t = 5

s, we find

s = 0 \times 5 + \frac{1}{2} \times 4 \times 5^2 = 50

85 km

95 km

60 km

25 km

Correct Answer: A

Solution:

The total distance covered is the sum of the distances from O to A and A to B, which is 60 km + 25 km = 85 km.

200 m

400 m

100 m

314 m

Correct Answer: D

Solution:

The distance covered is the circumference of the circle:

\pi \times 200 \approx 314 \text{ m}

60 km east

20 km east

120 km east

20 km west

Correct Answer: A

Solution:

The total displacement is the net change in position. Starting from the initial point, the car moves 40 km east, then 30 km west, resulting in a net eastward displacement of 10 km. Finally, it moves 50 km east, resulting in a total displacement of 60 km east.

36.1 m

72.2 m

18.05 m

25.0 m

Correct Answer: A

Solution:

First, convert 52 km/h to m/s:

52 \times \frac{1000}{3600} = 14.44\, m/s

. Using the equation

v^2 = u^2 + 2as

, where

v = 0

u = 14.44\, m/s

a = \frac{-14.44}{5}

. Solving for

s

s = \frac{14.44^2}{2 \times \frac{-14.44}{5}} = 36.1\, m

0 meters

200 meters

100 meters

50 meters

Correct Answer: A

Solution:

Since the runner completes a full circle and returns to the starting point, the displacement is 0 meters.

20 m/s

10 m/s

15 m/s

25 m/s

Correct Answer: A

Solution:

Using the equation

v = u + at

, where initial velocity

u = 0

, acceleration

a = 4 \text{ m/s}^2

, and time

t = 5 \text{ s}

, we find

v = 0 + 4 \times 5 = 20 \text{ m/s}

100 m

200 m

300 m

400 m

Correct Answer: B

Solution:

The distance covered by the car can be calculated using the equation

s = ut + \frac{1}{2} at^2

. Here, initial velocity

u = 0

, acceleration

a = \frac{v-u}{t} = \frac{20-0}{10} = 2 \text{ m/s}^2

, and time

t = 10 \text{ s}

. Thus,

s = 0 \times 10 + \frac{1}{2} \times 2 \times 10^2 = 100 \text{ m}

. Therefore, the correct answer is 200 m.

The cyclist has zero acceleration.

The cyclist's velocity is constant.

The cyclist is in uniform circular motion with centripetal acceleration.

The cyclist's displacement is zero after one complete round.

Correct Answer: C

Solution:

In uniform circular motion, the speed is constant, but the direction of velocity changes, resulting in centripetal acceleration towards the center of the circle.

11 m/s

9 m/s

6 m/s

5 m/s

Correct Answer: A

Solution:

Using the equation

v = u + at

, where

u = 5

m/s,

a = 2

m/s², and

t = 3

s, we find

v = 5 + 2 \times 3 = 11

m/s.

100 m

200 m

0 m

314 m

Correct Answer: B

Solution:

The displacement is the straight line distance between the starting point and the endpoint, which is the diameter of the circle: 2 × 100 m = 200 m.

Distance: 1000 m, Displacement: 0 m

Distance: 1200 m, Displacement: 200 m

Distance: 1400 m, Displacement: 0 m

Distance: 1200 m, Displacement: 400 m

Correct Answer: A

Solution:

In 2 minutes 20 seconds (140 seconds), the athlete completes 3.5 rounds (since 140/40 = 3.5). The total distance covered is 3.5 times the circumference of the circle (

\pi \times 200

m), which equals 1000 m. Since the athlete ends at the starting point after completing full rounds, the displacement is 0 m.

50 km/h

40 km/h

60 km/h

30 km/h

Correct Answer: A

Solution:

Average speed is calculated as total distance divided by total time. Here, it is

\frac{400 \text{ km}}{8 \text{ h}} = 50 \text{ km/h}

200 m

314 m

400 m

628 m

Correct Answer: B

Solution:

The distance covered in one lap around a circular track is the circumference of the circle, which is calculated as

\pi \times \text{diameter} = 3.14 \times 200 = 628 \text{ m}

80 km

100 km

140 km

200 km

Correct Answer: A

Solution:

The car's path forms a right-angled triangle with legs of 80 km and 60 km. The displacement is the hypotenuse:

\sqrt{80^2 + 60^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100

km.

100 km

20 km

60 km

40 km

Correct Answer: A

Solution:

The total distance covered is the sum of the distances traveled in each direction: 60 km + 40 km = 100 km.

10 km

0 km

5 km

2.5 km

Correct Answer: B

Solution:

The displacement is the shortest distance from the initial to the final position. Since the person returns to the starting point, the displacement is 0 km.

It will continue in a circular path.

It will move radially inward.

It will move tangentially to the circular path.

It will move radially outward.

Correct Answer: C

Solution:

When released, the stone moves tangentially to the circular path due to inertia.

90 km

180 km

45 km

360 km

Correct Answer: B

Solution:

Distance = Speed × Time = 90 km/h × 2 h = 180 km.

0 m

100 m

314 m

157 m

Correct Answer: A

Solution:

Displacement is the shortest distance from the initial to the final position. After one full circle, the cyclist returns to the starting point, so the displacement is 0 m.

20 km

140 km

60 km

80 km

Correct Answer: A

Solution:

The displacement is the shortest distance from the initial to the final position, which is 80 km - 60 km = 20 km south.

3.14 m/s

6.28 m/s

4.71 m/s

9.42 m/s

Correct Answer: C

Solution:

The circumference of the circular track is

\pi \times 150 \text{ m} = 150\pi \text{ m}

. The total distance covered in two rounds is

2 \times 150\pi \text{ m} = 300\pi \text{ m}

. The time taken is 5 minutes or 300 seconds. Therefore, the average speed is

\frac{300\pi \text{ m}}{300 \text{ s}} = \pi \text{ m/s} \approx 4.71 \text{ m/s}

40 km north

40 km south

160 km north

160 km south

Correct Answer: A

Solution:

Displacement is the shortest distance from the initial to the final position. The car ends up 40 km north of point A.

True or False

Correct Answer: False

Solution:

The total path length can be greater than the magnitude of displacement, as displacement is the shortest distance between the initial and final positions.

Correct Answer: True

Solution:

Uniform motion is defined as motion where an object covers equal distances in equal intervals of time.

Correct Answer: True

Solution:

An object with constant acceleration can momentarily have zero velocity, such as when changing direction.

Correct Answer: True

Solution:

An object can have a constant acceleration, such as an object thrown upwards, which momentarily has zero velocity at its highest point before descending.

Correct Answer: True

Solution:

Displacement is the shortest distance from the initial to the final position of an object. If an object returns to its starting point, the displacement is zero, even though it has traveled a distance.

Correct Answer: True

Solution:

If an object returns to its initial position after traveling, its displacement is zero, but the distance traveled is not.

Correct Answer: True

Solution:

The perception of motion depends on the observer's frame of reference. For example, passengers in a moving bus see roadside trees moving backward, while a person standing outside sees the bus and its passengers moving.

Correct Answer: False

Solution:

Velocity is the displacement per unit time, not the distance covered.

Correct Answer: False

Solution:

Velocity is defined as the displacement per unit time, whereas speed is the distance covered per unit time.

Correct Answer: False

Solution:

We do not directly perceive the motion of the earth as it moves smoothly and at a constant speed, which does not provide any sensory cues like acceleration or changes in velocity.

Correct Answer: True

Solution:

Displacement is the shortest distance between the initial and final positions of an object. It can be zero if the object returns to its starting point, even if it has traveled a distance.

Correct Answer: False

Solution:

Displacement is the shortest distance between the initial and final positions of an object and can be less than the distance traveled, which is the total path length.

Correct Answer: True

Solution:

Motion is a change of position and can be described in terms of the distance moved or the displacement.

Correct Answer: False

Solution:

In uniform circular motion, the speed is constant, but the direction of velocity changes continuously, so the velocity is not constant.

Correct Answer: True

Solution:

An object moving in a circular path can have uniform speed but changing velocity due to the continuous change in direction.

Correct Answer: True

Solution:

The area under a velocity-time graph gives the displacement of the object. If the velocity is constant, this area represents the distance travelled.

Correct Answer: True

Solution:

Speed is a scalar quantity that measures how much distance an object covers per unit time, whereas velocity is a vector quantity that measures displacement per unit time.

Correct Answer: True

Solution:

An object can have zero velocity at an instant while accelerating, such as when it changes direction. For example, a ball thrown upwards has zero velocity at its peak before descending.

Correct Answer: False

Solution:

The motion of the Earth is not directly perceived; it is inferred through phenomena like sunrise and sunset.

Correct Answer: True

Solution:

Uniform circular motion is defined as motion in a circular path with constant speed.

Correct Answer: True

Solution:

In uniform circular motion, the object moves at a constant speed along a circular path. However, its velocity changes because the direction of motion is continuously changing.

Correct Answer: False

Solution:

Speed is a scalar quantity representing how fast an object is moving, while velocity is a vector quantity that includes direction. They can differ if the direction of motion changes.

Correct Answer: True

Solution:

Speed is defined as the distance covered per unit time, which is a scalar quantity.

Correct Answer: True

Solution:

Uniform circular motion refers to an object moving in a circular path at a constant speed, although its velocity changes due to the change in direction.

Correct Answer: False

Solution:

The speed of light is much faster than the speed of sound. Light travels at approximately

3 \times 10^8

m/s, while sound travels at about 346 m/s in air.

Correct Answer: True

Solution:

In uniform circular motion, while the speed remains constant, the direction of motion continuously changes, which is why the object moves along a circular path.

Correct Answer: False

Solution:

The magnitude of displacement can be less than or equal to the distance travelled, but not always equal.

Correct Answer: True

Solution:

Displacement is the shortest path between two points, while distance is the total path covered. Therefore, displacement can never be greater than the distance.

Correct Answer: False

Solution:

Displacement is the shortest distance from the initial to the final position of an object and can be different from the total distance traveled, which is the path length.

Correct Answer: True

Solution:

To describe the position of an object, a reference point called the origin is needed. This helps in specifying the location of the object.

Motion

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CBSE Learning Objectives – Key Concepts & Skills You Must Know

Learning Objectives

CBSE Revision Notes & Quick Summary for Last-Minute Study

Chapter 7: Motion

7.1 Describing Motion

Types of Motion

7.2 Measuring the Rate of Motion

Key Equations

7.3 Graphical Representation of Motion

Example Data

7.4 Circular Motion

7.5 Applications of Motion

7.6 Exercises

Conclusion

CBSE Exam Tips, Important Questions & Common Mistakes to Avoid

Common Mistakes and Exam Tips

Common Pitfalls

Tips for Exam Preparation

Multiple Choice Questions

Q1.An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the Earth.

Correct Answer: B

Q2.A car travels from point X to Y, a distance of 40 km, and then from Y to Z, a distance of 30 km. What is the total distance covered by the car?

Correct Answer: A

Q3.A cyclist travels 5 km east, 3 km north, and then 5 km west. What is the magnitude of the cyclist's displacement?

Correct Answer: A

Q4.Which of the following statements is true about motion?

Correct Answer: B

Q5.A motorboat starting from rest accelerates uniformly at 2 m/s² for 10 seconds. How far does the boat travel during this time?

Correct Answer: B

Q6.If a stone is tied to a string and swung in a circular path, what happens when the string is released?

Correct Answer: B

Q7.If a person walks 3 km east and then 4 km north, what is the total distance traveled?

Correct Answer: B

Q8.An object moves 10 m north, then 10 m south. What is the magnitude of its displacement?

Correct Answer: A

Q9.An object moves along a straight line with a constant acceleration of 3 m/s². If its initial velocity is 2 m/s, what will be its velocity after 4 seconds?

Correct Answer: C

Q10.An athlete runs around a circular track of diameter 100 m. If the athlete completes two rounds in 4 minutes, what is the average speed of the athlete?

Correct Answer: B

Q11.A motorboat starting from rest accelerates uniformly at 3 m/s² for 8 seconds. How far does the boat travel during this time?

Correct Answer: A

Q12.A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m/s², with what velocity will it strike the ground?

Correct Answer: B

Q13.A car travels from point A to point B and back to point A along a straight path. If the total distance covered is 120 km and the displacement is 0 km, what is the distance between points A and B?

Correct Answer: A

Q14.An object moves along a straight line with a constant acceleration of 4 m/s². If its initial velocity is 5 m/s, what will be its velocity after 5 seconds?

Correct Answer: B

Q15.An athlete runs around a circular track of radius 100 m. If the athlete completes one and a half rounds, what is the total distance covered and the displacement?

Correct Answer: C

Q16.A car travels from point O to point A, covering a distance of 60 km, and then returns to point B, which is 25 km from point O. What is the magnitude of the car's displacement?

Correct Answer: D

Q17.A satellite is moving in a circular orbit around the Earth with a radius of 42,250 km. If it takes 24 hours to complete one revolution, what is its speed?

Correct Answer: A

Q18.A car accelerates uniformly from rest to a speed of 20 m/s in 10 seconds. What is the distance covered by the car during this time?

Correct Answer: A

Q19.An athlete runs around a circular track of diameter 200 m. What is the displacement after completing one full round?

Correct Answer: C

Q20.Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What is his average speed from A to C?

Correct Answer: B

Q21.A runner completes a 400 m circular track in 2 minutes. What is the displacement of the runner?

Correct Answer: C

Q22.Which of the following statements is true about displacement?

Correct Answer: B

Q23.A car moves 30 km north, 40 km east, and then 30 km south. What is the total distance traveled by the car?

Correct Answer: B

Q24.If a person walks 10 meters east and then 10 meters west, what is the magnitude of the displacement?

Correct Answer: A

Q25.A car travels 60 km north and then 40 km south. What is the total distance covered by the car?

Correct Answer: A

Q26.Consider a stone tied to a string being swung in a circular path. If the string is suddenly released when the stone is at the topmost point of the circle, in which direction will the stone move immediately after release?

Correct Answer: A

Q27.A car accelerates uniformly from rest to a speed of 20 m/s in 5 seconds. What is the acceleration of the car?

Correct Answer: A

Q28.A cyclist travels in a straight line for 10 km, turns back, and travels 4 km. What is the magnitude of the displacement?

Correct Answer: A

Q29.If a cyclist travels in a circular path and returns to the starting point, what is the displacement?

Correct Answer: B

Q30.A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m/s² for 8.0 seconds. How far does the boat travel during this time?