Question 1

Question 1An 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?

Accepted Answer

Given,Diameter of track (d) = 200 mCircumference of the track = πd = 200π mSo, distance covered in 40 seconds = 200π mDistance covered in 1 sec = 200π40\dfrac{200π}{40}40200π​Distance covered in 2 min 20 sec (i.e., 60 + 60 + 20 = 140 sec) = 200π40\dfrac{200π}{40}40200π​ x 140= 200×22×1407×40\dfrac{200 	imes 22 	imes 140}{7 	imes 40}7×40200×22×140​= 2200 mNumber of rounds completed in 140 sec = 14040\dfrac{140}{40}40140​ = 3.5∴ the final position (with respect to the initial position) is at the opposite end of the circular track. So, the net displacement will be equal to the diameter of the track = 200 m.Hence, the net distance covered by the athlete is 2200 m and the total displacement of the athlete is 200 m.

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 7: MotionWeb Content &mdash; Quick Reference
  
    Chapter 7: Motion &mdash; Quick Reference

distance
      displacement
      speed
      velocity
      acceleration
      equations of motion
      graphs
      circular motion

Quick Revision Points
      
        Distance (scalar, total path) vs Displacement (vector, shortest path)
Speed = distance/time (scalar); Velocity = displacement/time (vector)
Acceleration a = (v−u)/t; unit: m/s²
Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch07-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 2

Question 3Abdul, while driving to school, computes the average speed for his trip to be 20 kmh-1. On his return trip along the same route, there is less traffic and the average speed is 30 kmh-1. What is the average speed for Abdul's trip?

Accepted Answer

Distance travelled to reach the school = distance travelled to reach home = dTime taken to reach school = t1Time taken to reach home = t2Average speed = total distance travelledtotal time taken\dfrac{	ext{total distance travelled}}{	ext{total time taken}}total time takentotal distance travelled​On going to school :20 = dt1\dfrac{	ext{d}}{	ext{t}_1}t1​d​ ort1 = d20\dfrac{	ext{d}}{20}20d​On return trip :30 = dt2\dfrac{	ext{d}}{	ext{t}_2}t2​d​ ort2 = d30\dfrac{	ext{d}}{30}30d​Now, the average speed for the entire trip= d+dt1+t2\dfrac{	ext{d+d}}{	ext{t}_1 + 	ext{t}_2}t1​+t2​d+d​= 2dd20+d30\dfrac{	ext{2d}}{\dfrac{	ext{d}}{20} + \dfrac{	ext{d}}{30}}20d​+30d​2d​= 2d3d + 2d60\dfrac{	ext{2d}}{\dfrac{	ext{3d + 2d}}{60}}603d + 2d​2d​= 2d×605d\dfrac{	ext{2d} 	imes {60}}{5	ext{d}}5d2d×60​= 24 kmh-1∴ Abdul's average speed for the entire trip is 24 kmh-1.

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 7: MotionWeb Content &mdash; Quick Reference
  
    Chapter 7: Motion &mdash; Quick Reference

distance
      displacement
      speed
      velocity
      acceleration
      equations of motion
      graphs
      circular motion

Quick Revision Points
      
        Distance (scalar, total path) vs Displacement (vector, shortest path)
Speed = distance/time (scalar); Velocity = displacement/time (vector)
Acceleration a = (v−u)/t; unit: m/s²
Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch07-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 3

Question 4A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms-2 for 8.0 s. How far does the boat travel during this time?

Accepted Answer

Given,Initial velocity (u) = 0Acceleration (a) = 3 ms-2Time period (t) = 8 sDistance travelled (S) = ?According to the second equation of motion,S = ut + 12\dfrac{1}{2}21​at2Substituting we get,S = 0 + 12\dfrac{1}{2}21​ x 3 x 8 x 8= 96 m∴ The motorboat travels a distance of 96 m

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 7: MotionWeb Content &mdash; Quick Reference
  
    Chapter 7: Motion &mdash; Quick Reference

distance
      displacement
      speed
      velocity
      acceleration
      equations of motion
      graphs
      circular motion

Quick Revision Points
      
        Distance (scalar, total path) vs Displacement (vector, shortest path)
Speed = distance/time (scalar); Velocity = displacement/time (vector)
Acceleration a = (v−u)/t; unit: m/s²
Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch07-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 4

Question 6Fig below shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:(a) Which of the three is travelling the fastest?(b) Are all three ever at the same point on the road?(c) How far has C travelled when B passes A?(d) How far has B travelled by the time it passes C?

Accepted Answer

(a) As the slope of line B is the greatest, hence, B is travelling at the fastest speed.(b) No, as the three lines do not intersect at a single point.(c) As there are 7 unit areas of the graph between 0 and 4 on the Y axis, 1 graph unit equals 47\dfrac{4}{7}74​ km.As the initial point of object C is 4 graph units away from the origin, its initial distance from the origin is 4 x 47\dfrac{4}{7}74​ km = 167\dfrac{16}{7}716​ kmWhen B passes A, the distance between the origin and C is 8 km.∴ total distance travelled by C in this time = 8 – 167\dfrac{16}{7}716​ km = 5.71 km(d) Total distance travelled by B when it crosses C = 9 x 47\dfrac{4}{7}74​ = 5.14 km

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 7: MotionWeb Content &mdash; Quick Reference
  
    Chapter 7: Motion &mdash; Quick Reference

distance
      displacement
      speed
      velocity
      acceleration
      equations of motion
      graphs
      circular motion

Quick Revision Points
      
        Distance (scalar, total path) vs Displacement (vector, shortest path)
Speed = distance/time (scalar); Velocity = displacement/time (vector)
Acceleration a = (v−u)/t; unit: m/s²
Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch07-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 5

Question 7A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 ms-2, with what velocity will it strike the ground? After what time will it strike the ground?

Accepted Answer

Given,Initial velocity (u) = 0Distance travelled (s) = 20 mAcceleration (a) = 10 ms-2As per the third law of motion,v2 – u2 = 2asSubstituting we get,v2 - 0 = 2 x 10 x 20⇒ v2 = 400⇒ v = 400\sqrt{400}400​⇒ v = 20 ms-1Hence, ball strikes the ground with a velocity of 20 ms-1As per the first equation of motion,v = u + atort = v - ua\dfrac{	ext{v - u}}{	ext{a}}av - u​Substituting we get,t = 20−010\dfrac{20-0}{10}1020−0​ = 2 seconds∴ the ball reaches the ground after 2 seconds.

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 7: MotionWeb Content &mdash; Quick Reference
  
    Chapter 7: Motion &mdash; Quick Reference

distance
      displacement
      speed
      velocity
      acceleration
      equations of motion
      graphs
      circular motion

Quick Revision Points
      
        Distance (scalar, total path) vs Displacement (vector, shortest path)
Speed = distance/time (scalar); Velocity = displacement/time (vector)
Acceleration a = (v−u)/t; unit: m/s²
Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch07-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 6

Question 8The speed-time graph for a car is shown in Fig. below(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.(b) Which part of the graph represents uniform motion of the car?

Accepted Answer

(a) Distance travelled by the car in the first 4 seconds = area under the slope of the speed-time graph.To find this area, we will use the square counting method.On time axis (i.e., x axis) :5 squares = 2 units∴ 1 square = 25\dfrac{2}{5}52​ unitsOn speed axis (i.e., y axis) :3 squares = 2 units∴ 1 square = 23\dfrac{2}{3}32​ unitsArea of each square = 25\dfrac{2}{5}52​ x 23\dfrac{2}{3}32​ = 415\dfrac{4}{15}154​ sq units.We have fully filled squares, more than half filled squares, half filled squares and less than half filled squares in this area.Area of fully filled square = Area of 1 square = 415\dfrac{4}{15}154​ sq unitsArea of more than half filled square = Area of 1 square = 415\dfrac{4}{15}154​ sq unitsArea of half filled square = 12\dfrac{1}{2}21​ x Area of 1 square = 12×415\dfrac{1}{2} 	imes \dfrac{4}{15}21​×154​ = 215\dfrac{2}{15}152​ sq unitsArea of less than half filled square = 0No. of fully filled squares = 56 [Marked in graph with 'F']Area of fully filled squares = 56 x 415\dfrac{4}{15}154​ = 14.93 sq unitsNo. of more than half filled squares = 4 [Marked in graph with 'M']Area of more than half filled squares = 4 x 415\dfrac{4}{15}154​ = 1.06 sq unitsNo. of half filled squares = 3 [Marked in graph with 'H']Area of half filled squares = 3 x 415\dfrac{4}{15}154​ = 0.8 sq unitsNo. of less than half filled squares = 4 [Marked in graph with 'L']Area of less than half filled squares = 4 x 0 = 0 sq units∴ Total Area = 14.93 + 1.06 + 0.4 = 16.39 sq units∴ Total distance covered in first 4 seconds = Area under speed-time graph = 16.4 mHence, the car travels 16.4 m in first 4 seconds(b) From graph, the speed of the car does not change from the points (x = 6) to (x = 10). Hence, the car is said to be in uniform motion from the 6th to the 10th second.

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 7: MotionWeb Content &mdash; Quick Reference
  
    Chapter 7: Motion &mdash; Quick Reference

distance
      displacement
      speed
      velocity
      acceleration
      equations of motion
      graphs
      circular motion

Quick Revision Points
      
        Distance (scalar, total path) vs Displacement (vector, shortest path)
Speed = distance/time (scalar); Velocity = displacement/time (vector)
Acceleration a = (v−u)/t; unit: m/s²
Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch07-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 7

Question 9State which of the following situations are possible and give an example for each of these:(a) an object with a constant acceleration but with zero velocity(b) an object moving with an acceleration but with uniform speed.(c) an object moving in a certain direction with an acceleration in the perpendicular direction.

Accepted Answer

(a) It is possible;When an object is thrown up vertically into the air, it has constant acceleration due to gravity acting on it. However, when it reaches its maximum height, its velocity is zero.(b) It is possible;When an object moves on a circular path with uniform speed it is still under acceleration because the velocity changes due to continuous change in the direction of motion.(c) It is possible;When an object is accelerating in a circular trajectory, the acceleration is in the perpendicular direction.

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 7: MotionWeb Content &mdash; Quick Reference
  
    Chapter 7: Motion &mdash; Quick Reference

distance
      displacement
      speed
      velocity
      acceleration
      equations of motion
      graphs
      circular motion

Quick Revision Points
      
        Distance (scalar, total path) vs Displacement (vector, shortest path)
Speed = distance/time (scalar); Velocity = displacement/time (vector)
Acceleration a = (v−u)/t; unit: m/s²
Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch07-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 8

Question 10An 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.

Accepted Answer

Given,Radius of the orbit (r) = 42250 km∴ circumference of the orbit= 2 x 227\dfrac{22}{7}722​ x r= 2 x 227\dfrac{22}{7}722​ x 42250= 265571.42 kmTime taken to complete one revolution in the orbit = 24 hoursConvert hours into sec1 hour = 3600 secSo, 24 hour = 24 x 3600 secSpeed = DistanceTime\dfrac{	ext{Distance}}{	ext{Time}}TimeDistance​Substituting we get,Speed = 265571.4224×3600\dfrac{265571.42}{24 	imes 3600}24×3600265571.42​ = 3.07 kms-1∴ Speed of the artificial satellite = 3.07 kms-1.

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    .keyword-tags { display: flex; flex-wrap: wrap; gap: 6px; margin: 10px 0; }
    .keyword-tags span { background: var(--bt-primary-light); color: var(--bt-primary-dark); padding: 3px 10px; border-radius: 12px; font-size: 0.78rem; font-weight: 600; }

BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 7: MotionWeb Content &mdash; Quick Reference
  
    Chapter 7: Motion &mdash; Quick Reference

distance
      displacement
      speed
      velocity
      acceleration
      equations of motion
      graphs
      circular motion

Quick Revision Points
      
        Distance (scalar, total path) vs Displacement (vector, shortest path)
Speed = distance/time (scalar); Velocity = displacement/time (vector)
Acceleration a = (v−u)/t; unit: m/s²
Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch07-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 9

Question 2A 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?

Accepted Answer

Given,Time period of travel = 2 minutes 20 secondsConvert min into sec1 min = 60 secSo 2 min 20 sec = (2 x 60) sec + 20 sec = 120 + 20 = 140 secSide of the given square field = 10 mHence, the perimeter of a square = 4 x side = 4 x 10 = 40 mTime taken by the farmer to cover the boundary of 40 m = 40 sSo, in 1 s, the farmer covers a distance of = 4040\dfrac{40}{40}4040​ x 1 = 1 mHence, distance covered by the farmer in 140 sec = 140 mThe total number of rotations taken by the farmer to cover a distance of 140 m= total distanceperimeter\dfrac{\text{total distance}}{\text{perimeter}}perimetertotal distance​= 14040\dfrac{140}{40}40140​= 3.5If the farmer starts from point A, after 3.5 rounds he will be at point C of the field.∴ Displacement (from Pythagoras theorem)AC=(AB)2+(BC)2=(10)2+(10)2=100+100=200=102=10×1.414=14.14 m\text{AC} = \sqrt{(\text{AB})^2 + (\text{BC})^2} \[1em] = \sqrt{(10)^2 + (10)^2} \[1em] = \sqrt{100 + 100} \[1em] = \sqrt{200} \[1em] = 10\sqrt{2} \[1em] = 10 \times 1.414 \[1em] = 14.14 \text{ m}AC=(AB)2+(BC)2​=(10)2+(10)2​=100+100​=200​=102​=10×1.414=14.14 mHence, the magnitude of displacement = 14.14 m

Web Content: Motion | Bright Tutorials

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 7: MotionWeb Content &mdash; Quick Reference
  
    Chapter 7: Motion &mdash; Quick Reference

distance
      displacement
      speed
      velocity
      acceleration
      equations of motion
      graphs
      circular motion

Quick Revision Points
      
        Distance (scalar, total path) vs Displacement (vector, shortest path)
Speed = distance/time (scalar); Velocity = displacement/time (vector)
Acceleration a = (v−u)/t; unit: m/s²
Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch07-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 10

Question 5A stone is thrown in a vertically upward direction with a velocity of 5 ms-1. If the acceleration of the stone during its motion is 10 ms-2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Accepted Answer

Given, initial velocity (u) = 5 ms-1Terminal velocity (v) = 0Acceleration = -10 ms-2 (as it is in upward direction)According to the third equation of motion,v2 – u2 = 2as⇒s=v2−u22a\Rightarrow 	ext{s} = \dfrac{	ext{v}^2 - 	ext{u}^2}{2	ext{a}}⇒s=2av2−u2​Substituting we get,s = 0−522×−10=−25−20\dfrac{0 - {5}^2 }{2 	imes -10} = \dfrac{-25}{-20}2×−100−52​=−20−25​ = 1.25 m.Hence, distance covered = 1.25 mAccording to the first equation of motion,v = u + at∴ time taken by the stone to reach a position of rest (maximum height) = v−ua\dfrac{	ext{v} - 	ext{u}}{	ext{a}}av−u​Substituting we get,0−5−10\dfrac{0-5}{-10}−100−5​ = 0.5 sec∴ The stone reaches a maximum height of 1.25 m in 0.5 sec.

Web Content: Motion | Bright Tutorials

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 7: MotionWeb Content &mdash; Quick Reference
  
    Chapter 7: Motion &mdash; Quick Reference

distance
      displacement
      speed
      velocity
      acceleration
      equations of motion
      graphs
      circular motion

Quick Revision Points
      
        Distance (scalar, total path) vs Displacement (vector, shortest path)
Speed = distance/time (scalar); Velocity = displacement/time (vector)
Acceleration a = (v−u)/t; unit: m/s²
Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch07-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Motion

Exercises

Intext Questions 1

Intext Questions 5