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In the example of an athlete running back and forth on a straight track (Fig. 4.4), when will the displacement of the athlete be zero? What will be the total distance travelled in that case?
The displacement of the athlete will be zero when she returns to her starting point O, because her initial and final positions are the same. She runs from O to A, covering 100 m and then returns from A to O, covering another 100 m. Therefore, the total distance travelled is 100 + 100 = 200 m, whileRead more
The displacement of the athlete will be zero when she returns to her starting point O, because her initial and final positions are the same. She runs from O to A, covering 100 m and then returns from A to O, covering another 100 m. Therefore, the total distance travelled is 100 + 100 = 200 m, while the displacement is 0 m.
For more NCERT Solutions of Class 9 Science Exploration Chapter 4 Describing Motion Around Us Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-4/
See lessMy father went to a shop from home which is located at a distance of 250 m on a straight road. On reaching there, he discovered that he forgot to carry a cloth bag. He came home to take it, went to the shop again, bought provisions and came back home. How much was the total distance travelled by him? What was his displacement from home?
The shop is 250 m from his home. He travelled from home to the shop, returned home to collect the bag, went to the shop again and finally returned home. Thus, total distance travelled = 250 × 4 = 1000 m. Since his final position is the same as his starting position, his displacement from home is 0 mRead more
The shop is 250 m from his home. He travelled from home to the shop, returned home to collect the bag, went to the shop again and finally returned home. Thus, total distance travelled = 250 × 4 = 1000 m. Since his final position is the same as his starting position, his displacement from home is 0 m.
For more NCERT Solutions of Class 9 Science Exploration Chapter 4 Describing Motion Around Us Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-4/
See lessA student runs from the ground floor to the fourth floor of a school building to collect a book and then comes down to their classroom on the second floor. If the height of each floor is 3 m, find: (i) the total vertical distance travelled and (ii) their displacement from the starting point.
The height of each floor is 3 m. From the ground floor to the fourth floor, the student travels 4 × 3 = 12 m upward. From the fourth floor to the second floor, they travel 2 × 3 = 6 m downward. Therefore, total vertical distance travelled = 12 + 6 = 18 m. The final position is 6 m above the startingRead more
The height of each floor is 3 m. From the ground floor to the fourth floor, the student travels 4 × 3 = 12 m upward. From the fourth floor to the second floor, they travel 2 × 3 = 6 m downward. Therefore, total vertical distance travelled = 12 + 6 = 18 m. The final position is 6 m above the starting point, so displacement = 6 m upward.
For more NCERT Solutions of Class 9 Science Exploration Chapter 4 Describing Motion Around Us Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-4/
See lessDownload Class 9 Science Exploration Notes for First Term Exam (2026-27)
Download Class 9 Science Exploration Notes for First Term Exam 2026-27, along with NCERT Solutions PDF, question answers, important questions, key concepts, definitions, formulas, diagrams and exam-focused revision material. These Class 9 Science notes are based on the new Exploration textbook introRead more
Download Class 9 Science Exploration Notes for First Term Exam 2026-27, along with NCERT Solutions PDF, question answers, important questions, key concepts, definitions, formulas, diagrams and exam-focused revision material. These Class 9 Science notes are based on the new Exploration textbook introduced for the 2026-27 academic session and cover Physics, Chemistry, Biology and Earth Science topics. Use the PDF notes and NCERT solutions for revision, homework, concept clarity and first-term exam preparation. Download and study anytime.
Pdf Download Class 9 Science Exploration Notes
https://www.tiwariacademy.com/ncert-solutions/class-9/science/
See lessHow would you use the following figure to justify the statement that the sum of the opposite angles of a cyclic quadrilateral is 180°?
In the figure, OA, OB, OC and OD are equal radii. Therefore, the triangles formed by these radii are isosceles. Since BOD is a diameter, the angles around the centre give: p + v = 90° and similarly, q + u = 90°. Now, ∠A = p + v = 90° and ∠C = q + u = 90° in the corresponding construction, so the oppRead more
In the figure, OA, OB, OC and OD are equal radii. Therefore, the triangles formed by these radii are isosceles. Since BOD is a diameter, the angles around the centre give:
p + v = 90°
and similarly,
q + u = 90°.
Now, ∠A = p + v = 90° and ∠C = q + u = 90° in the corresponding construction, so the opposite angles together give 180°. Thus, the sum of the opposite angles of a cyclic quadrilateral is 180°.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-5/
See less