PTB 9th Class Physics Unit 2 “Kinematics” Solved Questions and Numericals Problems

View and Download the solved questions, solved numerical problems or 9th Class Physics notes of Chapter 2 “Kinematics” of Punjab Textbook Board. These 9th class physics notes of 2nd Chapter are according to the new syllabus (2013) of Punjab Boards including Lahore Board, Gujranwala Board, Sahiwal Board, Rawalpindi Board, Faisalabad Board, Multan Board, Bahawalpur Board, Federal Board, DG Khan Board and Sargodha Board. Students of 9th Class of all these board can facilitate from the physics notes given below. You can also download 9th Class Physics Chapter 2 Solved Questions and Numericals Problems.

Unit 2: Kinematics

After studying this unit, the students will be able to: Describe using examples how objects can be at rest and in motion simultaneously. Identify different types of motion i.e. translatory, (linear, random, and circular); rotatory and vibratory motions and distinguish among them. Differentiate with examples between distance and displacement, speed and velocity.Differentiate with examples between scalar and vector quantities. Represent vector quantities by drawing. Define the terms speed, velocity and acceleration. Plot and interpret distance-time graph and speed-time graph. Determine and interpret the slope of distance-time and speed-time graph. Determine from the shape of the graph, the state of a body including at rest, moving with constant speed and moving with variable speed.

9th Class Physics Chapter 2 Short Question Answers

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9th Class Physics Chapter 2 Solved Numericals

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Correction and Comments about Numericals and Short Question Answers

• Numerical 2.9
  Question Asked by Maria Khan : The answer of numerical 2.9 of 2nd part where the distance is asked to find is, not coming correct.so please check this.your answer is 266.53m but through calculator is 296.14m.plz solve this query.
  Answer : Actually, the value of 266m comes when you do not round off the value of acceleration (means a= -0.3331666875 m/s*s) and all other velocities. So your answer is also correct because you round off the value of acceleration. In exams I would say that write the value of distance 296m because this is the value you computed after rounding off and it is correct also.
• Numerical 2.8
  Problem : Incomplete Numerical and missing page number 14.
  Remaining Part:
  0 = 6vi – 180
  vi = 30m/s
  Now calculating height reached by the ball
  using formula 2as = vf2 – vi2
  putting values in the above formula
  s = (vf2 – vi2)/2a
  s = (0 – 30*30)/2*9.8
  s = 45m approx
  Hence the height reached by the ball is 45 meters and the initial velocity is 30 m/s

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