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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For Problem 2.9, you used acceleration as 0.33m/s.I’d say you should use o.3332m/s because I solved and the answer was coming out all wrong.

You did this wrong too:

vf= vi + at

0= 26.667 – 0.3332t

0.3332t= 26.667

t= 26.667/0.3332=80.03 which rounded off will be 80 because 80.80 makes 81.

Thank you.

your effort is really awesome sir.All stuff is very nice n quite easy to understand.

sir kindly upload other chapters of physics 9th, chap 6,7,8,,, thanks for your courtesy sir.

Question 2.8 is not completely solved.

AOA,the solution of P2.8 was incomplete this numerical problem was done by wrong method so plz correct it,because i tally this frm book and these ans are, in step 1

the answer was 45m, and in 2nd step the ans was 30m/s

r the notes gone :'( ? i cant see them today ~ mathar

Thanks 4 your notes.

ur site is sooo awsome….all numaricals r so eassy then my teacher tought to me….i really thank u for my hart….thank u

hmmm ur notes r really awesome and they r really helpful for me

thanks its very helpful.plz baki chapters and subjects k bhi notes provide ker dain.

Its good .