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## Physics Numerical Class 9 Motion with Solutions

Motion class 9 numericals with solutions 2022.

Here are some Class 9 motion numericals extra questions. Practice these Physics Numerical Class 9 Motion to further clear your concepts and build a better concept.

We are also providing you with the solution of Class 9 Motion NCERT Numericals. So, do check these also but firstly try to solve by yourself.

How do you solve numerical motion in Class 9?

To solve the numerical first read the question carefully, and note down all the given data. You also need to write down the related information. Now try to apply these physics formulas or the motion formulas given below.

Note- Take care of ‘u’ (Initial Velocity) and ‘v’ (Final Velocity).

## All formulas of Physics Class 9 Motion

What are the three equations of motion in class 9?

Here are the three equations of motion. You must apply these three equations of motion to solve the numerical related to class 9 motion.

Newtons Equation of Motions

- s = ut + ½ × a × t²
- v² = u² + 2as

Here u = initial velocity

v = final velocity

a = acceleration

Distance traveled in nth second

If a question is asked about finding the distance in the nth second use this formula.

S_{n}=u+\frac{a}{2}(2n-1)

## Questions on Motion Class 9 Numericals with Answers

Before solving the Questions, I request you to check the Class 9 Motion Notes also.

- Read More: Chapter 8 Motion Class 9 Notes

1. A scooter traveling at 10 m/s speeds up to 20 m/s in 4 sec. Find the acceleration of the scooter.

Initial velocity (u) = 10m/s

Final velocity (v) = 20 m/s

Time (t) = 4s

Use the First Equation of Motion

v = u + at = a = (v-u)/t

or, a = (20-10)/4 =10/4 =2.5 m/s²

2. An object dropped from a cliff falls with a constant acceleration of 10 m/s 2 . Find its speed 5 s after it was dropped.

Apply, v = u + a t

v = 0 + a t

3. A ball is thrown upwards and it goes to the height of 100 m and comes down

a) What is the net displacement?

b) What is the net distance?

a) What is the net displacement = 0 m

b) What is the net distance = 100 + 100 = 200 m

Also Check this Video Solution for Q1, Q2, Q3

SUBSCRIBE ALSO: Click Here

4. A car starts from rest and acquires a velocity of 54 km/h in 2 sec. Find

a) the acceleration

b) distance traveled by car (assume the motion of the car is uniform)

Initial velocity = u = 0 m/s

Final velocity = v = 54 km/hr = 15 m/s

Time = t = 2 sec

Equation of motion:

a = 15/2 = 7.5 m/s²

s = distance traveled

s = ut + 1/2 at² = 0 + 1/2 × 7.5 × 2² = 15 m

5. A bullet hits a Sandbox with a velocity of 20 m/s and penetrates it up to a distance of 6 cm. Find the deceleration of the bullet in the sandbox.

Distance = 6cm = 0.06 m

v² = u²+2as

Then v²=2as

20×20=2×a×0.06

a=-3333.33 m/s2

Another Method:

Apply the concept of ΔKE = Work Done

- Also Read- Gravitation Class 9 Numerical with Solution

6. An object moves along a straight line with an acceleration of 2 m/s2. If its initial speed is 10 m/s, what will be its speed 2 s later?

Here u=10m/s, a=2m/s², t = 2 sec

v=10+2×2 =10+4 =14m/s

- Read More- Force and Laws of Motion Class 9 Numerical
- Read More- Zoology Objective Questions and Answers

7. A particle experiences constant acceleration for 20 seconds after starting from rest. If it travels a distance D 1 in the first 10 seconds and a distance D 2 in the next 10 seconds then

(a) D 2 = D 1

(b) D 2 = 2D 1

(c) D 2 = 3D 1

(d) D 2 = 4D 1

We know that

distance (D) = ut + ½at²

Initial velocity u = 0 m/sec

Since the particle starts from rest

D 1 =1/2×a×(10)²

D 1 = 50a ——————– (1)

Now initial velocity for D 2 is the final velocity after covering D 1 distance

Final velocity of D 1 = Initial velocity of D 2

so, v=u +at

v=at (Final velocity after covering D 1 distance)

v=10a (because here u=0 and t=10)

This v is equal to u for D 2

D 2 = ut+1/2a(t²)

D 2 = (10a)10 + ½a10²

D 2 = 100a+50a

D 2 = 150a ——————- (2)

D 1 /D 2 = 50a/150a

You can also Watch the video for Solution (Q4, Q5, Q6, & Q7)

SUBSCRIBE ALSO for More Questions: Click Here

8. An artificial satellite is moving in a circular orbit of radius 42250 km (approx.). Calculate its velocity if it takes 24 hours to revolve around the earth.

The radius of the circular orbit along which the satellite is moving = 42250 km.

Time taken by the satellite to complete one revolution around earth = 24 hr.

Distance = 2πr

Time = 24 hr

Distance = speed × time

Solve and get Speed = 11065.47 km/h

9. Two cars A and B race each other. Car A ran for 2 minutes at a speed of 7.5 km/h, slept for 56 minutes and again ran for 2 minutes at a speed of 7.5 km/h. find the average speed of car A in the race.

d1 = 7.5×2/60

d1 = 0.25 km

d2 = 7.5×2/60

d2 = 0.25 km

Total distance = d1+d2 = 0.5 km Total time = 2+2+56 = 60min = 1 hr

Average speed = 0.5/1 = 0.5 km/hr

a) What is the distance traveled by the SUV in the first 2 seconds?

b) What is the braking force at the end of 5 seconds to bring the SUV to a stop within one second?

(a) Find the area of the triangle ABE

where, a = {final velocity (at D)– initial velocity (at C)}/ time (FD) = (0-15)/1 = – 15 m/s²

11. A car travels 30 km at a uniform speed of 40 km/hr and the next 30 km at a uniform speed of 20 km/hr. Find its average speed.

12. A body A starts from rest with an acceleration a1. After two seconds, another body B starts from rest with an acceleration a2. If they travel an equal distance in the 5th second after the start of A, then find the ratio a1:a2.

13. A particle moving in a straight line covers half the distance with a speed of 3 m/s. The other half of the distance is covered in two equal time intervals with a speed of 4.5 m/s and 7.5 m/s respectively. What is the average speed of the particle?

14. A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/h in 10 minutes. Find its acceleration.

15. Show that 1 radian = 57.3 degree

For Solution. Watch the Video below, and Please Support the Channel by SUBSCRIBING

- Read More:- CBSE NCERT Solutions & Revision Notes

Solution of Q11.

Average speed = \frac { x+y }{ \frac { x }{ u } +\frac { y }{ v } }

Solve and get Average speed = 26.6 km/h

Solution of Q12.

Distance traveled by A in 5th second

Solution of Q13.

Average Velocity = S/(t1 + t2)

t1 = s/2 ÷ 3

⇒ s/2 = 4.5× t2/2 +7.5× t2/2

⇒ t2 = s/12

Put the values, and solve

Average Velocity = 4 m/s.

Solution of Q14.

Here we have,

Initial velocity, u = 0,

Final velocity, v = 40km/h = 11.11m/s

Time (t) = 10 minute = 60×10 = 600 s

Acceleration (a) = 11.11/600 = 0.0185 m/s2

Solution of Q15.

π radian = 180

So, 1 radian = 180/π = 57.3 degrees

16. A train travels at a speed of 60 km/h from station A to B and then comebacks with a speed of 80 km/h from station B to A. Find the average speed and the average velocity of the train.

17. An Odometer of a bike reads 5000 km at 10:15 AM and it again reads 5075 km at 11:10 AM. Find the average speed of the bike.

18. Salman swims in a 90 meter long pool. He covers 160 m in 2 minutes by swimming from one end to the other and back along the same straight path. Find the average speed and the average velocity of Salman.

19. Raju walks 1 km towards the east in 15 minutes and then he turns north and runs 2 km in 10 minutes. Find the average speed and the average velocity of Raju in m/s.

20. If a car travels 20 km south, 15 km west, and 14 km north in 40 minutes. Find the average speed and the average velocity of the car.

21. A train 120 m long is moving with a velocity of 75 km/hr. Find the time it will take to cross a bridge that is 3 km long.

A solution will be provided very soon. Meanwhile do check Our Youtube Channel and Subscribe for free materials

## Motion Class 9 NCERT Exercise Numericals’ Solution

(Question No. 1) An 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?

(Question No. 2) 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 are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?

(Question No. 3) Abdul, while driving to school, computes the average speed for his trip to be 20 km/h. On his return trip along the same route, there is less traffic and the average speed is 30 km/h. What is the average speed for Abdul’s trip?

(Question No. 4) A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m/s 2 for 8.0 s. How far does the boat travel during this time?

(Question No. 7) A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m/s 2 , with what velocity will it strike the ground? After what time will it strike the ground?

(Question No. 10) An artificial satellite is moving in a circular orbit with a radius of 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

If you face any problem or a detailed solution, please comment on our Youtube Channel . We will surely try to provide you with a solution.

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## 48 thoughts on “Physics Numerical Class 9 Motion with Solutions”

SIR PLEASE add english audio in doubt video

Nice but I need more questions

There is no question number 16

Yes u r right i also want the answer of that que but it is not available there

Thanks a lot for these questions and please upload more questions for class 9

Yes, we need more questions for solving these equations. I am enjoying these questions and solving physics numerical is one of my favourite

I can’t understand q 12 please examplain it more better🙏🙏🙏🙏

Plz do check the youtube channel. CBSE Portal

SIR PLEASE if your putting solutions video in only hindi how come the students will understand please add english or make it available all the languages of the india

Hi there! for that question, calculate the 2 distances and then put them in ratio ie. S1:S2 you can reply if you want any further help :))

I have doubt in question no. 13 can u please make a video regarding its solution

Sure will talk to the team, Keep visiting CBSE Portal

Provide q 15 soloution

in question 2nd i think it is wrong could you please notice it and i hope that you may solve it thankyou …..

Yes it maybe wrong as when it falls v should be 0

In Q2 it is asking you the velocity after 5 seconds, not when it touches the ground. You don’t know if it touches the ground or not in 5 seconds. For more details do follow the YouTube Channel mentioned CBSE Portal

I like this website b’coz it is not very complicated as it is very easy to use and questions are very simple. But the problem is that there are so many advertisements. The answers are very nice 👍🏾😊 and I liked it…ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤthe best website ever seen on Google platform. ㅤㅤㅤㅤㅤㅤniceㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

We are sorry for the inconvenience dear. These help us maintain the website, and also pay the contributors. So, that we could provide you the materials at free of cost.

I got a doubt regarding a question from my syllabus . Kindly help me with it A train travels with a speed of 60km/hr from station A to B and then comes back with a speed of 80km/hr from station B to A. Find : i)the average speed ii)the average velocity of the train .

We have provided the solution for You Do Check here- https://youtu.be/6-hh9Zg0ez8 Don’t Forget to SUBSCRIBE

This is very easy question on my openion b’coz it is given in the refresher as well as guide

AVERAGE velocity is 0 Average speed is 68.57

Formula for this question (2S1)in to (S2 )upon S1 +S2

both answers are 70km/hr as average speed and velocity= u+v/2 so 80+60/2 = 70

plz upload more questions for practice

I have a doubt in question no. 8 I think the answer would be 11055.42 km/hr

Your answer is also correct, since you have taken pi = 3.14. If you take 22/7 = 3.14285714 it will give 11065.4762

Please, how can I calculate net distance

If the satellite strats from point A , complete it revolution and comes to point A again in 24 hrs, then net displacement is 0

Therefore, velocity =. Displacement /Time 0/Time = 0 Therefore the velocity should be 0 m/s

Please check ✔️

Tooo easy for me Good questions Thank u No doubt

I want more questions for practice

Sure, Will upload. keep visiting

That’s it

All the questions were easiest Please upload some tricky questions according to class IX.

Sure, we will update. And thanks for letting us know.

Good questions

Could you please tell me your doubt?

can you show slove in this question

Sure, Please Keep Visiting the website and the Youtube Channel, CBSE Portal

Î7 in the doubt

Ur name is not good.

Ur an bad boy/girl

I have a doubt in 9

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High School Physics

## Physics Numericals Class 9 Motion & other Chapters solved

Last updated on July 21st, 2023 at 11:01 am

Solved Numerical Problems in Physics Class 9 Motion & other chapters – This page is based on Physics Numericals Class 9 . Find 101+ solved numerical problems, Questions & answers from class 9 motion & other chapters, for CBSE, ICSE, & UPSC.

This page contains solved numerical problems from Motion Physics (kinematics), and other topics and chapters as per CBSE & ICSE syllabus for class 9 physics. This superset of physics questions includes 3 worksheets to solve [links given at the end of this post] as well.

## Solved Numerical Problems in Physics class 9 Motion | Motion Numericals class 9

Motion numerical problems – on force, momentum & acceleration, gravitation – solved questions, force, & acceleration questions – solved, motion numerical in physics class 9, gravitation numerical (set 2), vertical motion numerical, physics sum for class 9, class 9th – numericals from pressure chapter – cbse, icse, numerical problems in physics class 9 – on conservation of momentum, pendulum problems.

- Relative density Numerical of class 9 physics | practice problems – set 1
- Motion questions physics – practice problems – set 2
- Mirrors and images numerical problems Physics | practice problems – set 3

The next set of numerical questions (set 1) covers questions from the Motion chapter (kinematics) of class 9 physics and these can be easily solved using motion equations ( Suvat equations ).

1) A speedboat has an acceleration of 2 m/s 2 . What would the final velocity of the speedboat be after 5 seconds if the initial velocity of the speedboat is 4 m/s?

a = 2 m/s 2 u = 4 m/s t = 5 s final velocity = v v = u + at = 4 + 2×5 = 14 m/s

2) A vehicle has an acceleration of 5 m/s 2 . What would the final velocity of the vehicle be after 10 seconds if the initial velocity of the vehicle is 20 m/s? What is the displacement of the vehicle during the 10 s time interval?

a = 5 m/s 2 u = 20 m/s t = 10 s final velocity = v v = u + at = 20 + 5×10 = 70 m/s final velocity = 70 m/s Displacement during the 10 s time interval = s s = ut + (1/2)at 2 => s = 20×10 + (½)x5x10 2 => s= 200 + 250 = 450 m Displacement during the 10 s time interval = 450 m

3) A child on a toboggan starts from rest and accelerates down a snow-covered hill at 0.8 m/s 2 . How long does it take the child to reach the bottom of the hill if it is 25.0 m away?

Solution: initial velocity u = 0 acceleration a = 0.8 m/s 2 distance = s = 25 m time taken t = ? s= ut + (1/2)at 2 As here, u = 0, so we can write: s= (1/2)at 2 => t 2 =(2 s /a ) = (2×25)/0.8 = 62.5 time t = 7.9 sec

4. A car accelerates uniformly from a velocity of 21.8 m/s to a velocity of 27.6 m/s. The car travels 36.5 m during this acceleration.

a) What was the acceleration of the car?

b) Determine the time interval over which this acceleration occurred.

a) final velocity v = 27.6 m/s initial velocity u = 21.8 m/s distance traveled s=36.5 m We will use this equation: v 2 = u 2 + 2as a = ( v 2 – u 2 ) /(2s) = (27.6 2 – 21.8 2 )/(2×36.5) = 3.92 m/s 2 Acceleration = 3.92 m/s 2

b) time interval over which this acceleration occurred = t v = u + at => t = (v-u)/a = (27.6-21.8) / 3.92 = 1.48 seconds.

5. A motorcycle starts from rest and accelerates at +2.50 m/s 2 for a distance of 150.0 m. It then slows down with an acceleration of –1.50 m/s 2 until the velocity is +10.0 m/s. Determine the total displacement of the motorcycle.

First part: displacement = s = 150 m We need to find out the final velocity(v) of this accelerating part. v 2 = u 2 + 2as => v 2 = 0 + 2×2.5×150 = 750 => v = 27.4 m/s This velocity becomes the initial velocity of the second part of the journey.

Second part:

u = 27.4 m/s a = -1.5 m/s 2 v= 10 m/s v 2 = u 2 + 2as 10 2 = 27.4 2 + 2.(-1.5).s => 100 = 750 – 3s s= 650/3 = 216.7 m Second part displacement = 216.7 m Considering a straight-line journey without a change of direction, total displacement = (150 + 216.7) m = 366.7 m

## Physics Numericals class 9

So let’s start with sets of Physics Numericals Class 9 which contains Numerical problems & questions from various chapters of Class 9 physics with solutions & answers.

1) A force of 1000 Newton is applied on a 25 kg mass for 5 seconds. What would be its velocity?

Solution: F =1000 N m=25kg t= 5 sec u=0 v=?

Acceleration = a = F/m = 1000/25 = 40 m/s 2

v = u + at =0 + 40X5 = 200 m/s

2) A force is applied to a mass of 16 kg for 3 seconds. As the force is removed the mass moves 81 meters in 3 seconds. What was the value of the force applied?

First 3 secs: Force is present, and in the next 3 secs Force is not there, so uniform velocity for the last 3 seconds is mentioned.

In the last 3 seconds, the mass travels with uniform velocity (as there is no force that means no acceleration).

So the velocity in the next 3 secs: V = 81/3 = 27 m/s

This velocity was attained in the first 3 seconds when the force was present.

So in the first 3 seconds, velocity changes from 0 to 27m/s.

Therefore the acceleration in the first 3 seconds caused by the force present = (27-0)/3 = 9 m/s².

So the force applied on the mass for first 3 secs= mass x acc = 16 x 9 = 144 N

3) A mass of 50 kg was moving with a velocity of 400 m/s. A force of 40000 N is applied to the mass and its velocity is reduced to 50 m/s after some time. What is the distance traveled by the mass during this period?

Mass = 50 kg u = 400 m/s v= 50 m/s F = 40000 N t = unknown Distance traveled in time t =?

Acceleration = a = F/m = 40000/50 = 800 m/s 2

Again, acceleration a = change of velocity /t or, t = change of velocity/a = 350/800 sec

Distance travelled: s = ut – (1/2)a t 2

= 400 * (350/800) – (1/2) 800. (350/800) 2 = 175 – 76.5 = 98.5 m

4) A ball of 150 g mass moves with 12 m/s and bounces back after hitting a wall with 20 m/s after a small duration of 0.01 seconds. What was the force applied to the ball when it hits the wall?

Physics problems for class 9 with solutions – physics problems solved

Mass = 150 gram = 150/1000 kg = 0.15 kg

Change of velocity = 20 – (-12) = 32 m/s

T = 0.01 sec

Acceleration = change in velocity / time = 32/0.01 = 3200 m/s 2

Force = mass X acc = 0.15 X 3200 N= 480 N.

5) Momentum of an object changes from 100 kg m/s to 200 kg m/s in 2 seconds. What is the force applied to it?

Note: Force = ma = m(v-u)/t = (mv – mu)/t = change of momentum /time = rate of change of momentum

Force= change of momentum /time = (200-100)/2 = 100/2 = 50 N

6) A force of 200 N is applied to a body and its velocity changes from 5m/s to 10m/s in a second. What is the mass of the body?

mass = force/acceleration here acceleration =a = (10-5)/1 = 5 m/s 2 so, mass = 200/5 kg = 40 kg

7) A mass of 1 kg is moving from east to west with a velocity of 10 m/s. A force is applied to it for 2 seconds and its velocity becomes 5 m/s. What is the value and direction of the force?

Mass = 1kg velocity = 10 m/s (east to west) Because of the force, the velocity reduces to 5 m/s in 2 secs. acceleration = change of vel/time=5/2 = 2.5 m/s 2

Force = mass X acceleration= 1 X 2.5 = 2.5 N. As the velocity reduces because of force, (negative acc i.e. retardation) hence the direction of the force would be opposite to the initial direction. So the direction of the force will be west to east.

8) If the distance between 2 objects is doubled then how will the gravitational force between them change?

F1 = GMm/r 2 F2 = GMm/(2r) 2 = GMm/4r 2 so F2/F1 = ¼ , so the final gravitational force will be ¼th of the initial gravitational force

9) If the distance between 2 objects is halved then how will the gravitational force between them change?

F1 = GMm/r 2 F2 = GMm/(r/2) 2 =4 GMm/r 2 so F2/F1 = 4/1 so the final gravitational force will be 4 times the initial gravitational force

10) If the masses of the 2 objects are doubled then how will the gravitational force between them change?

F1 = GMm/r 2 F2 = (G2M2m)/(r) 2 =4 GMm/r 2 so F2/F1 = 4/1 so the final gravitational force will be 4 times the initial gravitational force

[ solutions: for Q11 to Q71 – solution links are just beside the question above. Click that link to reach the page with a solution. ] This set is again based on force and acceleration.

11) A force acts for 10 s on a stationary body of mass 100 kg after which the force ceases to act. The body moves through a distance of 100 m in the next 5 s. Calculate (i) the velocity acquired by the body, (ii) the acceleration produced by the force, and (iii) the magnitude of the force. [ Solution: click this link for solution Q11 ]

12) A force acts for 0.1 sec on a body of mass 2 kg initially at rest. The force is then withdrawn and the body moves with a velocity of 2 m/s. Find the magnitude of the force. Solution: click this link for solution Q12

13) A cricket ball of mass 100 g moving at a speed of 30 m/s is brought to rest by a player in 0.03 s. find the average force applied by the player. Solution: click this link for solution Q13

14) A body of mass 500 g, initially at rest, is acted upon by a force that causes it to move a distance of 4 m in 2 sec. Calculate the force applied. Solution: click this link for solution Q14

15) A force causes an acceleration of 10 m/s^2 in a body of mass 500 g. What acceleration would be caused by the same force in a body of mass 5 kg. Solution: click this link for solution Q15

16) A bullet of mass 50 g moving with an initial velocity of 100 m/s, strikes a wooden block and comes to rest after penetrating a distance 2 cm in it. Calculate: (i) Initial momentum of the bullet (ii)final momentum of the bullet (iii) retardation caused by the wooden block, and (iv) resistive force exerted by the wooden block Solution: click this link for solution Q16

Find another set of Physics Numerical for Class 9 Motion.

17) From the velocity-time graph , find out the distance traveled in each section A, B, C and D. Also find the total distance traveled.

Solution: click this link for solution Q17

18) A car starts from rest and accelerates uniformly over a time of 7 seconds for a distance of 98 m. Determine the acceleration of the car. Solution: click this link for solution Q18

19) A car traveling at 20 m/s skids to a stop in 3 s. Determine the skidding distance of the car (assume uniform acceleration). Solution: click this link for solution Q 19 20) An airplane accelerates down a runway at 4 m/s^2 for 30 s until it finally lifts off the ground. Determine the distance traveled before takeoff. Solution: click this link for solution Q 20

21) A feather is dropped on the moon from a height of 10 meters. The acceleration of gravity on the moon is one-sixth of that on the earth. Determine the time for the feather to fall to the surface of the moon. Solution: click this link for solution Q 21

22) If a rocket-powered sled is accelerated to a speed of 400 m/s in 2 seconds, then what is the acceleration, and what is the distance that the sled travels? Solution: click this link for solution Q 22

23) An engineer is designing the runway for an airport. Of the planes that will use the airport, the lowest acceleration rate is likely to be 3 m/s 2 . The takeoff speed for this plane will be 65 m/s. Assuming this minimum acceleration, what is the minimum allowed length for the runway? Solution: click this link for solution Q 23

24) Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.60 seconds, what will be his final velocity and how far will he fall? Solution: click this link for solution Q 24

25) A bike accelerates uniformly from rest to a speed of 10 m/s over a distance of 30 m. Determine the acceleration of the bike. Solution: click this link for solution Q25

26) A race car accelerates uniformly from 25 m/s to 50 m/s in 5 seconds. Determine the acceleration of the car and the distance traveled. Solution: click this link for solution Q 26

27) A bullet leaves a rifle with a muzzle velocity of 600 m/s. While accelerating through the barrel of the rifle, the bullet moves a distance of 0.8 m. Determine the acceleration of the bullet (assume uniform acceleration). Solution: click this link for solution Q 27

28) A bullet is moving at a speed of 350 m/s when it embeds into a lump of moist clay. The bullet penetrates for a distance of 0.05 m. Determine the acceleration of the bullet while moving into the clay. (Assume a uniform acceleration.) Solution: click this link for solution Q 28

29) A plane has a takeoff speed of 100 m/s and requires 1500 m to reach that speed. Determine the acceleration of the plane and the time required to reach this speed. Solution: click this link for solution Q 29

30)A baseball is popped straight up into the air and has a hang-time of 6 s. Determine the height to which the ball rises before it reaches its peak. (Hint: the time to rise to the max height is one-half the total hang time.) Solution: click this link for solution Q30

31) If a box is pushed horizontally on the floor with a force of 10 N and it moves 5 meters along the line of action of the force, then what is the work done by the Gravity or earth’s gravitational pull on that box? Solution: click this link for solution Q31

Related & Suggested Reading for the readers: 1) physics Questions & Answers for class 9 CBSE, ICSE, state 2) Class 9 Numericals set 2 on Laws of Motion and Force – hand-picked problems, Author Name Tweet

32) A train is moving at a velocity of 25 ms-1. It is brought to rest by applying the brakes which produce uniform retardation of 0.5 ms-2. Calculate (i) the velocity of the train after 10 s (ii) If the mass of the train is 20000 kg then calculate the force required to stop the train Solution: click this link for solution Q3 2

33) A spring balance is used to find the weight of a body X on the surface of the moon. The mass of the body X is 2 kg and its weight is recorded as 3.4 N. The weight of another body Y recorded by the same balance is found to be 7.65 N. Calculate the mass of the body Y. Solution: click this link for solution Q3 3

34) Calculate the force of gravitation between two objects of masses 50 kg and 120 kg respectively kept at a distance of 10 m from one another. (Gravitational constant, G = 6.7 × 10 -11 Nm^2/kg^2 ) Solution: click this link for solution Q34

35) What is the force of gravity on a body of mass 150 kg lying on the surface of the earth? (Mass of earth = 6 × 10^24 kg; Radius of earth = 6.4 × 10^6 m; G = 6.7 × 10 -11 Nm^2/kg^2) Solution: click this link for solution Q35

36) The mass of sun is 2 × 10^30 kg and the mass of earth is 6 × 10^24 kg. If the average distance between the sun and the earth is 1.5 × 10^8 km, calculate the force of gravitation between them. Solution: click this link for solution Q36

The following set on physics numerical for class motion specifically covers vertical motion questions.

37) A piece of stone is thrown vertically upwards. It reaches the maximum height in 3 seconds. If the acceleration of the stone is 9.8 m/s^2 directed towards the ground, calculate the initial velocity of the stone with which it is thrown upwards. Solution: click this link for solution Q37

38) A stone falls from a building and reaches the ground 2.5 seconds later. How high is the building? ( g = 9.8 m/s^2) Solution: click this link for solution Q3 8

39) A stone is dropped from a height of 20 m.( i ) How long will it take to reach the ground? ( ii ) What will be its speed when it hits the ground? ( g = 10 m/s^2) Solution: click this link for solution Q3 9

40) A stone is thrown vertically upwards with a speed of 20 m/s. How high will it go before it begins to fall? ( g = 9.8 m/s^2) Solution: click this link for solution Q 40

41) When a cricket ball is thrown vertically upwards, it reaches a maximum height of 5 meters. ( a ) What was the initial speed of the ball? ( b ) How much time is taken by the ball to reach the highest point? ( g =10 m/ s^2) Solution: click this link for solution Q 41

42) A force of 20 N acts upon a body whose weight is 9.8 N. What is the mass of the body and how much is its acceleration? ( g = 9.8 m/ s^2). Solution: click this link for solution Q42

43) A stone resting on the ground has a gravitational force of 20 N acting on it. What is the weight of the stone? What is its mass? ( g = 10 m/s^2). Solution: click this link for solution Q43

44) (a) An object has a mass of 20 kg on Earth . What will be its ( i ) mass, and ( ii ) weight, on the moon ?( g on moon = 1.6 m/s^2). ( b ) Calculate the acceleration due to gravity on the surface of a satellite having a mass of 7.4 × 10^22 kg and a radius of 1.74 × 10^6 m ( G = 6.7 × 10 -11 Nm^2/kg^2). Solution: click this lin k for solution Q44

45) The mass of a planet is 6 × 10^24 kg and its diameter is 12.8 × 10^3 km. If the value of the gravitational constant is 6.7 × 10 -11 Nm2/kg2, calculate the value of acceleration due to gravity on the surface of the planet. Solution: click this link for solution Q45

46) The values of g at six distances A, B, C, D, E, and F from the surface of the earth are found to be 3.08 m/s^2, 9.23 m/s^2, 0.57 m/s^2, 7.34 m/s^2, 0.30 m/s^2 and 1.49 m/s^2, respectively. ( a ) Arrange these values of g according to the increasing distances from the surface of the earth (keeping the value of g nearest to the surface of the earth first) ( b ) If the value of distance F be 10000 km from the surface of the earth( hint: check if possible or not ), state whether this distance is deep inside the earth or high up in the sky. Give the reason for your answer of the place of observation Solution: click this link for solution Q4 6

47) A pebble is dropped freely in a well from its top. It takes 20 s for the pebble to reach the water surface in the well. Taking g = 10 m/s^2 and speed of sound = 330 m/s, find (i) the depth of the water surface, and (ii) the time when the echo is heard after the pebble is dropped. **[answer Hint: 2000 m ; 26.06 secs ] Solution: click this link for solution Q4 7

48) A ball is thrown vertically upwards from the top of a tower with an initial velocity of 19·6 m/s. The ball reaches the ground after 5 s. Calculate: (i) the height of the tower, (ii) the velocity of ball on reaching the ground. Take g = 9·8 m/s^2 **[answer Hint: 24.5 m ; 29.4 m/s ] Solution: click this link for solution Q4 8

49) A hammer exerts a force of 1·5 N on each of the two nails A and B. The area of the cross-section of the tip of nail A is 2 mm^2 while that of nail B is 6 mm^2. Calculate the pressure on each nail in Pascal. Solution: click this link for solution Q4 9

50) A block of iron of mass 7·5 kg and of dimensions 12 cm x 8 cm x 10 cm is kept on a tabletop on its base of side 12 cm x 8 cm. Calculate (a) thrust and (b) pressure exerted on the tabletop. Take 1 kgf=10 N. Solution: click this link for solution Q50

51) A vessel contains water up to a height of 1·5 m. Taking the density of water 1()^3 kg /m^3, acceleration due to gravity 9·8 m/ S^2, and area of the base of vessel 100 cm^2, calculate: (a) the pressure and (b) the thrust, at the base of the vessel. Solution: click this link for solution Q51

52) The area of the base of a cylindrical vessel is 300 cm^2. Water (density = 1000 kg /m^3) is poured into it up to a depth of 6 cm. Calculate (a) the pressure and (b) the thrust of water on the base. (g = 10 m/ s^2) . Solution: click this link for solution Q52

53)Calculate the height of a water column that will exert on its base the same pressure as the 70 cm column of mercury. The density of mercury is 13.6 g/cm^3. Solution: click this link for solution Q5 3

54) A boy of mass 50 kg running at 5 m/s jumps onto a 20 kg trolley traveling in the same direction at 1.5 m/s. What is their common velocity? Solution: click this link for solution Q54

55) A girl of mass 50 kg jumps out of a rowing boat of mass 300 kg onto the bank, with a horizontal velocity of 3 m/s. With what velocity does the boat begin to move backward? Solution: click this link for solution Q5 5

56)A truck of mass 500 kg moving at 4 m/s collides with another truck of mass 1500 kg moving in the same direction at 2 m/s. What is their common velocity just after the collision if they move off together? Solution: click this link for solution Q5 6

57) A ball X of mass 1 kg traveling at 2 m/s has a head-on collision with an identical ball Y at rest. X stops and Y moves off. Calculate the velocity of Y after the collision. Solution: click this link for solution Q57

58) A heavy car A of mass 2000 kg traveling at 10 m/s has a head-on collision with a sports car B of mass 500 kg. If both cars stop dead on colliding, what was the velocity of car B? Solution: click this link for solution Q58

59) A man wearing a bulletproof vest stands still on roller skates. The total mass is 80 kg. A bullet of mass 20 grams is fired at 400 m/s. It is stopped by the vest and falls to the ground. What is then the velocity of the man? Solution: click this link for solution Q59

60) A metallic sphere of mass 2 kg and volume 2.5 x 10 -4 m 3 is completely immersed in water. Find the buoyant force exerted by water on the sphere. Density of water = 1000 kg/m 3 Solution: click this link for solution Q60

61) Two simple pendulums A and B have equal lengths , but their bobs weigh 50 gf and l00 gf respectively. What would be the ratio of their time periods ? Give a reason for your answer. Solution: click this link for solution Q61

62) Two simple pendulums A and B have lengths of 1·0m and 4·0 m respectively at a certain place. Which pendulum will make more oscillations in 1 minute? Explain your answer. Solution: click this link for solution Q62

63)A simple pendulum completes 40 oscillations in one minute. Find its (a) frequency, (b) time period. Solution: click this link for solution Q6 3

64) The time period of a simple pendulum is 2 s. What is its frequency? What name is given to such a pendulum? Solution: click this link for solution Q6 4

65) A seconds’ pendulum is taken to a place where the acceleration due to gravity falls to one-fourth. How is the time period of the pendulum affected, if at all? Give reason. What will be its new time period? Solution: click this link for solution Q6 5

66) Find the length of a seconds’ pendulum at a place where g = 10 m/s 2 (Take ∏= 3·14). Solution: click this link for solution Q6 6

67) A pendulum completes 2 oscillations in 5 s. (a) What is its time period? (b) If g = 9·8 m /S^2, find its length. Solution: click this link for solution Q6 7

68) Compare the time periods of two pendulums of length 1 m and 9 m Solution: click this link for solution Q6 8

69) The time periods of two simple pendulums at a place are in a ratio of 2:1. What will be the ratio of their length? Solution: click this link for solution Q6 9

70) It takes 0·2 s for a pendulum bob to move from the mean position to one end. What is the time period of the pendulum? Solution: click this link for solution Q70

71) How much time does the bob of a seconds’ pendulum take to move from one extreme of its oscillation to the other extreme? Solution: click this link for solution Q 71

Related: worksheets

Relative density worksheet

Motion worksheet (worksheet 2)

Mirrors & images numerical (worksheet 3)

We have 30 more numerical questions of this series for class 9 physics – these are listed as worksheets in the next 3 pages [links are given below]

## Numericals with Answers - Motion - Science Class 9

The study of Class 9 Science Chapter Motion is an essential part of physics, as it helps us understand how objects move and the factors that affect their movement. This chapter covers the basic concepts of motion, including types of motion, distance, displacement, speed, velocity, and acceleration. Let us see some Solved Numericals of the chapter.

## Solved Examples

Q1: A 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?

Speed of farmer = 40/40 = 1m/s Total distance covered = 1 × 140 = 140 m

1 round = 40 m

Therefore, number of rounds in 140 m = 140/40 = 3.5

Total rounds = = 3.5 round After every complete round, the farmer will be at the starting point, hence after 3.5 rounds farmer will be at the diagonal ends of the field. Hence, displacement = 10√2m

Q3 : In a long-distance race the athletes were expected to take four rounds of the track such that the line of finish was the same as the line of start. Suppose the length of the track was 200 m (i) What is the total distance to be covered by the athletes? (ii) What is the displacement of the athletes when they touch the finish line? (iii) Is the motion of the athlete's uniform or non-uniform. (iv) Is the displacement of an athlete and the distance moved by him at the end of the race equal?

Ans: (i) Total distance covered = 4 × 200m = 800 m (ii) As the athletes finish at the starting line Displacement = final position-initial position = 0 (iii) Motion is non-uniform as the direction of motion of the athlete is changing while running on the track. (iv) No, displacement and distance are not equal as total distance covered by athlete is 800m while displacement is zero as athlete comes back to its initial position. Q4 : On a 120 km track, a train travels the first 30 km with a uniform speed of 30 km/h. How fast must the train travel the next 90 km so as to average 60 km/h for the entire trip?

Ans: Total distance d = 120 km Average speed V av = 60 km/h Total time = t = ? Average speed = Total distance/Total time taken

V av = d/t

t = Putting the values

t = 120/60 = 2h .............(i)

Distance travelled in first part of trip d 1 = 30 km Speed in first part of the trip v 1 = 30 km/h Time taken in first part of trip t 1 = ? t 1 = putting the values t 1 = 30/30 = 1 h (as time = distance /speed )

Time taken left to complete second part of the trip t 2 = t - t 1 = 2 - 1 = 1h Distance to be covered in second part of the trip d 2 = 90 km reQues uired speed in second part v 2 = ? Now since, speed = d istance/time v 2 = d 2 /t 2 = 90 km/h Q5 : A bus decreases its speed from 80 km/h to 60 km/h is 5 seconds. Find the acceleration of the bus. Ans: u = 80 km/h = 80 × 5/18 = 22.22 m/s v = 60 km/h = 60 ×5/18 = 16.67 m/s time t = 5s Acceleration (a) = (V - U) / t

Ans: We draw perpendicular lines from the 10-minute point and the 30-minute point to the time-axis (fig.) The distance covered is equal to the area of the rectangle ABCD its value is ABCD = (30 - 10) × (10) = 20 × 10 = (20/60) × 10 = 10/3 km. Q12: Find the distance covered by a particle during the time interval t = 0 to t = 20 s for which the speed-time graph is shown in figure.

Ans: The distance covered in the time interval 0 to 20 s.

Area of the shaded triangle = 1/2 × base × height. = 1/2 × (20) × (20) = 200 m. Q13: Figure shows the velocity-time graphs for two objects, A and B, moving along the same direction. Which object has greater acceleration?

Ans: The slope of the velocity-time graph of B is greater than that for A, Thus, the acceleration of B is greater than that of A. Q14: A particle moves through a distance of 3 m due east and then 4m due north. (a) How much is the net distance traversed? (b) What is the magnitude of the net displacement? Ans: The situation is shown in the figure. The particle starts from O. It moves through a distance of 3m due east to reach A and then through a distance of 4m due north to reach B.

(a) The total distance moved is 3m + 4m = 7 m, (b) The magnitude of the net displacement is OB. In the right_angled triangle OAB, OB 2 = OA 2 + AB 2 = (3m) 2 + (4m) 2 = 9m 2 + 16m 2 = 25m 2

Therefore, OB = 5 m, Q15: A car moves 30 km. in 30 min. and the next 30 km. in 40 min. calculate the average speed for the entire journey. Ans: Given, the total time taken is 30 min + 40 min. = 70 min. the total distance traversed = 30km + 30 km = 60 km. The average speed v av = 60km/70min - 60km/(70/60)h - 3600/70 km/hr = 51.4 km/h Q16: A boy runs for 10 min. at a uniform speed of 9 km/h. At what speed should he run for the next 20 min. So that the average speed comes to 12 km/h? Ans: Total time = 10 min 20 min = 30 min. The average speed is 12 km/h

using s = vt, the distance covered in 30 min = 12 km/h × 30 min = 12km/h x 1/2h = 6km The distance covered in the first 10 min = 9 km/h × 10 min. = 9km/h x 1/6h = 1.5 km Thus, he has to cover 6 km - 1.5 km = 4.5 km. in the next 20 min. The speed required is 4.5km/20min = 4.5km/(20/60)h = 13.5 km/h Q17: A particle was at rest from 9:00 am to 9:30 am. It moved at a uniform speed of 10 km/h from 9:30 am to 10:00 am. Find the average speed between (a) 9:00 am and 10:00 am (b) 9:15 am and 10:00 am Ans: (a) The distance moved by the particle between 9:30 am and 10:00 am is s = vt = 10km/h x 1/2h = 5 km. This is also the distance moved between 9:00 am and 10:00 am. Thus, the average speed during this interval is v av = s/t = 5km/1h = 5 km/h

(b) The distance moved between 9:30 am and 10:00 am is 5km. This is also the distance moved in the interval 9:15 am to 10:00 am. The average speed during this interval is _ v av = s/t = 5km / 45min = 6.67 km/h. Q18: It is estimated that the radio signal takes 1.27 seconds to reach the earth from the surface of the moon. Calculate the distance of the moon from the earth. Speed of radio signal = 3 × 10 8 ms -1 (speed of light in air). Ans: Here, time = 1.27 s speed = 3 × 10 8 ms -1 distance = ? Using distance = speed × time, we get distance = 3 × 10 8 × 1.27 = 3.81 × 10 8 km = 3.81 × 10 5 km

Q24: An object dropped from a cliff falls with a constant acceleration of 10 m/s2. Find its speed 5 s after it was dropped. Ans: v = u + at u = 0 v = 10 x 5 = 50m/s Q25: A ball is thrown upwards and it goes to the height 100 m and comes down (i) What is the net displacement? (ii) What is the net distance? Ans: As it comes down to the initial point (i) Net displacement is zero (ii) Net distance=200 m Go through detailed chapter notes of this chapter here .

You can find NCERT Solution of the chapter here .

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## Problems Based on Equations of Motion for Class 9 Science

## [PDF] Download Problems Based on Equations of Motion for Class 9 Science

EQUATIONS OF MOTION

When an object moves along a straight line with uniform acceleration, it is possible to relate its velocity, acceleration during motion and the distance covered by it in a certain time interval by a set of equations known as the equations of motion. There are three such equations. These are:

where u is the initial velocity of the object which moves with uniform acceleration a for time t, v is the final velocity, and s is the distance travelled by the object in time t.

Eq. (1) describes the velocity-time relation and Eq. (2) represents the position-time relation. Eq. (3), which represents the relation between the position and the velocity, can be obtained from Eqs. (1) and (2) by eliminating t. These three equations can be derived by graphical method.

Problems Based on Equations of Motion

Q.1. A car acquires a velocity of 72km/h in 10 seconds starting from rest. Find (a) the acceleration (b) the average velocity (c) the distance travelled in this time.

Q.2. A body is accelerating at a constant rate of 10m/s 2 . If the body starts from rest, how much distance will it cover in 2 seconds?

Q.3. The length of minutes hand of a clock in 5 cm. Calculate its speed.

Q.4. An object undergoes an acceleration of 8m/s 2 starting from rest. Find the distance travelled in 1 second.

Q.5. A moving train is brought to rest within 20 seconds by applying brakes. Find the initial velocity, if the retardation due to brakes is 2m/s 2 .

Q.6. A scooter acquires a velocity of 36km/h in 10 seconds just after the start. Calculate the acceleration of the scoter.

Q.7. A racing car has uniform acceleration of 4m/s 2 . What distance will it cover in 10 seconds after start?

Q.8. A car accelerates uniformly from 18km/h to 36 km/h in 5 seconds. Calculate (i) acceleration and (ii) the distance covered by the car in that time.

Q.9. A body starts to slide over a horizontal surface with an initial velocity of 0.5 m/s. Due to friction, its velocity decreases at the rate of 0.05 m/s 2 . How much time will it take for the body to stop?

Q.10. A car increases its speed from 20 km/h to 50 km/h in 10 seconds. What is its acceleration?

Q.11. A ship is moving at a speed of 56km/h. One second later, it is moving at 58km/h. What is its acceleration?

Q.12. A train starting from the rest moves with a uniform acceleration of 0.2 m/s 2 for 5 minutes. Calculate the speed acquired and the distance travelled in this time.

Q.13. A bus was moving with a speed of 54 km/h. On applying brakes, it stopped in 8 seconds. Calculate the acceleration and the distance travelled before stopping.

Q.14. A train starting from rest attains a velocity of 72 km/h in 5 minutes. Assuming that the acceleration is uniform, find (i) the acceleration and (ii) the distance travelled by the train for attaining this velocity.

Q.15. Calculate the speed of the tip of second’s hand of a watch of length 1.5 cm.

Q.16. A cyclist goes once round a circular track of diameter 105m in 5 minutes. Calculate his speed.

Q.17. A cyclist moving on a circular track of radius 50m complete revolution in 4 minutes. What is his (i) average speed (ii) average velocity in one full revolution?

Q.18. A car starts from rest and moves along the x-axis with constant acceleration 5m/s 2 for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest?

Q.19. A motor cycle moving with a speed of 5 m/s is subjected to an acceleration of 0.2 m/s 2 . Calculate the speed of the motor cycle after 10 seconds and the distance travelled in this time.

Q.20. The brakes applied to a car produce an acceleration of 6 m/s 2 in the opposite direction to the motion. If the car takes 2 seconds to stop after the application of brakes, calculate the distance it travels during this time.

## Download Numericals on Equation of Motion for Class 9 PDF

Numericals on Equation of Motion for Class 9 is an invaluable resource for students seeking to excel in their physics studies. Download the PDF today and embark on a journey of discovery and mastery in the fascinating world of motion and equations! Join our Telegram Channel for PDF Download.

## Motion Class 9 Numericals Related Posts

Importance of solving numericals for motion class 9.

Solving numerical problems in Class 9 plays a crucial role in the learning process and is essential for several reasons:

- Understanding concepts: Numerical problems require students to apply the theoretical concepts they have learned in practical situations. By solving numericals, students gain a deeper understanding of the subject matter and its real-world applications.
- Problem-solving skills: Numerical problems often involve critical thinking and analytical skills. When students tackle these problems, they develop their problem-solving abilities, which are valuable not only in academics but also in various aspects of life.
- Application of formulas: Numericals help students apply mathematical formulas and equations to real-life scenarios. This strengthens their mathematical foundation and improves their ability to use formulas effectively.
- Confidence boost: Successfully solving numerical problems can boost a student’s confidence in their academic abilities. It provides a sense of accomplishment and motivates them to take on more challenging tasks.
- Exam preparation: Many examinations, including Class 9 assessments, include numerical questions. Practicing numerical problems prepares students for these exams and helps them perform well.
- Building logical thinking: Numerical problems often require a logical approach to arrive at the correct solution. By practicing such problems, students enhance their logical reasoning skills.
- Retention of concepts: Actively engaging with numerical problems reinforces the concepts learned in class, improving retention and recall during exams and beyond.
- Connecting theory with practice: Numerical problem-solving bridges the gap between theoretical knowledge and its practical applications. This connection helps students grasp the real significance of the concepts they are studying.
- Facilitating higher-level learning: As students progress to higher classes, numerical problem-solving becomes more complex and advanced. Building a strong foundation in Class 9 prepares them for the challenges ahead.
- Career and academic success: In many fields, such as science, technology, engineering, and mathematics (STEM), problem-solving skills are highly valued. Excelling in numerical problem-solving in Class 9 can pave the way for future success in these fields.

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