# 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, firstly 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

1. v = u + at
2. s = ut + ½ × a × t²
3. v² = u² + 2as

Here u = initial velocity

v = final velocity

t = time

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.

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/s2. Find its speed 5 s after it was dropped.

Apply, v = u + a t

v = 0 + a t

v = 10 × 5

v = 50 m/s

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

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 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:

v = u + at

a = 15/2 = 7.5 m/s²

s = distance traveled

Equation of motion:

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

v= 20 m/s

u=0 m/s

v² = u²+2as

Here u = 0

Then v²=2as

20×20=2×a×0.06

a=-3333.33 m/s2

Another Method:

Apply the concept of ΔKE = Work Done

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?

v =u+at

Here u=10m/s, a=2m/s², t = 2 sec

v=10+2×2 =10+4 =14m/s

7. A particle experiences constant acceleration for 20 seconds after starting from rest. If it travels a distance D1 in the first 10 seconds and distance D2 in the next 10 seconds then

(a) D2 = D1

(b) D2 = 2D1

(c) D2 = 3D1

(d) D2 = 4D1

We know that

distance (D) = ut + ½at²

Initial velocity u = 0 m/sec

Since the particle starts from rest

D1=½at²

D1=1/2×a×(10)²

D1 = 50a ——————– (1)

Now initial velocity for D2 is the final velocity after covering D1 distance

Final velocity of D1 = Initial velocity of D2

so, v=u +at

v=at (Final velocity after covering D1 distance)

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

This v is equal to u for D2

Now

D2 = ut+1/2a(t²)

D2 = (10a)10 + ½a10²

D2 = 100a+50a

D2 = 150a ——————- (2)

Now

D1/D2 = 50a/150a

D2 = 3D1

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

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

We Know,

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.

Distance = speed × time

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

10. The velocity-time graph of SUV is given below. The mass of the SUV is 1500 kg.

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 triangle ABE

(b) F = ma

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

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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

${ S }_{ n }=u+\frac { a }{ 2 } \left( 2n-1 \right)$ ${ S }_{ A }=0+\frac { { a }_{ 1 } }{ 2 } \left( 2\times 5-1 \right) =\frac { 9{ a }_{ 1 } }{ 2 }$ ${ S }_{ B }=0+\frac { { a }_{ 2 } }{ 2 } \left( 2\times 3-1 \right) =\frac { 5{ a }_{ 2 } }{ 2 }$ ${ S }_{ A }:{ S }_{ B }=9:5$

Solution of Q13.

Average Velocity = S/(t1 + t2)

t1 = s/2 ÷ 3

t2 = s’/V’

⇒ s/2 = 4.5× t2/2  +7.5× t2/2

s =12t2

⇒ 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.

So, 1 radian = 180/π = 57.3 degrees

16. A train travels with 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 m 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/s2 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/s2, 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 of a radius of 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

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

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

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

1. Nice but I need more questions

2. There is no question number 16

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

3. 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

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

• Plz do check the youtube channel. CBSE Portal

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

6. Provide q 15 soloution

7. 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

8. 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.
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• 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.

9. 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 .

• 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

I THINK

• 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

10. plz upload more questions for practice

11. 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

12. 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

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

• Yes I also

13. Tooo easy for me
Good questions
Thank u
No doubt

14. Good questions

• can you show slove in this question