Physics Numerical Class 9 Motion with Solutions

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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.

motion class 9 numericals

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

  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.

Click for Answer/Explanation

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.

Click for Answer/Explanation

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?

Click for Answer/Explanation

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

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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)

Click for Answer/Explanation

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.

Click for Answer/Explanation

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?

Click for Answer/Explanation

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 a distance D2 in the next 10 seconds then

(a) D2 = D1

(b) D2 = 2D1

(c) D2 = 3D1

(d) D2 = 4D1

Click for Answer/Explanation

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)

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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.

Click for Answer/Explanation

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.

Click for Answer/Explanation

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 the SUV is given below. The mass of the SUV is 1500 kg.class 9 physics numerical

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?

Click for Answer/Explanation

(a) Find the area of the 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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Click for Answer/Explanation

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.

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.

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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 with a radius of 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

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

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

      Reply

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

    Reply

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  10. 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 .

    Reply

    • 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 ✔️

      Reply

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