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.

motion class 9 numericals

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 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 nth second use this formula.

Sn = u + 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 travelling at 10 m/s speed 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 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


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

a) the acceleration

b) distance travelled by car (assume 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


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 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 = 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 the u for D2


D2 = ut+1/2a(t²)

D2 = (10a)10 + ½a10²

D2 = 100a+50a

D2 = 150a ——————- (2)


D1/D2 = 50a/150a

D2 = 3D1

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


Click on Page 2 for More Questions

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