Motion Class 9 Numericals with Solutions 2020

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

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 formula 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 numericals 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.

{ S }_{ n }=u+\frac { a }{ 2 } \left( 2n-1 \right)

Questions on Motion Class 9 with Answers

1. A scooter traveling 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

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)

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

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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

D = ut + ½at²

Initial velocity u=0 m/sec

Since the particle starts from rest

D1=½at²

D1=1/2a10²

D1 = 50a ——————– (1)

Now initial velocity for D2 is the final velocity of s1

Let final velocity of D1=initial velocity of D2=v

so, v=u +at

v=at

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

Now

D2 = ut+1/2a(t²)

D2 = (10a)10 + ½a10²

D2 = 100a+50a

D2 = 150a ——————- (2)

Now

D1/D2 = 50a/150a

D2 = 3D1

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