Equations of Motion Class 9 Extra Questions and Answers Keralla Syllabus Physics Chapter 2

The comprehensive approach in Kerala Syllabus 9th Standard Physics Notes Pdf Chapter 2 Equations of Motion Notes Extra Questions and Answers ensures conceptual clarity.

Kerala Syllabus Std 9 Physics Chapter 2 Equations of Motion Extra Questions and Answers

Question 1.
When an object moves at a constant speed and in a straight line, what kind of motion is it?
Answer:
Uniform motion

Question 2.
What is the SI unit of displacement?
Answer:
The SI unit of displacement is meter (m).

Question 3.
A body moves in a circle of radius 2R. What is the distance covered and displacement of the body after two complete rounds?
Answer:
Distance covered after 2 complete rounds
= 2 × circumference
= 2 × 2π (2R) = 8π R
After 2 complete rounds, the body comes back to its initial position. So, Displacement = 0

Question 4.
Define the term velocity. What is its SI unit’?
Answer:
Velocity is a measure of the rate of change of an object’s position with respect to time. Its SI unit is meters per second (m/s).

Question 5.
Is it a scalar or vector quantity?
Answer:
Velocity is a vector quantity as it includes both magnitude (speed) and direction.

Question 6.
When a both said to have uniform velocity?
Answer:
A body is said to have uniform velocity when it maintains a constant speed and direction over time. This means that it covers equal displacements in equal intervals of time in a consistent direction.

Question 7.
What is the SI unit of acceleration?
Answer:
The SI unit of acceleration is meters per second squared (m/s²).

Question 8.
What is the acceleration of a body moving with uniform velocity?
Answer:
The acceleration of a body moving with uniform velocity is zero.

Question 9.
What do you mean by the term retardation or negative acceleration?
Answer:
Retardation, or negative acceleration, occurs when the velocity of a body decreases overtime. Examples of such motion include a car decelerating to stop at a red light or a ball thrown upwards slowing down due to gravity.

Question 10.
State the relationship connecting u, y, a and t for an accelerated motion. Give an example of motion in which acceleration is uniform.
Answer:
The relationship connecting initial velocity (u), final velocity (y), acceleration (a), and time (t) for an accelerated motion is given by the equation:
y = u+at. An example of uniform acceleration is an object in free fall near the surface of the Earth due to gravity.

Question 11.
Impossible position – time graph is

Answer:
B

Question 12.
Draw position – time graph

Time (s)	0	1	2	3	4
Position (m)	0	5	10	15	20

Answer:

Question 13.
a) Draw the velocity – time graph based on the table.

Time (s)	0	5	10	15	20
Velocity (m/s)	20	20	20	20	20

b) Find out the displacement between 10 s and 20s from the graph.
Answer:

b) Displacement = Area of ABCD
= AB × AD
=20 × 10 = 200 m

Question 14.
Write the equations of motion. What does each letter indicate?
Answer:
v = u + at
s = ut + 1/2 at²
v² = u² + 2as
u- Initial velocity, y- final velocity, a- acceleration, t- time, s-Distance

Question 15.
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?
Answer:
Initial velocity, u = O
Time taken, t =8 s, Acceleration, a =3 m/s²
∴ Distance covered during the given time,
s = ut + \(\frac{1}{2}\) at² [from second equation of motion]
= 0 × 8 + \(\frac{1}{2}\) × 3 × (8)² = \(\frac{1}{2}\) × 3 × 64 = 96 m

Question 16.
What does the equation v = u + at represent?
Answer:
Relationship between velocity and time.

Question 17.
Which of the following represents a positive uniform acceleration?

Answer:
(a)

Question 18.
Is velocity a scalar or a vector quantity?
Answer:
Velocity is a vector quantity.

Question 19.
Can displacement be zero even if the distance travelled is not zero?
Answer:
Yes, displacement can be zero if the object returns to its initial position.

Question 20.
What is displacement?
Answer:
Displacement is the shortest distance from the initial to the final position of an object.

Question 21.
An object is starting from rest and moving with an acceleration of 2 m/s². What will be its velocity after 10s?
Answer:
v = u + at
=0 + 210 = 20 m/s²

Question 22.
A car is moving with a uniform velocity of 20 m/s. It is brought to rest by applying brake uniformly for 5 s.
a) What is the retardation of the car?
b) What is the displacement of the car during this time interval?
Answer:
a) u = 20 m/s, t = 5 s, v = 0 m/s
a = \(\frac{v-u}{t}\) = \(\frac{0-20}{5}\) = \(\frac{-20}{5}\) = -4m/s²
So retardation is 4 m/s²

b) s = ut + \(\frac{1}{2}\)at² = 20 × 5 + \(\frac{1}{2}\)(-4 × 5²) = 100 – 50 = 50m

Question 23.
A car starts from rest and covers 25m in 5s. Calculate the acceleration of the car.
Answer:
u = 0 m/s, s = 25 m,t = 5 s
s = ut + \(\frac{1}{2}\) at²
25 = 0 × 5 + \(\frac{1}{2}\) × a × 5²
25 = \(\frac{25 a}{2}\), a = 2 m/s²

Question 24.
What is velocity? How does it differ from speed?
Answer:
Velocity is a vector quantity that describes the rate of change of displacement of an object and includes direction. Speed is a scalar quantity that only measures how fast an object is moving, regardless of direction.

Question 25.
Describe a situation where an object can have a constant speed but a changing velocity.
Answer:
An object moving in a circular path at a constant speed has a changing velocity because the direction of its motion is continually changing even though its speed remains constant.

Question 26.
A car travelling with 20 m/s velocity comes to rest when brake is applied for 8 s.
a) What is the final velocity of the car?
b) Find the retardation of the car.
Answer:
u = 20 m/s, v = 0, t = 8s
a) Final velocity, v = 0
b) a = \(\frac{v – u}{t}\) = \(\frac{0 -20}{5}\) = \(\frac{-20}{5}\) = -2.5 m/s²

Question 27.
What does a negative acceleration indicate about an object’s motion?
Answer:
Negative acceleration, also known as deceleration, indicates that the object is slowing down. This means that the velocity of the object is decreasing over time. For instance, when a car applies brakes, it experiences negative acceleration as its speed reduces.

Question 28.
Describe a scenario where displacement is zero but distance is not.
Answer:
If a person walks 5 meters east and then 5 meters west, their displacement is zero because they end up at their starting point. However, the distance travelled is 10 meters as it is the total path length covered. This illustrates how displacement considers direction, while distance does not.

Question 29.
Match the following.

Answer:

Question 30.
Motion of a car is shown below using a diagram.

a) What is the distance covered by the car between B and C?
b) Find the velocity of the car during this period.
c) Among the following which are the time interval where car travels in uniform velocity (0s → 2s, 2s → 4s, 4s → 6s)?
Answer:
a) 20m
b) Velocity = \(\frac{Displacement}{Time}\)
= \(\frac{20}{2}\) = 10 m/s
c) 0s → 2s, 4s → 6s

Question 31.
a) How is velocity different from speed?
b) If a car travels 60 km north in 2 hours, what is its average velocity?
Answer:
a) Speed is a scalar quantity that measures how fast an object is moving, regardless of direction. Velocity however, includes both the
speed and the direction of the object’s motion.
b) Average velocity = \(\frac{Displacement}{Time}\)
Here, displacement =60km north and time = 2 hours. So, average velocity 60 km / 2
hours = 30km/h north.

Question 32.
Explain how velocity and acceleration are related.
a) What happens to velocity if acceleration is zero?
b) A car moving at 10 m/s increases its speed to 30 m/s in 5 seconds. Calculate the acceleration.
Answer:
a) If acceleration is zero, it means the velocity remains constant
b) Acceleration = (Final velocity – Initial velocity) / Time
= (30m/s – 10 m/s) / 5s
= 20 m/s / 5s – 4 m/s².

Question 33.
Observe the given graphs

a) Which among the graphs represents the body at rest.
b) Other graph represents _(uniform acceleration / uniform velocity)
c) What is the velocity of the both shoe n in graph (i)
d) How much will be the acceleration for the body represented by graph (ii)
Answer:
a)graph(i)
b) uniform velocity
e) O (because the body is at rest)
d) O (because the body is in uniform velocity, it has no acceleration)

Question 34.
The area under velocity time graph gives.
(Velocity, displacement, distance, time)
Answer:
Displacement

Question 35.
Position – time graph of a car is given.

The distance covered by the car in 6s is …. (10 m, 40 m, 30m, 50 m)
Answer:
30m

Question 36.
What does the area of the shaded portion of the graph represent
(Acceleration, Force, Velocity, Displacement)

Answer:
Displacement

Question 37.
Complete the table given below based on the graph.

Time (s)	2	–	–	–	–
Velocity (m/s)	10	–	–	–	–

Answer:

Time (s)	2	4	6	8	10
Velocity (m/s)	10	20	30	40	50

Question 38.
A graph is a two-dimensional picture that can be drawn by relating two quantities to each other. Write down two uses of graphs.
Answer:
Gaining information using a graph is more simple than obtaining it by mathematical calculation. The equations can be formulated from graph.

Question 39.
The velocity-time graph of an object with non uniform velocity is given below.

Observe the graph and answer the following questions.
a) What is the velocity of the object in the 4^th second?
b) Find the instant at which it attains maximum velocity.
Answer:
a) 10 m/s
b) 6^th second

Question 40.
The velocity time graph of a moving object is given below. Answer the following questions based on the graph.

a) The motion of the object is ……… (uniform/non uniform)
b) Find acceleration of the body from O to A.
Answer:
Non uniform
b) u = 0, v = 2m/s, t = 4s
a = \(\frac{v-u}{t}\) = \(\frac{2-0}{4}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\) = 0.5 m/s²

Question 41.
The velocity-time graph of a uniformly accelerated body is shown below. Answer the following questions from the graph.

a. What is initial velocity of the body?
b. Calculate displacement of the body in first 4s.
Answer:
a. Initial velocity u = 0
b. Displacement = Area of the triangle OAB
= 1/2 × base × altitude
= 1/2 × 4 × 40
= 80 m

Question 42.
The details regarding the travelling of a car with uniform velocity are tabulated.
a) Draw the velocity-time graph using the data.
(The graph paper given in the question paper can be used to answer. Attach it to the answer sheet)

Time (s)	0	2	4	6	8	10
Position (m/s)	10	10	10	10	10	10

b) Calculate the dLspbcenwnt of the car from the 3^rd second to the 10^th second.
Answer:

b) Displacement = Area of the rectangle portion between 3^rd and 10^th = 10 × 7 = 70 m

Question 43.
The position of a moving car at various times is given in the following table.
a. Write the appropriate scale to draw the position- time graph using the above table.
X – Axis 1 cm = ______
Y – Axis 1 cm = ______
b. Draw the position-time graph (Use the graph sheet attached to the question paper. Attach it to the answer sheet).
c. Calculate the distance travelled by the ear within 6 second.

Time (s)	0	2	4	6	8	10
Position (m)	10	15	20	25	30	35

Answer:
a) X – Axis 1 cm = 2s
Y – Axis 1 cm = 10m

c) 25 m

Question 44.
Velocity-time graph of a moving car of 1000kg is given.

a. Find out the acceleration of the car?
b. What is the force experienced by the car?
Answer:
a) u = 0
v = 20 m/s
t = 10 s
a = \(\frac{v-u}{t}\) = \(\frac{20-0}{10}\) = 2 m/s²
b)F = ma = 1000 × 2 = 2000N

Question 45.
The position of an object in motion at various times is given in the following table

Time (s)	0	2	4	6
Position (m/s)	0	4	8	12

a) Using the data in the table suggest a suitable scale for drawing the position- time graph
b) Draw the position -time graph in the graph paper given
c) Analyse the graph and write down the characteristics of motion of the object. (Uniform motion/Non uniform motion)
Answer:
a) X axis 1 cm = 2s
Y axis 1 cm = 4m

c) Uniform motion

Question 46.
What is the first equation of motion?
Answer:
v = u + at

Question 47.
In the first equation of motion v= u + at, what does y = u + at represent?
Answer:
Initial velocity

Question 48.
If a car starts from rest, what is its initial velocity (u)?
Answer:
O m/s

Question 49.
Which equation of motion can be used to find the final velocity directly if the initial velocity, acceleration, and displacement are known?
Answer:
The third equation of motion, v² = u² +2 as

Question 50.
An object starts from rest and travels with an acceleration of 5 m/s²
a. Write down an equation that helps to find the velocity after a given time.
b. What ¡s the velocity of the car at a time t = 3s.
Answer:
a) v = u + at
b) u = 0, a = 5 m/s², t = 3s
v = u + at = 0 + 5 × 3 = 15 m/s

Question 61.
A car starting from rest travels 100 m in 5 s with uniform acceleration. Find the acceleration of the car?
Answer:
u = 0 m/s
t = 5s
s = 100m, We know, s = Ut + \(\frac{1}{2}\)at²
100 = 0 × 5 + 1/2 × a × 5 × 5
100 = 1/2 × 25 × a
a = \(\frac{200}{25}\) = 8 m/s²

Question 62.
Write down any two equations of motion. What do each letter in these equations indicate?
Answer:
v = u + at
v² = u² + 2as
Where, u = Initial velocity
v = Final velocity
a = Acceleration
s = Displacement
t = Time

Question 63.
A car starting from rest attains a velocity of 20 m/s within 4 s.
a. Calculate acceleration of the car.
b. Find displacement of the car with in 4 s.
Answer:
a) u = 0 m/s,v = 20 m/s, t = 4s
a = \(\frac{v-u}{t}\) = \(\frac{20-0}{4}\) = \(\frac{20}{4}\) = 5 m/s²

b) s = ut + \(\frac{1}{2}\) at²
= 0 × 4 + \(\frac{1}{2}\) × 5 × 4²
= 0 + \(\frac{1}{2}\) × 5 × 16
= 5 × 8 = 40 m

Question 64.
A ball thrown vertically upward reached a maximum height of 20 m .
a) What was the velocity of the stone at the instant of throwing up?
b) How much time did the ball take to reach the height 20 m ?
Answer:

a) v² = u² + 2as
u² = v² – 2as
= 0² – 2 × -10 × 20 = 400
u = 20 m/s

b) v = u + at
t = \(\frac{v-u}{a}\) = \(\frac{0-20}{-10}\) = 2s

Question 65.
If a train traveling at a speed of 20 m/s comes to rest in 4s after breaking.
a) calculate acceleration of the train.
b) find the distance (raelIed by the train after applying the brakes.
Answer:
a) u = 20m/s
v = 0 m/s
t = 4s
a = (v – u)/t
= (0.20)/4 = -5 m/s²

b) s = ut + 1/2 at²

Question 66.
If a hall falls vertkaIl downward from the top of a building of 40 m height with a velocity of 10 m/s (g = 10 m/s²),
a) find the eIoclly of the ball at 1s.
b) find the velocity of the ball just before reaching the ground.
Answer:
a) s = 40m, u = 10 m/s, a = 10 m/s², t = 1 s
v = u + at = 10 + 10 × 1 = 20 m/s

b) v² + u² = 10² + 2 × 10 × 40 = 900
v = 30m / s

Question 67.
A car came to rest when break was applied for 4 seconds to get a retardation of 3 m/s². Calculate how far the car would have travelled after applying the break.
Answer:
Here, a = -3 m/s²
t = 4s,v = 0 : u = ? s = ?
v = u + at
0 = u + (- 3 × 4)
u = 12 m/s
v² = u² + 2as
0 = (12 × 12) + 2 × (-3) × s
6s = 144
s = 144/6 = 24 meter
The distance travelled by the car = 24 m

Question 68.
A car moving with uniform acceleration and changes its velocity from 10 m/s to 50 m/s in 4 seconds, Then.
a) Find out the acceleration of car?
b) Find out the displacement in this time?
Answer:
a) a = \(\frac{v-u}{4}\)
a = \(\frac{50 – 10}{4}\) = 10 m/s²

b) s = ut + \(\frac{1}{2}\) at²
= 10 × 4 + \(\frac{1}{2}\) (10 × 4²)
= 40 + 80 = 120 m

Question 69.
A car is moving with a uniform velocity of 20 m/s. it is brought to rest by applying brake uniformly for 5 s.
a) What is the retardation of the car?
b) What is the displacement of the car during this time interval?
Answer:
a) u = 20 m/s, t = 5 s, y = 0 m/s .
a = \(\frac{v-u}{t}\) = \(\frac{0-20}{5}\) = \(\frac{-20}{5}\) = -4 m/s²

b) s = ut + \(\frac{1}{2}\) × -4 × 5²
=20 × 5 + \(\frac{1}{2}\) × – 4 × 25 = 100 – 50 = 50m

Question 70.
A ball thrown vertically upward reached a maximum height of 20 m.
a) What was the velocity of the stone at the instant of throwing up?
b) How much time did the ball take to reach the height 20 m?
Answer:
a) v² = u² + 2as
u² = v² – 2as
= 0² – 2 × -10 × 20 = 400
u = 20 m/S

b) v = u + at
t = \(\frac{V-u}{a}\) = \(\frac{0-20}{-10}\) = 2s

Question 71.
a) Write down the equation showing position time relation.
b) Write down the equation showing velocity time relation.
c) A car starting from rest travels 200 m in 5 s, with uniform acceleration. Find the acceleration of the car.
Answer:
a) s = ut + 1/2 at²
b) v = u + at
c) u = 0 m/s, t = 5s,s = 200 m
s = ut + \(\frac{1}{2}\) at²
200 = 0 × 5 + \(\frac{1}{2}\) × a × 5²
200 = \(\frac{25 a}{2}\)
a = \(\frac{200 × 2}{25}\) = 16 m/s²