Understanding Motion through Experience Class 9 Question Answer Advanced Science Chapter 2

Chapter-wise Class 9 Advanced Science Solutions and NCERT Class 9 Advanced Science Chapter 2 Understanding Motion through Experience Question Answer are useful for focused study.

Understanding Motion through Experience Class 9 Questions and Answers

Understanding Motion through Experience Question Answer Class 9

Quick Test

Question 1.
Why do we feel pushed backward when a bus suddenly starts moving?
Answer:
When the bus starts suddenly, our body wants to stay at rest (because of inertia). But the bus moves forward, so we feel pushed backward.

Question 2.
Can an object be at rest for one observer but moving for another?
Answer:
Yes, imagine you are sitting in a moving train.

• For your friend outside the train – you are moving.
• For your friend sitting with you in the train – you are at rest.

So, motion and rest are relative (it depend on the observer).

Question 3.
How do athletes decide the best angle to throw a ball?
Answer:
They throw the ball at an angle (usually around 45°) so that it travels the maximum distance. This is studied in projectile motion.

Question 4.
Can we measure motion using simple tools available in the classroom?
Answer:
Yes, we can measure motion with simple tools such as

• use a meter scale or measuring tape to measure distance.
• use a stopwatch to measure time.

Then we can find speed = Distance/Time.

Check Your Understanding

Question 1.
Define a frame of reference in your own words.
Answer:
A frame of reference is a system of coordinate axes (like x,y,z axes) attached to an observer or an object, which is used to describe the position,
motion and other physical quantities of another object. It helps us to measure how something is moving relative to that observer.

Question 2.
Give two real-life examples where motion depends on the observer.
Answer:
Example 1:
A passenger sitting inside a moving train appears to be at rest to another passenger sitting in the same train (both have the same frame of reference). But to a person standing outside on the platform, the same passenger appears to be moving with the speed of the train.

Example 2:
When you are sitting in a moving car and look at the trees outside, the trees appear to be moving backwards (from your frame). But a person standing on the road sees the trees at rest and the car moving forward.

Question 3.
Why does a person sitting in a moving train appear at rest to another passenger?
Answer:
Because both persons (the one sitting and the observer) are in the same frame of reference (the moving train). They have no relative motion with respect to each other. Both are moving with the same velocity relative to the ground, so relative velocity between them is zero. Hence, the person appears at rest to the other passenger.

Question 4.
Classify the following as scalar or vector quantities:
speed, velocity displacement, distance acceleration and mass.
Answer:

Quantity	Type
Speed	Scalar
Velocity	Vector
Displacement	Vector
Distance	Scalar
Acceleration	Vector
Mass	Scalar

Question 5.
Explain the difference between distance and displacement with an activity diagram.
Answer:

Distance	Displacement
It is actual path length travelled by a body.	It is the shortest straight line distance between the initial and final position.
It is a scalar quantity.	It is a vector quantity.

Activity:
A student walks in a straight line from point A to point B, covering 5 m distance. He then turns around and walks back from B to A, covering another distance 5 m.

Activity diagram:

Observation:

• Total distance covered = AB + BA = 5 m + 5 m = 10 m (actual path length)
• Net displacement = 0 m (since the final position coincides with the initial position A)

This clearly shows that when a body returns to its starting point, the distance travelled is not zero, but the displacement becomes zero.

Question 6.
Give two everyday examples of vector quantities.
Answer:
Velocity and force

Question 7.
Draw two vectors of 4 units East and 3 units North and find the resultant using the triangle method.
Answer:

Magnitude of resultant R = \(\sqrt{(4)^2+(3)^2}\)
= \(\sqrt{16+9}\)
= \(\sqrt{25}\)
= 5 unit

Question 8.
Explain how vector subtraction is performed graphically.
Answer:
Suppose A and B are two vectors and vector B is to be subtracted from vector A. The subtraction of B from A is same as addition of – B to A

i.e. A – B = A + (- B)

Question 9.
Draw two opposite vectors of equal magnitude. Calculate its resultant.
Answer:

Let A = 5 units and B = 5 units where B is opposite in direction to A.
So, B = – A
Resultant = A + (- B)
= 5 + (- 5) = 0
The resultant has zero magnitude and no direction.

Question 10.
A body starts from rest and accelerates at 4 m/s². Find the distance travelled in the 6th second.
Answer:
Given, u = 0 and
a = 4 m/s²
The distance travelled in nth second is given by
s_n = u + \(\frac{a}{2}\) (2n – 1)
= 0 + \(\frac{4}{2}\) (2 × 6 – 1)
= 2 (12- 1) = 22 m

Question 11.
A car with initial velocity 8 m/s accelerates at 2 m/s². Find the distance covered in 5th second.
Answer:
Given, u = 8 m/s and
a = 2 m/s²
s_n = u + \(\frac{a}{2}\) (2n – 1)
= 8 + \(\frac{2}{2}\) (2 × 5 – 1)
= 8 + (10 – 1)
= 17 m

Understanding Motion through Experience Class 9 MCQ

Question 1.
Which of the following statements is true for a car moving on the road?
(A) With respect to the frame of reference attached to the ground, the car is at rest.
(B) With respect to the frame of reference attached to the person sitting in the car, the car is at rest.
(C) With respect to the frame of reference attached to the person outside the car, the car is at rest.
(D) None of the above
Answer:
(B) With respect to the frame of reference attached to the person sitting in the car, the car is at rest.

For a car in motion, if we describe this event w.r.t. a frame of reference attached to the person sitting inside the car, the person will appear to be at rest as the person inside the car (observer) is also moving with same velocity and in the same direction as car.

Question 2.
The displacement of a car is given as – 240 m. Here, negative sign indicates
(A) direction of displacement.
(B) negative path length.
(C) position of car is at point whose coordinate is – 120.
(D) no significance of negative sign.
Answer:
(A) direction of displacement.
The displacement of a car is given as – 240 m. Here, negative sign indicates the direction of displacement.

Question 3.
A car is moving with a velocity of 30 ms^-1. On applying the brakes, the velocity decreases to 15 ms^-1 in 2 s. The acceleration of the car is
(A) + 7.5 ms^-2
(B) – 7.7 ms^-2
(C) – 7.5 ms^-2
(D) + 15 ms^-2
Answer:
(C) – 7.5 ms^-2

Question 4.
A and B are two inclined vectors. R is their sum. Choose the correct figure for the given description.
(A)
(B)
(C)
(D)
Answer:
(B)

Here, v = 15 m/s,
u =30 ms^-1 and
t = 2s
Using relation, v = u + at
⇒ 15 = 30 + a × 2
Acceleration of the car,
a = \(\frac{(15-30) \mathrm{ms}^{-1}}{2 \mathrm{~s}}\)
=- \(\frac{15}{2}\) ms^-2
= – 7.5 ms^-2

Assertion-Reason Questions

Directions (Question Nos. 1 and 2): In each of the following questions, a statement of Assertion is given by the corresponding statement of Reason. Mark the correct answer as
(A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(B) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(C) Assertion is true but Reason is false.
(D) Assertion is false but Reason is true.

Question 1.
Assertion (A): Displacement and velocity are vector quantities.
Reason (R): Displacement and velocity have both magnitude and direction.
Answer:
(A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
Quantities having both magnitude and direction are called vector quantities.
e.g. Displacement, velocity, force

Question 2.
Assertion (A): Two vectors are said to be the negative of each other if their magnitudes are equal but directions are opposite.
Reason (R): When two vectors act in mutually opposite directions, the magnitude of their resultant vector will be equal to the difference of their magnitudes, |A – B|.
Answer:
(B) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Understanding Motion through Experience Class 9 Extra Questions and Answers

Case/Source Based Questions

Question 1.
Sita lives in a city where her house, bus stop, traffic signal and.school are located at specific point given below

Points A at (1, 1), flat (3, 1),C at (3, 5) and D at (4, 5) (all the values mentioned in the graph are in km) represent Sita’s house, bus stop, traffic signal and school respectively. In the morning Sita travels from A to B on foot, the B to D via C in the school bus. (all the values mention in the graph are in km), Then calculate
(i) Distance travelled by Sita on foot.
(ii) Distance travelled by Sita by the school bus.
(iii) Magnitude of total displacement of Sita from her house to the school.
Answer:
(i) Distance AB = 3 – 1 = 2 km

(ii) Distance BC = 5 – 1 = 4 km
Distance CD = 4 – 3 = 1 km
Total distance BC + CD
= 4 + 1 = 5 km

(iii) Change ∆x = 4 – 1 = 3 km (East)
Change ∆y = 5 – 1 = 4 km (North)
Displacement magnitude = \(\sqrt{\Delta x^2+\Delta y^2}\)
= \(\sqrt{(3)^2+(4)^2}\)
= \(\sqrt{9+16}\)
= \(\sqrt{25}\) = 5 km

Very Short Answer Type Questions

Question 1.
Give the example of uniformly accelerated linear motion.
Answer:
Motion of a body under gravity.

Question 2.
What is the non-inertial frame?
Answer:
A frame that is accelerating (speeding up, slowing down or changing direction) is called a non-inertial frame.

Question 3.
Define the scalar quantities?
Answer:
Those physical quantities that have magnitude only is called the scalar quantities.

Question 4.
Write the equation of motion which describes the distance in the n th second?
Answer:
s = u + \(\frac{a}{2}\) (2n – 1)

Short Answer Type Questions

Question 1.
Explain the meaning of motion. Why do we need a reference point to describe whether an object is in motion or at rest?
Answer:
Motion is the change in the position of an object with respect to time. Motion can be slow or fast, straight or curved, uniform or non-uniform. We observe motion in our daily life everywhere-a moving car, flying birds, flowing river, etc.

To describe motion, we must specify a reference point (also called origin). Without a fixed reference point, it is impossible to say whether an object is moving or at rest. Rest and motion are relative terms. The same object can appear at rest for one observer and in motion for another.

Question 2.
What is a frame of reference? Differentiate between inertial frame and non-inertial frame of reference with examples. Why are inertial frames important in physics?
Answer:
A frame of reference is a point or an object with respect to which the position and motion of another object is observed and measured. It acts as a reference point or origin for describing the state of motion of any body.

There are mainly two types of frames of reference

(i) Inertial Frame of Reference:
A frame that is at rest or moving with constant velocity (uniform motion in a straight line) is called an inertial frame. In such frames, Newton’s laws of motion hold true without any modification. For example, a train moving with constant speed on a straight track or a laboratory fixed on the ground can be considered approximately inertial frames.

(ii) Non-inertial Frame of Reference:
A frame that is accelerating (speeding up, slowing down or changing direction) is called a non-inertial frame. In non-inertial frames, Newton’s laws do not hold directly and we need to introduce pseudo (fictitious) forces to explain the motion. For example, a bus that suddenly starts moving or applies brakes. A person standing inside the bus feels pushed backward when the bus accelerates forward. This is due to the non-inertial nature of the bus frame.

Long Answer Type Questions

Question 1.
State and describe the parallelogram law of vector addition with the help of neat labelled diagram.
Answer:
Parallelogram law of vector addition states that if two vectors acting on a particle at the same time are represented in magnitude and direction by two adjacent sides of a parallelogram drawn from a point, then their resultant vector is represented in magnitude and direction both by the diagonal of the parallelogram drawn from the same point.

Consider two vectors A and B that lie in a plane as shown in Fig. (a). From a common point O, draw a vector OP equal and parallel to A and vector OR equal and parallel to B. Complete the parallelogram OPQR as shown in Fig. (b). Then, the resultant vector is given by OQ.

According to parallelogram law of vector addition,
OQ = OP + OR or
Resultant vector, R = A + B.

Question 2.
Derive the expression for the distance travelled by a uniformly accelerated body during the nth second i.e, s_n = u + \(\frac{a}{2}\) (2n – 1).
Answer:
From the second equation of motion
s = ut + \(\frac{1}{2}\) at²
where,
u = initial velocity
a = acceleration
t = time and
s = displacement in time t
Distance travelled in n seconds,
s_n = un + ½ an²
Distance travelled in (n – 1) second
S_n-1 = u(n – 1) + ½ a(n – 1)²
Distance travelled in the nth second = s_n – s_n-1
= [½ an²] – [u (n – 1) + ½ a (n – 1)²]
Solving,
= un + ½ an² – un + u – ½ a(n² – 2n + 1)
= u + ½ a [2n – 1]
s = u + \(\frac{a}{2}\) (2n – 1)

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