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1. Distance and Displacement

Practice Questions:

Question 1: easy

The numerical ratio of distance to magnitude of displacement is :

1. Always equal to one

2. Always less than one

3. Always greater than one

4. Equal to or more than one

View Answer

The numerical ratio of distance to the magnitude of displacement depends on the type of motion:

1. For straight-line motion in one direction, the distance and displacement are the same, so the ratio is:
\[
\frac{\text{Distance}}{\text{Displacement}} = 1
\]

2. For any other type of motion (like a curved path or circular motion), the distance is generally greater than or equal to the displacement, making the ratio:
\[
\frac{\text{Distance}}{\text{Displacement}} \geq 1
\]
The ratio is greater than 1 because distance is the total path travelled, while displacement is the shortest straight line between the start and end points.

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Question 2: easy

A body covered a distance of 5 m along a semicircular path. The ratio of distance to displacement is :

1. 11 : 7

2. 12 : 5

3. 8 : 3

4. 7 : 5

View Answer

1. Distance covered: \(d = 5 \, \text{m}\) (along the semicircular path).

2. Displacement: The displacement is the straight-line distance from the starting point to the endpoint. For a semicircle with a radius \(r\):

\[
\text{Diameter} = 2r
\]
Since the distance covered is the semicircle's arc length:
\[
\text{Arc length} = \frac{1}{2}(2\pi r) = \pi r
\]

Therefore, if \(d = 5\):
\[
r = \frac{5}{\pi}
\]
So, the displacement (which is the diameter) is:
\[
\text{Displacement} = 2r = 2 \cdot \frac{5}{\pi} = \frac{10}{\pi} \, \text{m}
\]

3. Ratio of distance to displacement:
\[
\text{Ratio} = \frac{d}{\text{Displacement}} = \frac{5}{\frac{10}{\pi}} = \frac{5 \pi}{10} = \frac{\pi}{2}
\]

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Question 3: easy

Which of the following statement is incorrect ?

1. Displacement is independent of the choice of origin of the axis

2. Displacement may or may not be equal to the distance travelled

3. When a particle returns to its starting point, its displacement is not zero

4. Displacement does not tell the nature of the actual motion of a particle between the points

View Answer

When particle comes back to initial position displacement becomes zero.

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Question 4: easy

A particle moves in x-y plane according to rule \(x = a \sin \omega t\) and \(y = a \cos \omega t\). The particle follows :

1. An elliptical path

2. A circular path

3. A parabolic path

4. A straight line path equally inclined to x and y-axes

View Answer

Squaring and adding the coordinates: \(x^2 + y^2 = a^2 \sin^2\omega t + a^2 \cos^2\omega t = a^2\). This is the standard equation of a circle of radius \(a\).

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Question 5: easy

Assertion (A): In any interval, the magnitude of displacement is always less than or equal to the distance travelled.

Reason (R): For a particle travelling in a straight line with constant acceleration, the magnitude of the change in the velocity during any interval is always less than or equal to the change in the speed during that interval.

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)

2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)

3. (A) is true but (R) is false

4. Both (A) and (R) are false

View Answer

Displacement is the straight-line distance, so its magnitude is always \(le\) distance. When a particle reverses its direction of motion, the magnitude of change in velocity can be greater than the change in speed, so R is false.

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Question 6: easy

Assertion (A): In any interval, the magnitude of displacement is always less than or equal to the distance travelled.

Reason (R): For a particle travelling in a straight line with constant acceleration, the magnitude of the change in the velocity during any interval is always less than or equal to the change in the speed during that interval.

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)

2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)

3. (3) (A) is true but (R) is false

4. (4) Both (A) and (R) are false

View Answer

Assertion (A) is true because displacement is the shortest path, so its magnitude is always less than or equal to the distance travelled. Reason (R) is false. For example, if velocity changes from \(-5 \text{ m/s}\) to \(+5 \text{ m/s}\), change in velocity magnitude is \(10 \text{ m/s}\), but change in speed is \(0 \text{ m/s}\). Thus, (A) is true, (R) is false.

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Question 7: easy

Assertion (A): When a particle is observed from two different inertial reference frames the general shape of the trajectory of particle is same.

Reason (R): The position vector of a particle and its velocity are frame independent quantities.

1. (1) Both (A) \& (R) are true and the (R) is the correct explanation of the (A)

2. (2) Both (A) \& (R) are true but the (R) is not the correct explanation of the (A)

3. (3) (A) is true but (R) is false

4. (4) Both (A) and (R) are false

View Answer

The general shape of a particle's trajectory is invariant across inertial reference frames. However, position vectors and velocities are frame-dependent quantities, changing with the relative motion of frames. Therefore, assertion (A) is true, but reason (R) is false.

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Question 8: easy

Assertion (A): Displacement of a body is vector sum of the area under velocity-time graph.

Reason (R): Displacement is a vector quantity.

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)

2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)

3. (3) (A) is true but (R) is false

4. (4) Both (A) and (R) are false

View Answer

The area under a velocity-time graph indeed represents displacement, considering areas above the time axis as positive and below as negative, which is a vector sum.

Displacement is a vector quantity, meaning it has both magnitude and direction. This vector nature directly explains why the signed area (vector sum) under the velocity-time graph yields displacement.

Both (A) and (R) are true, and (R) correctly explains (A).

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Question 9: easy

Assertion (A): If a body moves on a straight line, magnitude of its displacement and distance covered by it must be same.

Reason (R): Along a straight line, a body can move only in one direction.

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)

2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)

3. (A) is true but (R) is false

4. Both (A) and (R) are false

View Answer

Solution: (A) is false; if a body moves forward and then reverses on a straight line, distance will be greater than magnitude of displacement. (R) is false; a body can change its direction of motion while staying on a straight line (e.g., moving forward, then backward). Since both are false, option (4) is correct.

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