Practice NEET 1. Distance and Displacement Questions
Question 1:
easy
The numerical ratio of distance to magnitude of displacement is :
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.
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:
\[
\text{Ratio} = \frac{d}{\text{Displacement}} = \frac{5}{\frac{10}{\pi}} = \frac{5 \pi}{10} = \frac{\pi}{2}
\]
A particle moves in x-y plane according to rule \(x = a \sin \omega t\) and \(y = a \cos \omega t\). The particle follows :
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 \(a\).
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.
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.
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.
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.
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.
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.
Assertion (A): Displacement of a body is vector sum of the area under velocity-time graph.
Reason (R): Displacement is a vector quantity.
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).
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.
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.