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.