Assertion (A): For uniformly accelerated motion along straight line, the position versus time graph is a straight line.
Reason (R): For uniformly accelerated motion the position in equal intervals of time changes by same amount.
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 false.
For uniformly accelerated motion, position \( x \) is related to time \( t \) by \( x = x_0 + v_0 t + \frac{1}{2} a t^2 \). This is a parabolic equation, so the position-time graph is a parabola, not a straight line.
Reason (R) is false. In uniformly accelerated motion, velocity changes by equal amounts in equal time intervals, but position does not. Position changes by increasing amounts (if starting from rest).
Therefore, both the Assertion and the Reason are false.
Assertion (A): In one dimensional motion, area under velocity-time graph gives change in position i.e., displacement.
Reason (R): In one dimensional motion, area under acceleration-time graph gives final velocity.
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
Area under v-t graph = displacement; Area under a-t graph = change in velocity.
Solution: (A) is true as area under v-t graph is displacement. (R) is false as area under a-t graph gives change in velocity, not final velocity. So, (A) is true and (R) is false.
Assertion (A): A body dropped from a height of \(10 \text{ m}\) from the ground will have the velocity \(5 \text{ m/s}\) at the height of \(5 \text{ m}\).
Reason (R): At the height of \(5 \text{ m}\) from the ground, the acceleration due to gravity is \(5 \text{ m/s}^2\).
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
Formula: \(v^2 = u^2 + 2gs\), \(g \approx 9.8 \text{ m/s}^2\).
Solution: (A) is false; for a fall of \(5 \text{ m}\) from rest, \(v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 5} = \sqrt{98} \approx 9.9 \text{ m/s}\), not \(5 \text{ m/s}\). (R) is false; acceleration due to gravity is approximately \(9.8 \text{ m/s}^2\) or \(10 \text{ m/s}^2\), not \(5 \text{ m/s}^2\).
Assertion (A): A particle moves in a straight line with constant acceleration. The average velocity of this particle can not be zero in any time interval.
Reason (R): For a particle moving in straight line, the average velocity in a time interval is always ((frac{u+v}{2})), where (u) and (v) are initial and final velocities of the particle in given time 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
Solution: (A) is false; average velocity can be zero if displacement is zero (e.g., object returns to start with constant acceleration). (R) is false; the formula \(v_{avg} = \frac{u+v}{2}\) is only valid for constant acceleration, not 'always' for any straight line motion.
Assertion (A): At any instant, acceleration of a body can change its direction without any change in the direction of velocity.
Reason (R): At any instant, direction of acceleration is same as that of direction of change in velocity vector at that instant.
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
Concept: Relationship between acceleration and velocity.
Formula: \(vec{a} = \frac{d\vec{v}}{dt}\).
Solution: (A) is true. For example, a car moving straight can accelerate forward, then brake (accelerate backward) while its velocity direction remains forward. (R) is true; acceleration is defined as the rate of change of velocity, so its direction is the same as the direction of the change in velocity. (R) correctly explains (A) by providing the fundamental definition.
Assertion (A): For motion from rest with constant acceleration distance time graph is a parabola, always with increasing slope.
Reason (R): Speed of the body starting from rest with constant acceleration always increases linearly with time.
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
Formula: For (u=0), \(s = \frac{1}{2}at^2\) and (v = at).
Solution: (A) is true; \(s = \frac{1}{2}at^2\) is a parabola, and its slope (velocity (v=at)) increases with time. (R) is true; (v=at) shows speed increases linearly from rest. (R) correctly explains (A) because the linearly increasing speed implies an increasing slope for the distance-time graph.
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.
Assertion (A): An object moving with a velocity of magnitude \(10 \text{ m/s}\) is subjected to a uniform acceleration \(2 \text{ m/s}^2\) at right angle to the initial motion. Its velocity after \(5s\) has a magnitude nearly \(14 \text{ m/s}\).
Reason (R): The equation \(\vec{v} = \vec{u} + \vec{a}t\) can be applied to obtain \(\vec{v}\) if \(\vec{a}\) is constant.
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
Assertion (A): Given \(u = 10 \text{ m/s}\), \(a = 2 \text{ m/s}^2\), \(t = 5 \text{ s}\). Since \(\vec{u}\) and \(\vec{a}\) are perpendicular, the final velocity magnitude is \(|\vec{v}| = \sqrt{u^2 + (at)^2} = \sqrt{10^2 + (2 \times 5)^2} = \sqrt{100+100} = \sqrt{200} \approx 14.14 \text{ m/s}\). So (A) is true.
Reason (R): The equation \(\vec{v} = \vec{u} + \vec{a}t\) is valid when acceleration \(\vec{a}\) is constant. So (R) is true.
(R) correctly explains (A) as the formula is used due to constant acceleration.
Assertion (A): A coin is allowed to fall in a train moving with constant velocity. Its trajectory is a straight line as seen by observer attached to the train.
Reason (R): An observer on ground will see the path of coin as a parabola.
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
Assertion (A): From the train's frame of reference (inertial, moving with constant velocity), the coin only has vertical motion under gravity, thus appearing as a straight line. So (A) is true.
Reason (R): From the ground frame, the coin has an initial horizontal velocity (that of the train) and vertical acceleration due to gravity, resulting in a parabolic path. So (R) is true.
However, (R) describes a different frame of reference and does not explain why the path is a straight line in the train's frame.
Assertion (A): Two particles start moving with velocities \(vec{v}_1\) and \(vec{v}_2\) respectively in a plane. They can meet only if component of their velocities perpendicular to line joining them are equal.
Reason (R): Relative velocity of a body w.r.t. other body is calculated along the line joining two bodies.
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
Assertion (A): For particles to meet, their relative perpendicular velocity component must be zero, meaning their perpendicular velocities must be equal. Otherwise, they would move apart perpendicular to the line joining them. So (A) is true.
Reason (R): Relative velocity is a vector difference and can be calculated in any direction, not exclusively along the line joining two bodies. So (R) is false.