Assertion (A): If initial velocity is negative but acceleration is positive then displacement of a particle can never be positive.
Reason (R): If initial velocity is negative and acceleration is positive then motion must be retarded throughout.
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) is false. If initial velocity is negative and acceleration is positive, the particle can eventually move in the positive direction, leading to a positive displacement (e.g., \(s = v_0 t + \frac{1}{2}at^2\) can be positive for large \(t\)).
Reason (R) is false. Motion is initially retarded, but as velocity becomes positive (due to positive acceleration), the motion becomes accelerated.
Assertion (A): \(|\Delta v| / \Delta t\) and \(\Delta |v| / \Delta t\) are same if particle is moving in one dimension.
Reason (R): In one dimensional motion there is no component of acceleration perpendicular to 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
Assertion (A) is true if the particle does not reverse its direction of motion; otherwise, it is generally false. Assuming this condition for 'moving in one dimension' for the purpose of the question.
Reason (R) is true; in one dimension, velocity and acceleration are collinear.
R is not a correct explanation for A, as the absence of perpendicular acceleration components does not directly imply the equality of magnitude of average acceleration and average rate of change of speed when velocity changes direction.
Thus, both A and R are true, but R is not the correct explanation of A.
Assertion (A): If velocity of a particle moving in a straight line is zero at a point, its acceleration will be zero at that point.
Reason (R): Wherever \(a = v \frac{dv}{dx}\) holds, \(v = 0 \Rightarrow a = 0\).
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) is false. For example, a ball thrown vertically upwards has zero velocity at its highest point, but its acceleration is \(g\).
Reason (R) is false. While the formula \(a = v \frac{dv}{dx}\) is correct, the implication \(v = 0 \Rightarrow a = 0\) from this formula is physically incorrect as \(dv/dx\) itself might be undefined or lead to physically inconsistent conclusions when \(v=0\). Physically, \(a = dv/dt\), which can be non-zero when \(v=0\).
Assertion (A): A body is thrown vertically upwards with an initial speed \( 25 \text{ m/s} \) from a position 1. It falls back to position 1 after some time. During this time duration, total change of velocity of the body is zero.
Reason (R): Average acceleration of the body during this time is zero.
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. Initial velocity is \( +25 \text{ m/s} \). Final velocity at the same position is \( -25 \text{ m/s} \). The change in velocity is \( \Delta \vec{v} = (-25) - (+25) = -50 \text{ m/s} \). Reason (R) is false. Since the change in velocity \( \Delta vec{v} \) is not zero, and \( \Delta t \) is a finite time, the average acceleration \(\ vec{a}_{avg} = \frac{\Delta \vec{v}}{\Delta t} \) is also not zero. It is \( -g \).
Therefore, both the Assertion and the Reason are false.
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.