Assertion (A): If initial velocity is negative and acceleration is positive then motion is retarded (initially).
Reason (R): If initial velocity is negative but acceleration is positive then displacement of a particle can never be positive.
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 if velocity and acceleration have opposite signs (negative velocity, positive acceleration), the object is slowing down (retarding) initially.
Reason (R) is false. An object with negative initial velocity and positive acceleration can eventually reverse direction and achieve positive displacement (e.g., if it starts at \(x=0\), it will eventually cross \(x=0\) and move to positive \(x\)).
Thus, (A) is true, (R) is false.
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): 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): A particle has positive acceleration it means that its speed always increases.
Reason (R): Acceleration is the rate of change of speed with respect to 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
Assertion (A) is false because positive acceleration doesn't always mean increasing speed; it depends on the direction of velocity. Speed increases only when \(vec{a}\) and \(vec{v}\) are in the same direction. Reason (R) is false because acceleration is the rate of change of velocity, not speed.
A car starts from rest, accelerates uniformly at \( 2 \, \text{m/s}^2 \). The distance travelled by the car in fourth second is
1. 2.25 m
2. 14 m
3. 7 m
4. Zero
View Answer
The distance travelled in the \( n^{\text{th}} \) second is \( s_n = u + \frac{a}{2}(2n - 1) \). Substituting \( u = 0 \), \( a = 2 \, \text{m/s}^2 \), and \( n = 4 \) yields \( s_4 = 0 + \frac{2}{2}(2(4) - 1) = 7 \, \text{m} \).
A body starts from rest, what is the ratio of the distance travelled by the body during the \(4^{\text{th}}\) and \(3^{\text{rd}}\) second?
(1993)
1. 7/5
2. 5/7
3. 7/3
4. 3/7
View Answer
Concept: Distance covered in the \(n^{text{th}}\) second for uniformly accelerated motion.
Formula: \(S_n = u + \frac{a}{2}(2n - 1)\). Since it starts from rest, \(u=0\).
Solution: \(S_4 = \frac{a}{2}(2(4) - 1) = \frac{7a}{2}\), \(S_3 = \frac{a}{2}(2(3) - 1) = \frac{5a}{2}\). Ratio \(S_4:S_3 = 7a/2 : 5a/2 = 7:5\).
If a car at rest accelerates uniformly to a speed of \(144 \text{ km/h}\) in \(20 \text{ sec}\), it covers a distance of:
(1997)
1. \(1440 \text{ cm}\)
2. \(2980 \text{ cm}\)
3. \(20 \text{ m}\)
4. \(400 \text{ m}\)
View Answer
Given (u=0), \(v = 144 \text{ km/h} = 144 \times \frac{5}{18} = 40 \text{ m/s}\), \(t = 20 \text{ s}\). Using \(S = \frac{u+v}{2}t), we get \(S = \frac{0+40}{2} \times 20 = 20 \times 20 = 400 \text{ m}\).