NEET Equations of Motion MCQs & PYQs

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) 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): 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): 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) 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.

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 \( s_4 = 0 + \frac{2}{2}(2(4) - 1) = 7 \, \text{m} \).

Concept: Kinematics equations involving differentiation.
Formula: Velocity \(v = \frac{ds}{dt}\) and acceleration \(a = \frac{dv}{dt}\).
Solution: Given \(s = t^3 - 6t^2 + 3t + 4\). Then \(v = 3t^2 - 12t + 3\) and \(a = 6t - 12\). Setting \(a=0\) gives \(t=2\text{ s}\). Substituting \(t=2\text{ s}\) into \(v\) gives \(v = 3(2)^2 - 12(2) + 3 = 12 - 24 + 3 = -9\text{ m/s}\).

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 \(S_4:S_3 = 7a/2 : 5a/2 = 7:5\).

Concept: Equations of motion under uniform acceleration.
Formula: \(v^2 = u^2 + 2as\).
Solution: Let \(u_P=30\), \(v_Q=40\) and distance be \(s\). \(v_Q^2 = u_P^2 + 2as\) gives \(40^2 = 30^2 + 2as\) => \(1600 = 900 + 2as\) => \(2as = 700\) => \(as = 350\). For the midway point, \(v_m^2 = u_P^2 + 2a(s/2) = u_P^2 + as = 30^2 + 350 = 900 + 350 = 1250\). So, \(v_m = sqrt{1250} = 25\sqrt{2}\text{ km/h}\).

Concept: Galileo's law of odd numbers for free fall.
Formula: Distance in \(n^{text{th}}\) second is \(S_n = \frac{g}{2}(2n - 1)\).
Solution: \(S_1:S_2:S_3:S_4 = (2(1)-1):(2(2)-1):(2(3)-1):(2(4)-1) = 1:3:5:7\).