A particle moving with a uniform acceleration travels \(24\) m and \(64\) m in the first two consecutive intervals of \(4~\text{s}\) each. Its initial velocity is:
1. \(1~\text{m/s}\)
2. \(10~\text{m/s}\)
3. \(5~\text{m/s}\)
4. \(2~\text{m/s}\)

The position of a particle moving in the XY plane at any time \(t\) is given by \(𝑥 = ( 3 𝑡^ 2 − 6 𝑡 )\) metres, \(y=(t^2-2t)\) metres. Select the correct statement about the moving particle from the following.

1. The acceleration of the particle is zero at \(t = 0\) second
2. The velocity of the particle is zero at \(t = 0\) second
3. The velocity of the particle is zero at \(t = 1\) second
4. The velocity and acceleration of the particle are never zero

Two cars \(A\) and \(B\) are travelling in the same direction with velocities \(v_1\) and \(v_2 (v_1>v_2)\). When the car \(A\) is at a distance \(d\) behind car \(B\), the driver of the car \(A\) applied the brake producing uniform retardation \(a\). There will be no collision when:
1. \(d< \dfrac{(v_1-v_2)^2}{2a}\)

2. \(d< \dfrac{v^2_1-v^2_2}{2a}\)

3. \(d> \dfrac{(v_1-v_2)^2}{2a}\)

4. \(d> \dfrac{v^2_1-v^2_2}{2a}\)

A body moves from rest with a constant acceleration of \(5\ \text{m/s}^2\). Its instantaneous speed (in m/s) at the end of \(10\ \text{s}\) is  

1. \(50\)
2. \(5\)
3. \(2\)
4. \(0.5\)

The acceleration \(a\) in m/s² of a particle is given by $a=3t^2+2t+2$ where t is the time. If the particle starts out with a velocity, \(u=2\) m/s at t = 0, then the velocity at the end of \(2\) seconds will be:
1. \(12\) m/s
2. \(18\) m/s
3. \(27\) m/s
4. \(36\) m/s

The displacement of a particle starting from rest (at \(t = 0\)) is given by \(𝑠 = 6 𝑡^ 2 − 𝑡^ 3\). The time in seconds at which the particle will attain zero velocity again is:  

1. \(2\)
2. \(4\)
3. \(6\)
4. \(8\)

A body is moving according to the equation \(𝑥 = 𝑎 𝑡 + 𝑏 𝑡^ 2 − 𝑐 𝑡^ 3\) where \(x\) is the displacement, and \(a,\ b\) and \(c\) are constants. The acceleration of the body is:

1. \(𝑎 + 2 𝑏 𝑡\) 
2. \(2 𝑏 + 6 𝑐 𝑡  \)
3. \(2 𝑏 − 6 𝑐 𝑡  \)
4. \(3 𝑏 − 6 𝑐 𝑡^ 2  \)

The relation \(3t = \sqrt{3x} + 6\) describes the displacement of a particle in one direction where \(x\) is in metres and \(t\) in seconds. The displacement, when velocity is zero, is: 

The average velocity of a body moving with uniform acceleration travelling a distance of \(3.06\ \text{m}\) is \(0.34\ \text{ms}^{–1}\). If the change in velocity of the body is \(0.18\ \text{ms}^{–1}\) during this time, its uniform acceleration is:

1. \(0.01\ \text{ms}^{–2}\)

2. \(0.02\ \text{ms}^{–2}\)

3. \(0.03\ \text{ms}^{–2}\)

4. \(0.04\ \text{ms}^{–2}\)

The displacement of a particle is proportional to the cube of the time elapsed. How does the acceleration of the particle depends on time obtained?

1. \(𝑎 ∝ t^ 2 \)
2. \(𝑎 ∝ 𝑡^ 4 \)
3. \(𝑎 ∝ 𝑡 ^3\)
4. \(𝑎 ∝ 𝑡  \)