The displacement is given by \(𝑥 = 2 𝑡^ 2 + 𝑡 + 5 ,\) the acceleration at \(𝑡 = 2 \ \text{s}\) is:

1. \(4\ \text{m/s}^2\)
2. \(8\ \text{m/s}^2\)
3. \(10\ \text{m/s}^2\)$\mathrm{}$
4. \(15\ \text{m/s}^2\)$\mathrm{}$

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. \(0.5\)

If a car at rest accelerates uniformly to a speed of \(144\ \text{km/h}\) in \(20\ \text{s}\). Then it covers a distance of:

1. \(20\ \text{m}\)
2. \(400\ \text{m}\)
3. \(1440\ \text{m}\)
4. \(2880\ \text{m}\)

If a train travelling at \(72\ \text{km/h}\) is to be brought to rest in a distance of \(200\) metres, then its retardation should be:

1. \(20\ \text{ms}^{–2}\)
2. \(10\ \text{ms}^{–2}\)
3. \(2\ \text{ms}^{–2}\)
4. \(1\ \text{ms}^{–2}\) 

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. \(8\)

Two cars \(A\) and \(B\) are at rest at the same point initially. If \(A\) starts with a uniform velocity of \(40\ \text{m/s}\) and \(B\) starts in the same direction with a constant acceleration of \(4\ \text{m/s}^2\), then \(B\) will catch \(A\) after how much time?

1. \(10\ \text{s}\)
2. \(20\ \text{s}\)
3. \(30\ \text{s}\)
4. \(35\ \text{s}\)

The motion of a particle is described by the equation \(𝑥 = 𝑎 + 𝑏 𝑡^ 2\) where \(a = 15\ \text{cm}\) and \(b = 3\ \text{cm/s}^2\). Its instantaneous velocity at time \(3\) seconds will be:

1. \(36\ \text{cm/s}\)
2. \(18\ \text{cm/s}\)
3. \(16\ \text{cm/s}\)
4. \(32\ \text{cm/s}\)

Consider the acceleration, velocity and displacement of a tennis ball as it falls to the ground and bounces back. Directions of which of these changes in the process ?

1. Velocity only

2. Displacement and velocity

3. Acceleration, velocity and displacement

4. Displacement and acceleration

The displacement of a particle moving in a straight line is given by \(𝑠 = 2 𝑡^2 + 2 𝑡 + 4\) where \(s\) is in meters and \(t\) in seconds. The acceleration of the particle is:

1. \(2\ \text{m/s}^2\)
2. \(4\ \text{m/s}^2\)
3. \(6\ \text{m/s}^2\)
4. \(8\ \text{m/s}^2\)

The velocity of a bullet is reduced from \(200 \ \text{m/s}\) to \(100 \ \text{m/s}\) while travelling through a wooden block of thickness \(10\ \text{cm}\). The retardation, assuming it to be uniform, will be: 

1. \(10×10^4\ \text{m/s}^2\)

2. \(12×10^4\ \text{m/s}^2\)$\mathrm{}$

3. \(13.5×10^4\ \text{m/s}^2\)$\mathrm{}$

4. \(15×10^4\ \text{m/s}^2\)$\mathrm{}$