The motion of a particle is given by the equation \(S = \left(3 t^{3} + 7 t^{2} + 14 t + 8 \right) \text{m} ,\) The value of the acceleration of the particle at \(t=1~\text{s}\) is:

The position \(x\) of a particle varies with time \(t\) as \(x=at^2-bt^3\). The acceleration of the particle will be zero at time \(t\) equal to:

A body is moving along a straight line according to the equation of motion, \(x= t^{2} - 3 t + 4\), where \(x\) is in metre and \(t\) is in seconds. What is the acceleration of the body when it comes to rest?
1. zero
2. \(2~\text{m/s}^2\)
3. \(\frac{3}{2}~\text{m/s}^2\)
4. \(1~\text{m/s}^2\)

A particle moves along a straight line such that its displacement at any time \(t\) is given by \(S = t^{3} - 6 t^{2} + 3 t + 4\) metres. The velocity when the acceleration is zero is:

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

When the velocity of a body is variable, then:

If the velocity of a particle is given by $v={(180-16x)}^{1/2}$ m/s, then its acceleration will be: 
1. zero
2. \(8\) m/s²
3. \(-8\) m/s²
4. \(4\) m/s² 

A particle moves a distance \(x\) in time \(t\) according to equation \(x = (t+5)^{-1}\). The acceleration of the particle is proportional to: 

If a particle has negative velocity and negative acceleration, its speed: