The relation between time and distance is given by $$\(t=\alpha x^2+\beta x,\) where \(\alpha\) and \(\beta\) are constants. The retardation, as calculated based on this equation, will be (assume \(v\) to be velocity):
1. $\mathrm{}$\(2\alpha v^3\)
2. \(2\beta v^3\)$\mathrm{}{\mathrm{}}^{}$
3. \(2\alpha\beta v^3\)$\mathrm{}{\mathrm{}}^{}$
4. \(2\beta^2 v^3\)$\mathrm{}{\mathrm{}}^{}$

The displacement of a particle is given by \(y = a + bt + ct^{2} - dt^{4}\). The initial velocity and acceleration are, respectively:

An elevator car, whose floor to ceiling distance is equal to \(2.7~\text{m}\), starts ascending with constant acceleration of \(1.2~\text{ms}^{-2}\). \(2\) sec after the start, a bolt begins falling from the ceiling of the car. The free fall time of the bolt is: 
1. \(\sqrt{0.54}~\text{s}\)
2. \(\sqrt{6}~\text{s}\)
3. \(0.7~\text{s}\)
4. \(1~\text{s}\)

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

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 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:

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: 

A student is standing at a distance of \(50\) metres from the bus. As soon as the bus begins its motion with an acceleration of \(1\) ms^–2, the student starts running towards the bus with a uniform velocity \(u\). Assuming the motion to be along a straight road, the minimum value of \(u\), so that the student is able to catch the bus is:
1. \(5\) ms^–1
2. \(8\) ms^–1
3. \(10\) ms^–1
4. \(12\) ms^–1

If the velocity of a particle is given by \(v = (180-16x)^{1/2}~\text{m/s}\), then its acceleration will be: 

Two trains, each \(50\) m long, are travelling in the opposite direction with velocities \(10\) m/s and \(15\) m/s. The time of crossing is:
1. \(10\) sec
2. \(4\) sec
3. \(2\sqrt{3}\) sec
4. \(4\sqrt{3}\) sec