The displacement \(x\) of a particle moving in one dimension under the action of a constant force is related to time t by the equation $t$ $=$ $\sqrt{x}$ $+$ $3$, where \(x\) is in metres and \(t\) is in seconds. What is the displacement of the particle from t = 0 s to t = 6 s?

1. \(0\)

2. \(12\) m

3. \(6\) m

4. \(18\) m

The acceleration \(a\) (in ${\mathrm{ms}}^{-2}$) of a body, starting from rest varies with time \(t\) (in \(\mathrm{s}\)) as per the equation \(a=3t+4.\) The velocity of the body at time \(t=2\) \(\mathrm{s}\) will be:
1. \(10~\text{ms}^{-1}\)
2. \(18~\text{ms}^{-1}\)
3. \(14~\text{ms}^{-1}\)
4. \(26~\text{ms}^{-1}\)

A point moves in a straight line under the retardation \(av^2\)${\mathrm{}}^{}$. If the initial velocity is \(u,\) the distance covered in \(t\) seconds is:
1. \((aut)\)
2. \(\frac{1}{a}\mathrm{ln}(aut)\)
3. \(\frac{1}{a}\mathrm{ln}(1+aut)\)
4. \(a~\mathrm{ln}(aut)\)

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

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:

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