If v(t) = 3t-1 and x(2) = 1, then the original position function is: 
Hint: \(\left(v \left( t \right) = \frac{d s}{d t}\right)\)$$
1. $\frac{3}{2}{t}^2-t-3$

2. $\frac{1}{2}{t}^2-t-3$

3. $\frac{3}{2}{t}^2-2t-3$

4. None of the above

The position of a particle is given by \(s\left( t\right) = \dfrac{2 t^{2} + 1}{t + 1}\). Then, at \(t= 2\), its velocity is: \(\left(v_{inst}= \dfrac{ds}{dt}\right)\)
1. \(\dfrac{16}{3}\)
2. \(\dfrac{15}{9}\)
3. \(\dfrac{15}{3}\)
4. None of these

If acceleration of a particle is given as a(t) = sin(t)+2t. Then the velocity of the particle will be:
(acceleration $\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}$)
1. \(-\cos(t)+ \frac{t^2}{2}\)
2. \(-\sin(t)+ t^2\)
3. \(-\cos(t)+ t^2\)
4. None of these

If \(x= 3\tan(t)\) and \(y = \sec (t)\), then the value of \(\dfrac{d^{2} y}{d x^{2}}~\text{at}~t = \dfrac{\pi}{4}\) is:
1. \(3\)
2. \(\dfrac{1}{18\sqrt{2}}\)
3. \(1\)
4. \(\dfrac{1}{6}\)

A particle's position as a function of time is given by $\mathrm{x}=-{\mathrm{t}}^2+6\mathrm{t}+3$. The maximum value of the position co-ordinate of the particle is:
1. \(8\)

2. \(12\)

3. \(3\)

4. \(6\)

The equation of position \((x)\) of a particle is given by; \(x=(-3t^3+18t^2+5)~\text{m}.\) The maximum velocity of the particle is: (velocity is defined as \(v=\dfrac{dx}{dt}\))
1. \(+56\) m/s
2. \(+46\) m/s
3. \(+36\) m/s
4. \(+26\) m/s

The current in a circuit is defined as $I=\frac{dq}{dt}$. The charge (q) flowing through a circuit, as a function of time (t), is given by $q=5t^2-20t+3$. The minimum charge flows through the circuit at:
1. \(t = 4~\text{s}\)

2. \(t = 2~\text{s}\)

3. \(t = 6~\text{s}\)

4. \(t = 3~\text{s}\)

Work done by a force (\(F\)) in displacing a body by dx is given by W=$\int F\left(x\right).dx$. If the force is given as a function of displacement (\(x\)) by \(F \left(x\right) = \left( x^{2} - 2 x + 1\right) \text{N}\), then work done by the force from \(x=0\) to \(x=3\) m is:

1. \(3\) J

2. \(6\) J

3. \(9\) J

4. \(21\) J

The impulse due to a force on a body is given by \(I=\int Fdt\). If the force applied on a body is given as a function of time \((t)\) as \(F = \left(3 t^{2} + 2 t + 5\right) \text{N}\), then impulse on the body between \(t = 3~\text{s}\) to \(t =5~\text{s}\) is:
1. \(175\) kg-m/sec
2. \(41\) kg-m/sec
3. \(216\) kg-m/sec
4. \(124\) kg-m/sec

If $\mathrm{\theta }$ is the angle between vectors $\overrightarrow{\mathrm{A}}<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>\overrightarrow{\mathrm{B}}$, then which of the following is the unit vector perpendicular to $\overrightarrow{\mathrm{A}}<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>\overrightarrow{\mathrm{B}}$?

1. $\frac{\hat{\mathrm{A}}<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>\hat{\mathrm{B}}}{\mathrm{AB}<mspace\; width="1em"></mspace>\sin <mspace\; width="1em"></mspace>\mathrm{\theta }}$

2. $\frac{\overrightarrow{\mathrm{A}}<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>\overrightarrow{\mathrm{B}}}{\mathrm{AB}<mspace\; width="1em"></mspace>\cos <mspace\; width="1em"></mspace>\mathrm{\theta }}$

3. $\frac{\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}}{\mathrm{AB}<mspace\; width="1em"></mspace>\sin <mspace\; width="1em"></mspace>\mathrm{\theta }}$

4. $\frac{\overrightarrow{\mathrm{A}}<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>\overrightarrow{\mathrm{B}}}{\mathrm{AB}<mspace\; width="1em"></mspace>}$