If the force on an object as a function of displacement is \(F \left(x\right) = 3 x^{2} + x\)$$, what is work as a function of displacement \(w(x)\): \(\left(w= \int f\cdot dx\right)\) Assume \(w(0)= 0\) and force is in the direction of the object's motion.
1. \(\frac{3 x^{3}}{2} + x^{2}\)
2. \(x^{3} + \frac{x^{2}}{2}\)
3. \(6x+1\)
4. \(3 x^{2} + x\)
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Given velocity v(t) = $\frac{5}{2}\mathrm{t}+3$. Assume s(t) is measured in meters and t is measured in seconds. If s(0) = 0, the position s(4) at t = 4s is:  $\left(\mathrm{Given},<mspace\; width="1em"></mspace>\mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}\right)$

The acceleration of a particle is given by \(a=3t\) at \(t=0\), \(v=0\), \(x=0\). The velocity and displacement at \(t = 2~\text{sec}\) will be:
\(\left(\text{Here,} ~a=\frac{dv}{dt}~ \text{and}~v=\frac{dx}{dt}\right)\)
1. \(6~\text{m/s}, 4~\text{m}\)
2. \(4~\text{m/s}, 6~\text{m}\)
3. \(3~\text{m/s}, 2~\text{m}\)
4. \(2~\text{m/s}, 3~\text{m}\)

The acceleration of a particle starting from rest varies with time according to relation, $a=\alpha <mspace\; width="1em"></mspace>t+\beta$. The velocity of the particle at time instant \(t\) is: \(\left(\text{Here,}~ a=\frac{dv}{dt}\right)\)

1. $\alpha t^2+\beta t$

2. $\alpha t^2+\frac{\beta t}{2}$

3. $\frac{\alpha t^2}{2}+\beta t$

4. $2\alpha t^2+\beta t$

The area under the curve \(y=3x^2\) between the co-ordinates \(x = 0\) to \(x = 2\) is:
1. \(8\) m²
2. \(3\) m²
3. \(2\) m²
4. \(9\) m²

${\int }_0^Q\frac{q}{C}dq$, where C is a constant, can be expressed as:

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

The figure shows the graph of the function \(y=x^2. \) What is the area under the curve from \(x =0 \) to \(x=6, \) i.e., the area of the shaded region?