The acceleration of a particle varies with time according to the relation \(a=\alpha t+\beta.\) If the particle starts from rest, what is its velocity at time \(t\)? \(\left(\text{Here,}~ a=\dfrac{dv}{dt}\right)\)
1. \(\alpha t^2+\beta t\)
2. \(\alpha t^2+\beta t/2\)
3. \(\alpha t^2/2+\beta t\)
4. \(2\alpha t^2+\beta t\)

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. \(2~\text{m/s}, 3~\text{m}\)

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:  $\left(\mathrm{Given},<mspace\; width="1em"></mspace>\mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}\right)$

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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The current through a wire depends on time as \(i = (2+3t)~\text{A}\). The charge that crosses through the wire in \(10\) seconds is: \(\left(\text{Instantaneous current,}~i= \frac{dq}{dt} \right)\)
1. \(150~\text{C}\)
2. \(160~\text{C}\)
3. \(170~\text{C}\)
4. None of there

${\int }_0^Q\frac{q}{C}dq$, 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 work done by gravity exerting an acceleration of \(-10\) m/${\mathrm{s}}^2$ for a \(10\) kg block down \(5\) m from its original position with no initial velocity is: \(\left(F_{\text{gravitational}}= \text{mass}\times \text{acceleration and} ~w = \int^{b}_{a}F(x)dx \right)\)

1. \(250\) J

2. \(500\) J

3. \(100\) J

4. \(1000\) J

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

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