Please be patient as the PDF generation may take up to a minute.

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. \(6\)

If \(y = t^3+1\) and \(x = t^2+3,\) what is the value of \(\dfrac{dy}{dx}?\)
1. \(\dfrac{t^2}{3}\)
2. \(\dfrac{t}{2}\)
3. \(\dfrac{3t}{2}\)
4. \(t^2\)

A particle starts rotating from rest and its angular displacement is given by: \(\theta = \dfrac{t^2}{40}+\dfrac{t}{5}\). Then, the angular velocity \(\omega = \dfrac{d\theta}{dt}\) at the end of \(10~\text{s}\) will be:
1. \(0.7\)
2. \(0.6\)
3. \(0.5\)
4. \(0\)

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

If charge flown through a wire is given by q=3sin(3t), then-current flown through the wire at $\mathrm{t}=\frac{\mathrm{\pi }}{9}$ seconds is: $\left(\mathrm{i}=\frac{\mathrm{dq}}{\mathrm{dt}}\right)$
1. 4.5 Amp
2. 4.5$\sqrt{3}$ Amp
3. $\frac{\sqrt{3}}{2}$ Amp
4. 9 Amp

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

The area of a blot of ink, \(A\), is growing such that after \(t\) seconds, \(A=\left(3t^2+\frac{t}{5}+7\right)\text{m}^2\). Then the rate of increase in the area at \(t = 5~\text{s}\) will be:
1. \(30.1~\text{m}^2/\text{s}\)
2. \(30.2~\text{m}^2/\text{s}\)
3. \(30.3~\text{m}^2/\text{s}\)
4. \(30.4~\text{m}^2/\text{s}\)

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

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. \(2~\text{m/s}, 3~\text{m}\)

Current in a circuit is given by $i<mspace\; width="1em"></mspace>=3t^2+2t$. Find charge that crosses a cross-section from time t=0 to t=2 sec.
1. 12 C
2. 10 C
3. 8 C
4. 2 C

A force of $F\left(x\right)=2x^2+3$ N is applied to an object. How much work is done, in Joules, moving the object from x=1 to x=4 meters? $\left(work={\int }_a^{b}F\left(x\right)<mspace\; width="1em"></mspace>dx\right)$
1. $\frac{11}{3}<mspace\; width="1em"></mspace>J$
2. 51 J
3. $\frac{5}{3}<mspace\; width="1em"></mspace>J$
4. $\frac{164}{3}<mspace\; width="1em"></mspace>J$

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

A car has a certain displacement between 0 seconds and 2 seconds. If we defined its velocity as v(t)=6t-5, then the displacement in meters is: $\left(\mathrm{Here},<mspace\; width="1em"></mspace>\mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}\right)$
1. 1 m
2. 2 m
3. 3 m
4. 4 m

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

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

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 velocity of a rocket, in metres per second, \(t\) seconds after it was launched is modelled by \(v(t)=2\sqrt{t}\). What is the total distance travelled by the rocket during the first four seconds of its launch?
1. \(\frac{16}{3}~\text{m}\)
2. \(32~\text{m}\)
3. \(\frac{32}{3}~\text{m}\)
4. \(16~\text{m}\)

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