The relation between time t and distance x is $t=a{x}^2+bx$, where a and b are constants, the acceleration is $\left(here, v=\frac{dx}{dt} and a=\frac{d^{2}x}{d{t}^2}\right)$

(1) $2b v^3$

(2) $-2 ab v^2$

(3) $2 a{v}^2$

(4) $-2 a{v}^3$

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:  $\left(\mathrm{Given}, \mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}\right)$

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

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. \(30.4~\text{m}^2/\text{s}\)

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

The value of ${\int }_{x=\infty }^{x=R}\frac{GMm}{x^2} dx$ is:

1. $\frac{GMm}{R}$

2. $\frac{2GMm}{R}$

3. $\frac{-GMm}{R}$

4. $\frac{-2GMm}{R}$

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

1. $\frac{Q^2}{C}$

2. $\frac{-Q^2}{2C}$

3. $\frac{-Q^2}{C}$

4. $\frac{Q^2}{2C}$

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. \(3 x^{2} + x\)
$$

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 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