If the acceleration \(a(t)= 4t+6\), the velocity of a particle starting from rest is: \(\left(\text{here} ,  a = \frac{d v}{d t}\right)\)
1. \(2t+6\)
2. \(4\)
3. \(0\)
4. \(2t^2+6t\)

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? $\left(work={\int }_a^{b}F\left(x\right) dx\right)$

1. $\frac{11}{3} J$

2. 51 J

3. $\frac{5}{3} J$

4. $\frac{164}{3} J$

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

1. 1 m

2. 2 m

3. 3 m

4. 4 m

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