For the given figure, which of the following is true?

                          
1. \(\overrightarrow{B}= \overrightarrow{A}+\overrightarrow{C}\)
2. \(\overrightarrow{A}= \overrightarrow{B}+\overrightarrow{C}\)
3. \(\overrightarrow{C}= \overrightarrow{A}+\overrightarrow{B}\)
4. All of these

The value of the unit vector, which is perpendicular to both \(A=\hat{i} + 2\hat{j} + 3 \hat{k}\) and \(B= \hat{i} - 2\hat{j} - 3 \hat{k}\) $\mathrm{is\; equal\; to}$:

1. \(\frac{\hat{i}   +   2 \hat{j}   +   3 \hat{k}}{6}\)
2. \(\frac{6\hat{j} -4 \hat{k}}{\sqrt{52}}\)
3. \(\frac{6\hat{j} +4 \hat{k}}{\sqrt{52}}\)
4. \(\frac{2\hat{i} - \hat{j}}{\sqrt{5}}\)

The instantaneous velocity (defined as $v=\frac{ds}{dt}$) at time $t=\frac{\mathrm{\pi }}{2}$ of a particle, whose position equation is given as  s(t)=12 tan$\left(\frac{t}{2}+\mathrm{\pi }\right)$m, is:
1. 12 m/s
2. 12$\sqrt{2}$ m/s
3. 6 m/s
4. \(6\sqrt2\) m/s

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. \(30.4~\text{m}^2/\text{s}\)