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

Water flows into a container of 1000 L at a rate of (180+3t) gal/min for an hour, where t is measured in minutes. Find the amount of water that flows into the pool during the first 20 minutes.

1. 4000 gal

2. 2800 gal

3. 4200 gal

4. 3800 gal

If v(t) = 3t-1 and x(2) = 1, then the original position function is: 
Hint: \(\left(v \left( t \right) = \frac{d s}{d t}\right)\)$$
1. $\frac{3}{2}{t}^2-t-3$

2. $\frac{1}{2}{t}^2-t-3$

3. $\frac{3}{2}{t}^2-2t-3$

4. None of the above

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

A weight hanging from a spring is stretched down 3 cm beyond its rest position and released at time t=0 to bob up and down. Its position at any later time t is s=3cos(t). Then its velocity at time t is $\left(velocity =\frac{ds}{dt}\right)$

(1) cost

(2) 3cost

(3) 3sint

(4) -3sint

The position of a particle is given by \(s\left( t\right) = \frac{2 t^{2} + 1}{t + 1}\). Then, at \(t= 2\), its velocity is: \(\left(v_{inst}= \frac{ds}{dt}\right)\)
1. \(\frac{16}{3}\)
2. \(\frac{15}{9}\)
3. \(\frac{15}{3}\)
4. None of these

The instantaneous velocity at t=$\frac{\mathrm{\pi }}{2}$ of a particle whose positional equation is given by $x\left(t\right)=12 {\cos }^2\left(t\right)$ is $\left(v_{inst}=\frac{dx}{dt}\right)$ -

(1) 0

(2) -24

(3) 24

(4) $-12\sqrt{2}$

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

If \(x= 3\tan(t)\) and \(y = \sec (t)\), then the value of \(\frac{d^{2} y}{d x^{2}}~\text{at}~t = \frac{\pi}{4}\) is:
1. \(3\)
2. \(\frac{1}{18\sqrt{2}}\)
3. \(1\)
4. \(\frac{1}{6}\)