A particle moves in the XY plane and at time t is at the point whose coordinates are $\left({\mathrm{t}}^2, {\mathrm{t}}^3-2\mathrm{t}\right)$. Then at what instant of time, will its velocity and acceleration vectors be perpendicular to each other?

(1)  1/3 sec

(2)  2/3 sec

(3)  3/2 sec

(4)  never

A particle is moving along positive x-axis. Its position varies as $\mathrm{x}={\mathrm{t}}^3-3{\mathrm{t}}^2+12\mathrm{t}+20$, where x is in meters and t is in seconds.

Initial acceleration of the particle is

(A)  Zero

(B)  $1 \mathrm{m}/{\mathrm{s}}^2$

(C)  $-3 \mathrm{m}/{\mathrm{s}}^2$

(D)  $-6 \mathrm{m}/{\mathrm{s}}^2$

Two forces ${\overrightarrow{\mathrm{F}}}_1=2\hat{\mathrm{i}}+2\hat{\mathrm{j}} \mathrm{N}$ and ${\overrightarrow{\mathrm{F}}}_2=3\hat{\mathrm{j}}+4\hat{\mathrm{k}} \mathrm{N}$ are acting on a particle.

The resultant force acting on particle is:

(A)  $2\hat{\mathrm{i}}+5\hat{\mathrm{j}}+4\hat{\mathrm{k}}$

(B)  $2\hat{\mathrm{i}}-5\hat{\mathrm{j}}-4\hat{\mathrm{k}}$

(C)  $\hat{\mathrm{i}}-3\hat{\mathrm{j}}-2\hat{\mathrm{k}}$

(D)  $\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}$

$\mathrm{A}=4\mathrm{i}+4\mathrm{j}-4\mathrm{k}$ and $\mathrm{B}=3\mathrm{i}+\mathrm{j}+4\mathrm{k}$, then angle between vectors A and B is:

(1)  $180°$

(2)  $90°$

(3)  $45°$

(4)  $0°$

If a curve is governed by the equation y = sinx, then the area enclosed by the curve and x-axis between x = 0 and x =$\mathrm{ \pi }$ is (shaded region):

              
1. \(1\) unit
2. \(2\) units
3. \(3\) units
4. \(4\) units

The acceleration of a particle starting from rest varies with time according to relation, $a=\alpha t+\beta$. The velocity of the particle at time instant \(t\) is: \(\left(\text{Here,}~ a=\frac{dv}{dt}\right)\)

1. $\alpha t^2+\beta t$

2. $\alpha t^2+\frac{\beta t}{2}$

3. $\frac{\alpha t^2}{2}+\beta t$

4. $2\alpha t^2+\beta t$

The displacement of the particle is zero at \(t=0\) and at \(t=t\) it is \(x\). It starts moving in the \(x\)-direction with a velocity that varies as \(v = k \sqrt{x}\), where \(k\) is constant. The velocity will: (Here, \(v=\frac{dx}{dt}\))

The acceleration of a particle is given as $a=3x^2$.  At t = 0, v = 0 and x = 0. It can then be concluded that the velocity at t = 2 sec will be: (Here, \(a=v\frac{dv}{dx}\))

1.  0.05 m/s

2. 0.5 m/s

3. 5 m/s

4. 50 m/s 

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. \(2~\text{m/s}, 3~\text{m}\)

The 9 kg block is moving to the right with a velocity of 0.6 m/s on a horizontal surface when a force F, whose time variation is shown in the graph, is applied to it at time t = 0. Calculate the velocity v of the block when t= 0.4s. The coefficient of kinetic fricton is ${\mu }_k=0.3$. [This question includes concepts from Work, Energy & Power chapter]

1. 0.6 m/s

2. 1.2 m/s

3. 1.8 m/s

4. 2.4 m/s