If the distance 's' travelled by a body in time 't' is given by $\mathrm{s}=\frac{\mathrm{a}}{\mathrm{t}}+{\mathrm{bt}}^2$ then the acceleration equals

(1)  $\frac{2\mathrm{a}}{{\mathrm{t}}^3}+2\mathrm{b}$

(2)  $\frac{2\mathrm{s}}{{\mathrm{t}}^3}$

(3)  $2\mathrm{b}-\frac{2\mathrm{a}}{{\mathrm{t}}^3}$

(4)  $\frac{\mathrm{s}}{{\mathrm{t}}^2}$

The velocity of a particle moving on the x-axis is given by $\mathrm{v}={\mathrm{x}}^2+\mathrm{x}$ where v is in m/s and x is in m. Find its acceleration in $\mathrm{m}/{\mathrm{s}}^2$ when passing through the point x=2m.

1.  0

2.  5

3.  11

4.  30

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: (\(Here, a=\frac{dv}{dt}\))

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, \(v=\frac{dx}{dt}\))

1. vary with time.

2. be independent of time.

3. be inversely proportional to time.

4. be inversely proportional to acceleration.

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