The momentum is given by p=4t+1, the force at t=2s is-$[Given: F=\frac{dP}{dt}]$

(A)  4 N

(B)  8 N

(C)  10 N

(D)  15 N

If the momentum of a particle is given by P=(180-8t) kg m/s, then its force will be-$[Given: F=\frac{dP}{dt}]$

(1)  Zero

(2)  8 N

(3)  -8 N

(4)  4 N

The maximum value of the function \(7 + 6 x - 9 x^{2}\) is:
1. \(8\)
2. \(-8\)
3. \(4\)
4. \(-4\)

If \(f \left(x\right) = x^{2} - 2 x + 4\), then \(f(x)\) has:

A particle is moving along x-axis. The velocity v of particle varies with its position x as $\mathrm{v}=\frac{1}{\mathrm{x}}$. Find velocity of particle as a function of time t given that at t=0, x=1 .

1.  $\mathrm{v}=\sqrt{2\mathrm{t}+1}$

2.  $\mathrm{v}=\frac{1}{\sqrt{2\mathrm{t}+1}}$

3.  $\mathrm{v}=\frac{1}{\mathrm{t}}$

4.  None of these

A vector $\overrightarrow{\mathrm{A}}$ is directed along $30°$ west of north direction and another vector $\overrightarrow{\mathrm{B}}$ along $15°$ south of east. Their resultant cannot be in ____________ direction.

(1)  North

(2)  East

(3)  North-East

(4)  South

ABCD is a quadrilateral. Forces $\overrightarrow{\mathrm{BA}}, \overrightarrow{\mathrm{BC}}, \overrightarrow{\mathrm{CD}} \& \overrightarrow{\mathrm{DA}}$ act at a point. Their resultant is 

(A)  $2 \overrightarrow{\mathrm{AB}}$

(B)  $2 \overrightarrow{\mathrm{DA}}$

(C)  zero vector

(D)  $2 \overrightarrow{\mathrm{BA}}$

The maximum and minimum magnitude of the resultant of two vectors are 17 units and 7 units respectively. Then the magnitude of resultant of the vectors when they act perpendicular to each other is:

(1)  14

(2)  16

(3)  18

(4)  13

A vector $\overrightarrow{\mathrm{A}}$ makes an angle of $20°$ and $\overrightarrow{\mathrm{B}}$ makes an angle of $110°$ with the X-axis. The magnitudes of these vectors are 3 m and 4 m respectively. Find the magnitude of the resultant.

(A)  $1$

(B)  $5\sqrt{2}$

(C)  $5$

(D)  $7$

A particle is moving westward with a velocity ${\overrightarrow{\mathrm{v}}}_i=5 \mathrm{m}/\mathrm{s}.$ Its velocity changed to ${\overrightarrow{\mathrm{v}}}_f=5\mathrm{m}/\mathrm{s}$ northward. The change in velocity vector $\left(∆\overrightarrow{\mathrm{V}}={\overrightarrow{\mathrm{v}}}_f-{\overrightarrow{\mathrm{v}}}_i\right)$ is:

(1)  $5\sqrt{2} \mathrm{m}/\mathrm{s}$ towards north east

(2)  $5 \mathrm{m}/\mathrm{s}$ towards north west

(3)  zero

(4)  $5\sqrt{2} \mathrm{m}/\mathrm{s}$towards north west