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

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

If $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2-2\mathrm{x}+4$, 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 .

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.

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

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:

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 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:

Consider east as positive x-axis, north as positive y-axis and vertically upward direction as z-axis. A helicopter first rises up to an altitude of 100 m than flies straight in north 500 m and then suddenly takes a turn towards east and travels 1000 m east. What is position vector of helicopter. (Take starting point as origin)

A force $\overrightarrow{\mathrm{F}}=6\hat{\mathrm{i}}-8\hat{\mathrm{j}}+10\hat{\mathrm{k}}$ Newton produces acceleration $1 \mathrm{m}/{\mathrm{s}}^2$ in a body. The mass of the body is (in kg)

Given the vectors

$$\begin{gathered} \overrightarrow{\mathrm{A}}=2\hat{\mathrm{i}}+3\hat{\mathrm{j}}-\hat{\mathrm{k}} \\ \overrightarrow{\mathrm{B}}=3\hat{\mathrm{i}}-2\hat{\mathrm{j}}-2\hat{\mathrm{k}} \\ \& \overrightarrow{\mathrm{C}}=\mathrm{p}\hat{\mathrm{i}}+\mathrm{p}\hat{\mathrm{j}}+2\mathrm{p}\hat{\mathrm{k}} \end{gathered}$$

Find the angle between $\left(\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}\right) \& \overrightarrow{\mathrm{C}}$

(A)  $\mathrm{\theta }={\cos }^{-1}\left(\frac{2}{\sqrt{3}}\right)$

(B)  $\mathrm{\theta }={\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)$

(C)  $\mathrm{\theta }={\cos }^{-1}\left(\frac{\sqrt{2}}{3}\right)$

(D)  none of these

The vector having a magnitude of 10 and perpendicular to the vector $3\hat{\mathrm{i}}-4\hat{\mathrm{j}}$ is- 

A force $\overrightarrow{\mathrm{F}}=3\hat{\mathrm{i}}+\mathrm{c}\hat{\mathrm{j}}+2\hat{\mathrm{k}}$ acting on a particle causes a displacement $\overrightarrow{\mathrm{d}}=-4\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}}$. If the work done is 6J then the value of 'c' is- $[Given: W=\overrightarrow{F.}\overrightarrow{d}]$

A particle moving along a straight line according to the law $\mathrm{x}=\mathrm{At}+{\mathrm{Bt}}^2+{\mathrm{Ct}}^3$, where x is its position measured from a fixed point on the line and t is the time elapsed till it reaches position x after starting from the fixed point. Here A, B and C are positive constants.

If the velocity of a particle moving on x-axis is given by $\mathrm{v}=3{\mathrm{t}}^2-12\mathrm{t}+6$. At which time is the acceleration of particle zero?

Momentum of a body moving in a straight line is $\mathrm{p}=\left({\mathrm{t}}^2+2\mathrm{t}+1\right) \mathrm{kg} \mathrm{m}/\mathrm{s}$. Find the force acting on a body at t=2 sec

A particle moves along straight line such that at time t its position from a fixed point O on the line is $\mathrm{x}=3{\mathrm{t}}^2-2$. The velocity of the particle when t=2 is:

Coordinates of a moving particle are given by $\mathrm{x}={\mathrm{ct}}^2$ and $\mathrm{y}={\mathrm{bt}}^2$. The speed of the particle is given by

The x and y components of vector $\overrightarrow{\mathrm{A}}$ are 4m and 6m respectively. The x, y components of vector $\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}$ are 10m and 9m respectively. The length of $\overrightarrow{\mathrm{B}}$ is ______ and angle that $\overrightarrow{\mathrm{B}}$ makes with the x axis is given by _______.

A particle travels with speed 50 m/s from the point (3, 7) in a direction $7\hat{\mathrm{i}}-24\hat{\mathrm{j}}$. Find its position vector after 3 seconds.

If $\overrightarrow{\mathrm{a}}$ is a vector and x is a non-zero scalar, then which of the following is correct?

If $\mathrm{\theta }$ is the angle between two vectors $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$, and $|\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}|=\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}$, then $\mathrm{\theta }$ is equal to:

The vector $\overrightarrow{\mathrm{b}}$, which is collinear with the vector $\overrightarrow{\mathrm{a}}$=(2, 1, -1) and satisfies the condition $\overrightarrow{\mathrm{a}}$.$\overrightarrow{\mathrm{b}}$=3 is-

If a, b and c are three non-zero vectors such that $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=0$, then the value of $\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}}.\overrightarrow{\mathrm{a}}$ will be:

Two vectors $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ inclined at an angle $\mathrm{\theta }$ w.r.t. each other have a resultant $\overrightarrow{\mathrm{c}}$ which makes an angle $\mathrm{\beta }$ with $\overrightarrow{\mathrm{a}}$. If the directions of $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ are interchanged, then the resultant will have the same

Two vectors $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ lie in a plane. Another vector $\overrightarrow{\mathrm{C}}$ lies outside this plane. The resultant $\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}}$ of these three vectors

The vector sum of the forces of 10 N and 6 N can be

A set of vectors taken in a given order gives a closed polygon. Then the resultant of these vectors is a