Velocity

A particle of mass ${10}^{-2} \mathrm{kg}$ is moving along the positive $\mathrm{x}$-axis under the influence of a force $F\left(x\right)=\frac{-K}{\left(2x^2\right)}$ where $K={10}^{-2}{\mathrm{Nm}}^{-2}$. At time $t=0$ it is at $x=$$1.0 \mathrm{m}$ and its velocity is $v=0$ :
Find the time at which it reaches $x=0.25 \mathrm{m}$

A particle of mass ${10}^{-2} \mathrm{kg}$ is moving along the positive $\mathrm{x}$-axis under the influence of a force $F\left(x\right)=\frac{-K}{\left(2x^2\right)}$ where $K={10}^{-2}N{m}^{-2}.$ At time $t=0$ it is at $x=$$1.0 \mathrm{m}$ and its velocity is $v=0$ :
Find its velocity when it reaches $x=0.50 \mathrm{m}$

The displacement $x$ of particle moving in one dimension, under the action of a constant force is related to the time $t$ by the equation  $t=\sqrt{x}+3$ where $x$ is in meters and $t$ in seconds. Find the displacement of the particle when its velocity is zero.

In $1.0 \mathrm{s}$, a particle goes from point $A$ to point $B$, moving in a semicircle of radius $1.0 \mathrm{m}$ (see figure). The magnitude of the average velocity is

The distance travelled by an object in time $t$ is given by $s=\left(2.5\right)t^2$ . The instantaneous speed of the object at$t=5 \mathrm{s}$ will be :

As shown in the figure, a particle is moving with constant speed $\pi \mathrm{m} {\mathrm{s}}^{-1}$. Considering its motion from $A$ to $B$, the magnitude of the average velocity is:

A person travels $x$ distance with velocity $v_1$ and then $x$ distance with velocity $v_2$ in the same direction. The average velocity of the person is $v$, then the relation between $v, v_1$ and $v_2$ will be

If position of particle is changing with time as $r=t^2-2t \left(\mathrm{m}\right)$. Find the velocity at $t=2 \mathrm{s}.$

A particle moves from A to B via C with uniform speed $\pi \mathrm{m} {\mathrm{s}}^{-1}$. Average velocity during the journey is equal to

A particle is moving with a constant speed in a circle. What is the magnitude of average velocity after one fourth rotation?

A jet plane flying at a constant velocity $v$ at a height $h=8 \mathrm{km}$ is being tracked by a radar $R$ located at $O$ directly below the line of flight. If the angle is decreasing at the rate of $0.025 \mathrm{rad} {\mathrm{s}}^{-1}$, the velocity the plane when $\theta =60°$  is

A body is moving along square track $ABCD$ of arm $10 \mathrm{m}$ then find average velocity and average speed when body moves form $A\to B\to C$ in $10\sec$.

The position of an object moving in straight line is given as function of time by $x=\left(2+2 t\right)$, where $x$ is in meters and $t$ is in seconds. The average velocity of object between $t=0$ to $t=2$ seconds is

A car travelled half the distance with a velocity $V$ and it covered the remaining half distance with a velocity $\left(\frac{V}{2}\right)$ in the first half time and with velocity $2V$ in the second half time. The average velocity of the car for the whole journey is

The position of a particle is $\overrightarrow{r}=x\hat{\mathrm{i}}+y\hat{\mathrm{j}}$ where $x$ and $y$ are function of time $t$ and given as $x=\left(12+5t-t^2\right) \mathrm{m}$ and $y=\left(18-5t+t^2\right) \mathrm{m}$. At $t=1 \mathrm{s}$, the magnitude of the velocity vector of the particle is

A truck travelling with uniform acceleration crosses two points $A$ and $B$ with velocities $60 \mathrm{m} {\mathrm{s}}^{-1}$ and $40 \mathrm{m} {\mathrm{s}}^{-1}$ respectively. The speed of the body at the mid-point of $A$ and $B$ is nearest to:

The acceleration of an object is given by $a\left(t\right)=\cos \pi t \mathrm{m} {\mathrm{s}}^{-2}$ and its velocity at time $t=0$ is $\frac{1}{2\pi } \mathrm{m} {\mathrm{s}}^{-1}$ at origin. Its velocity at $t=\frac{3}{2} \mathrm{s}$ is

A body travelling along a straight line traversed one third of the total distance with a velocity $4{\mathrm{ms}}^{-1}$. The remaining part of the distance was covered with a velocity $1{\mathrm{ms}}^{-1}$ for half the time and with velocity $3{\mathrm{ms}}^{-1}$ for the other half of time. The mean velocity averaged over the whole time of motion is:

Velocity of a particle is given by $v=4x^3$. Average velocity of the particle averaged over displacement as it travels from $x_1=1 \mathrm{m}$ to $x_2=3 \mathrm{m}$ is