Acceleration

A body is moving with velocity  $30\frac{m}{s}$  towards east. After 10 seconds its velocity becomes  $40\frac{m}{s}$  towards north. The average acceleration of the body is

A particle is moving eastwards with a velocity of  $5\mathrm{m}{\mathrm{s}}^{-1}$. In $10 \mathrm{s}$, the velocity changes to  $5\mathrm{m}{\mathrm{s}}^{-1}$ northwards. The average acceleration in this time is

For a body moving on a straight line if $x$ is position coordinate and $t$ is time then acceleration of body is constant when -
(1) $x$ and velocity is linear
(2) $x$ and square of velocity is linear
(3) $t$ and velocity is linear
(4) $t$ and square of velocity is linear.

A particle of mass $10 \mathrm{g}$moves in a straight line with retardation $2x,$ where $x$ is the displacement in $\mathrm{SI}$ units. Its loss of kinetic energy for above displacement is ${\left(\frac{10}{x}\right)}^{-n}\mathrm{J}.$ The value of $n$ will be $\_\_\_\_\_\_\_\_.$

At any instant the velocity of a particle of mass $500 \mathrm{g}$ is $\left(2t \hat{\mathrm{i}} + 3t^2 \hat{\mathrm{j}}\right) \mathrm{m} {\mathrm{s}}^{-1}$ . If the force acting on the particle at $t=1 \mathrm{s}$ is $(\hat{\mathrm{i}} + x \hat{\mathrm{j}}) \mathrm{N}$ . Then the value of $x$ will be:

If position vector of a particle is given by $\overrightarrow{r}\left(t\right)=8t\hat{\mathrm{i}}+5t^2\hat{\mathrm{j}}+6\hat{\mathrm{k}}$, then the correct statement about the acceleration of the particle is

A bock of mass $100 \mathrm{g}$ is placed on smooth surface, moves with acceleration of $a= 2x$, then the change in kinetic energy can be given as $\left(\frac{x^n}{10}\right)$, find the value of $n$.

Velocity of a particle moving along x-axis varies with time $t$ as $v= (5t^2-7)$. Find the average acceleration of the particle from $t = 2 \mathrm{s}$ to $t = 4 \mathrm{s}$.

A particle moves in a straight line under an acceleration varying linearly with time. Its velocity-time graph is shown here. Tangent to point $A$ and $C$ makes an angle of $37°$ and $135°$ with positive $x$-axis respectively. Then

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At $t=0$, a particle at the origin has a velocity of $15 \mathrm{m} {\mathrm{s}}^{-1}$ at $37°$ above the horizontal $X$-axis. At $t=5 \mathrm{s}$, it is at $x=20 \mathrm{m}$ and $y=35 \mathrm{m}$ and its velocity is $30 \mathrm{m} {\mathrm{s}}^{-1}$ at $53°$ above the horizontal. Find

(a) its average velocity :

(b) its average acceleration.

A body moves with velocity $v=\sin \left(x\right) \mathrm{m} {\mathrm{s}}^{-1}$, where $x$ is its position. Acceleration of body is zero at

If $v=2x^2 \mathrm{m} {\mathrm{s}}^{-1}$ particle starts from rest at $x=0$. Find
(i) $a\left(x\right)$
(ii) $a$ at $x=1 \mathrm{m}$

Displacement as function of time is given as $s=3t^2-8t+24$. Find $t$ at which its acceleration is $0$.

Assertion: The zero velocity of a particle at any instant always implies zero acceleration at that instant

Reason: A body is momentarily at rest when reverses its direction of motion.

The correct option among the following is

Velocity of a particle moving along $x$-axis varies with time $t$ as $v=\left(5t^2-7\right){\mathrm{ms}}^{-1}$ where $t$ is in seconds. The average acceleration of the particle from $t=2 \mathrm{s}$ to $t=4 \mathrm{s}$ is

A car moves with constant speed around a horseshoe-shaped path as shown with the arrows in the figure. Which one of the following choices best describes the direction of the average acceleration of the car in travelling from W to X?

The displacement $x$ of a particle at time $t$ along a straight line is given by $x=\alpha -\beta t+\gamma t^2$. Find the acceleration of the particle.

The acceleration-time graph of a body starting from rest is shown below

The average acceleration of the body from $t= 0 \mathrm{s}$ to $t= 40 \mathrm{s}$ is

The acceleration $a$ of a body starting from rest varies with time following the equation $a= 8t+5$. The velocity of of the body at time $t= 2 \mathrm{s}$ is