A particle moves along a straight line such that its displacement at any time \\(t\\) is given by \\(s = t^3 - 6t^2 + 3t + 4\\) metres. The velocity when the acceleration is zero is:-

\\(S =t^3 - 6^2 +3t +4\\)\\(v= \dfrac{ds}{dt} \\)So \\(= 3t^2 -12 +3 =v \\)again differentiate ,\\(a = \dfrac{dv}{dt}\\)\\(6t -12 = 0\\)\\(t = 2 s\\)\\(v = 3(2)^2 - 12 \times 2 +3 = -9ms^{-1}\\)Hence (D) is correct answer.

The position x of a particle with respect to time t along x-axis is given by \\(x = 9t^2 - t^3\\) where x is in metres and t is in seconds. What will be the position of this particle when it achieves maximum speed along the + x direction?

Given that,\\(x = 9 t^2-t^3\\).....(I)We know that,\\(v=\dfrac{dx}{dt}\\)The maximum speed is\\(v= 18t-3t^2\\)\\(v\\) maximum, at \\(\dfrac{dv}{dt}=0\\)\\(\dfrac{dv}{dt}=18-6t\\)The time will be\\(0=18-6t\\)\\(t=3\ sec\\)The position of the particle at 3 secPut the value of t in equation (I)\\(x=9\times9-27\\)\\(x=81-27\\)\\(x= 54\ m\\)The position of the particle will be \\(54\ m\\).Hence, Option A is correct

The relation between time t and distance x is \\(t = a{x^2} + bx\\) where a and b are constants.The acceleration is

\\(t=ax^2+bx\\)Differentiate w.r.t time\\(1=2ax\dfrac{dx}{dt}+b\dfrac{dx}{dt}\\)\\(\dfrac{dx}{dt}=\dfrac{1}{2ax+b}\\)\\(V=\dfrac{1}{2ax+b}\\)...(1)\\(a=\dfrac{dV}{dt}=\dfrac{-(2a \dfrac{dx}{dt})}{(2ax+b)^2}\\)\\(a=\dfrac{-2av}{(2ax+b)^2}\\)....(2)Substiting \\(\dfrac{1}{2ax+b}=V\\) in (2)we will get, \\(a=-2av^3\\)

The position \\(x\\) of a particle varies with time as \\(x = at^{2} - bt^{3}\\). The acceleration of particle is zero at time \\(T\\) equal to

\\(x=at^2-bt^3\\)\\(\Rightarrow\\) Velocity \\(=\dfrac{dx}{dt}=2at-3bt^2\\)Acceleration\\(=\dfrac{d^2x}{dt^2}=\dfrac{d}{dt}(2at-3b^2)=2a-6bt\\)or \\(a=2a-6bt\\)Making \\(a=0\\)\\(\Rightarrow 2a=6bt\\)\\(\Rightarrow t=dfrac{2a}{6b}=\dfrac{a}{3b}\\).

The acceleration of a particle is increasing linearly with time \\(t\\) as \\(bt\\). The particle starts from the origin with an initial velocity \\({ v }_{ 0 }\\). The distance travelled by the particle in time \\(t\\) will be

here it is given that \\( a =bt \\)wkt, \\( a= \dfrac{change\ in\ velocity}{change\ in\ time} = \frac{dv}{dt} \\)\\( \therefore \dfrac{dv}{dt} = bt \\) now integrating this we will get \\( v \\)\\(v= \dfrac{bt^{2}}{2}+c\\)At \\( t =0\ s,\ v = v_{0},\ Thus\ v_{0} = c \\)\\( \therefore v= \dfrac{bt^{2}}{2}+v_{0} \\)Now. \\( v = \dfrac{change\ in\ position}{change\ in\ time} = \dfrac{ds}{dt} = \dfrac{bt^{2}}{2}+v_{0} \\)\\( \therefore ds = \left ( \dfrac{bt^{2}}{2} + v_{0}\right )dt \\)\\( \therefore \\) Integrating the above we get, \\( s = \dfrac{bt^{3}}{6} + v_{0}t+c^{'} at \ t = 0 ,\ s = 0 \therefore c{'} = 0 \\)i.e. \\( s = \dfrac{bt^{3}}{6} + v_{0}t \\)

The position of a particle is given by\\(\textbf r = 3.0t \hat{\textbf i} - 2.0t^2 \hat{\textbf j} + 4.0 \hat{\textbf k} \, m\\)where t is in seconds and the coefficients have the proper units for \\(\textbf r\\) to be in metres(a) Find the \\(\textbf v\\) and \\(\textbf a\\) of the particle?(b) What is the magnitude and direction of velocity of the particle at t = 2.0 s ?

(a) \\(\overrightarrow v(t) = d \overrightarrow r/dt \\) \\( =3.0 \hat i - 4.0t \hat j \\)\\(\overrightarrow a(t) = d \overrightarrow v(t) / dt \\) \\( = -4.0 \hat j \\)(b) \\(\overrightarrow v(2) = 3.0 \hat i - 8.0 \hat j\\)\\(v(2) = 8.54 m/s\\)\\(\angle v(2) = tan^{-1}(-8/3) = -69.44^\circ with \ x \ axis\\)

The \\(x\\) and \\(y\\) coordinates of the particle at any time are \\(x=5t-2t^2\\) and \\(y=10t\\) respectively, where \\(x\\) and \\(y\\) are in meters and \\(t\\) in seconds. The acceleration of the particle at \\(t=2\ s\\) is:

\\(x=5t-2t^2\\)\\(y=10t\\)\\(\dfrac{dx}{dt}=5-4t\\)\\(a_x=\dfrac{d^2x}{dt^2}=-4\\)\\(a_y=\dfrac{d^2y}{dt^2}=0\\)\\(a=a_x+a_y=-4+0\\)\\(a=-4 \ ms^{2}\\)

A particle starts from rest. Its acceleration (a) versus time (t) is as shown in the figure. The maximum speed of the particle will be:

Given,\\(u=0m/s\\) Particle initially at restAcceleration, \\(a=\dfrac{dv}{dt}\\)\\(dv=adt\\)\\(v=\int a dt\\). . . . . . . . .(1)From the above equation, velocity of the particle is equal to the area under the acceleration-time curve.From the given acceleration-time graph,\\(v_{max}=\dfrac{1}{2}\times base\times height\\)\\(v_{max}=\dfrac{1}{2}\times 11\times 10\\)\\(v_{max}=55m/s\\)The correct option is B.

A particle moving, eastwards with a velocity of \\(5 {m}/{s}\\). In \\(10 s\\), its velocity changes to \\(5 {m}/{s}\\) northwards. The average acceleration in this time is

Initial velocity, \\(u = 5 m/s\\) eastwardsFinal velocity, \\(v= 5m/s\\) northwards Now, change in the velocity \\(|\triangle v| = \sqrt {v^2 + u^2} = \sqrt{5^2 + 5^2} = 5\sqrt2 \\)And the direction of \\(\triangle v\\) is given by \\(tan \theta = \dfrac{v}{u} = \dfrac{-5}{5} = -1\\) (taking east direction as positive)\\(\theta = -\dfrac{\pi}{4}\\)Now, acceleration \\(a = \dfrac{\triangle v}{t}=\dfrac{5 \sqrt2}{10} = \dfrac{1}{\sqrt2} m/s^2\\) in north-west direction.

The initial velocity of a particle is \\(u\\) \\((at\;t=0)\\) and the acceleration is given by \\(f =at\\). Which of the following relations is valid?

As \\(\displaystyle f=at\;\;\;\;\Rightarrow \dfrac{dv}{dt}=at\\)\\(\displaystyle \Rightarrow \int_{u}^{v}dv=\int_{0}^{t}at\;\;dt\;\;\Rightarrow v-u=\dfrac{at^{2}}{2}\\)\\(\displaystyle \Rightarrow v=u+\dfrac{1}{2}at^{2}\\)

Basic Knowledge of Acceleration | Definition, Examples, Diagrams

definition

Acceleration

Acceleration is defined as the rate of change of velocity with time. It is a vector quantity. Its unit is

m / s^{2}

.
Constant speed does not guarantee that acceleration is zero. For example a body moving with constant speed in a circle changes its velocity every instant and hence its acceleration is not equal to zero.

definition

Retardation

Retardation (or deceleration) of a body whose velocity is decreasing with time is defined as the negative of acceleration. In other words, it is equal to the magnitude of acceleration of a body whose velocity is decreasing with time.
For a decelerating body, direction of velocity and acceleration is always opposite in direction.
Example: When a body is thrown vertically upwards, during the upward motion, its velocity decreases and at the highest point it becomes zero. This is a case of retardation.

definition

Unit of Acceleration

Acceleration is defined as the rate of change of velocity or velocity per unit time.Its unit is

m / s^{2}

, can also be expressed as

N / k g

.
Centripetal acceleration:Centripetal acceleration is the rate of change of tangential velocity.

Basic Knowledge of Acceleration

definition

Acceleration

definition

Retardation

definition

Unit of Acceleration

REVISE WITH CONCEPTS

Uniform and Non-uniform Acceleration

Advanced Knowledge of Distance/Displacement - Time Graph

Acceleration - Time Graph

Average and Instantaneous Acceleration

Advanced Knowledge of Acceleration

Average and Instantaneous Acceleration - Problem L1

Quick Summary With Stories

Acceleration

A particle moves along a straight line such that its displacement at any time $t$ is given by $s = t^{3} - 6 t^{2} + 3 t + 4$ metres. The velocity when the acceleration is zero is:-

The position x of a particle with respect to time t along x-axis is given by $x = 9 t^{2} - t^{3}$ where x is in metres and t is in seconds. What will be the position of this particle when it achieves maximum speed along the + x direction?

The relation between time t and distance x is $t = a x^{2} + b x$ where a and b are constants.The acceleration is

The position $x$ of a particle varies with time as $x = a t^{2} - b t^{3}$ . The acceleration of particle is zero at time $T$ equal to

The acceleration of a particle is increasing linearly with time $t$ as $b t$ . The particle starts from the origin with an initial velocity $v_{0}$ . The distance travelled by the particle in time $t$ will be

The $x$ and $y$ coordinates of the particle at any time are $x = 5 t - 2 t^{2}$ and $y = 10 t$ respectively, where $x$ and $y$ are in meters and $t$ in seconds. The acceleration of the particle at $t = 2 s$ is:

A particle starts from rest. Its acceleration (a) versus time (t) is as shown in the figure. The maximum speed of the particle will be:

A particle moving, eastwards with a velocity of $5 m / s$ . In $10 s$ , its velocity changes to $5 m / s$ northwards. The average acceleration in this time is

The initial velocity of a particle is $u$ $(a t t = 0)$ and the acceleration is given by $f = a t$ . Which of the following relations is valid?

definition

Acceleration

definition

Retardation

definition

Unit of Acceleration

REVISE WITH CONCEPTS

Uniform and Non-uniform Acceleration

Advanced Knowledge of Distance/Displacement - Time Graph

Acceleration - Time Graph

Average and Instantaneous Acceleration

Advanced Knowledge of Acceleration

Average and Instantaneous Acceleration - Problem L1

Quick Summary With Stories

Acceleration

A particle moves along a straight line such that its displacement at any time t is given by s=t3−6t2+3t+4 metres. The velocity when the acceleration is zero is:-

The position x of a particle with respect to time t along x-axis is given by x=9t2−t3 where x is in metres and t is in seconds. What will be the position of this particle when it achieves maximum speed along the + x direction?

The relation between time t and distance x is t=ax2+bx where a and b are constants.The acceleration is

The position x of a particle varies with time as x=at2−bt3. The acceleration of particle is zero at time T equal to

The acceleration of a particle is increasing linearly with time t as bt. The particle starts from the origin with an initial velocity v0​. The distance travelled by the particle in time t will be

The position of a particle is given byr=3.0ti^−2.0t2j^​+4.0k^mwhere t is in seconds and the coefficients have the proper units for r to be in metres(a) Find the v and a of the particle?(b) What is the magnitude and direction of velocity of the particle at t = 2.0 s ?

The x and y coordinates of the particle at any time are x=5t−2t2 and y=10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t=2 s is:

A particle starts from rest. Its acceleration (a) versus time (t) is as shown in the figure. The maximum speed of the particle will be:

A particle moving, eastwards with a velocity of 5m/s. In 10s, its velocity changes to 5m/s northwards. The average acceleration in this time is

The initial velocity of a particle is u (att=0) and the acceleration is given by f=at. Which of the following relations is valid?

A particle moves along a straight line such that its displacement at any time $t$ is given by $s = t^{3} - 6 t^{2} + 3 t + 4$ metres. The velocity when the acceleration is zero is:-

The position x of a particle with respect to time t along x-axis is given by $x = 9 t^{2} - t^{3}$ where x is in metres and t is in seconds. What will be the position of this particle when it achieves maximum speed along the + x direction?

The relation between time t and distance x is $t = a x^{2} + b x$ where a and b are constants.The acceleration is

The position $x$ of a particle varies with time as $x = a t^{2} - b t^{3}$ . The acceleration of particle is zero at time $T$ equal to

The acceleration of a particle is increasing linearly with time $t$ as $b t$ . The particle starts from the origin with an initial velocity $v_{0}$ . The distance travelled by the particle in time $t$ will be

The $x$ and $y$ coordinates of the particle at any time are $x = 5 t - 2 t^{2}$ and $y = 10 t$ respectively, where $x$ and $y$ are in meters and $t$ in seconds. The acceleration of the particle at $t = 2 s$ is:

A particle moving, eastwards with a velocity of $5 m / s$ . In $10 s$ , its velocity changes to $5 m / s$ northwards. The average acceleration in this time is

The initial velocity of a particle is $u$ $(a t t = 0)$ and the acceleration is given by $f = a t$ . Which of the following relations is valid?