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\\)

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}\\)

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

Initial velocity \\(u = 30\hat{i}\\)Final velocity \\(v = 40\hat{j}\\)Change in velocity \\(\Delta v = 40\hat{j}-30\hat{i}\\)Magnitude \\(|\Delta v| = \sqrt{30^2+40^2} =50 \\)Average acceleration \\(a = \dfrac{|\Delta v|}{\Delta t} = \dfrac{50}{10} = 5 \ m/s^2\\)

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to \\(v(x)=\beta x^{-2n}\\), where \\(\beta\\) and \\(n\\) are constants and \\(x\\) is the position of the particle. The acceleration of the particle as a function of \\(x\\), is given by

\\(v(x)=\beta x^{-2n}\\)\\(a(x)=v\dfrac{dv}{dx}= \beta x^{-2n}\dfrac{d\beta x^{-2n}}{dx}\\)\\(a(x)= \beta x^{-2n} x^{-2n} \times \beta (-2n)x^{-2n-1}\\)\\(a(x)= -2n\beta^2 x^{-4n-1}\\)

Advanced Knowledge of Distance/Displacement - Time Graph

result

Distance in nth second(distance-time graph)

Distance traveled in nth second can be found using distance time-graph. In the given figure, distance after

21 s

d_{2} = 21 m

and after

20 s

d_{1} = 20 m

. Hence, distance traveled in

2 1^{st}

sec is

d_{2} - d_{1} = 1 m

definition

Average velocity from a displacement-time graph

Average velocity equals the ratio of net displacement and the time taken.
Example:
In the given plot, for

t \in (0, 10)

,
Displacement

s = - 4 - 0 = - 4

Hence, average velocity is given by

v = s / t = - 4 / 10 = - 0.4 m / s

definition

Calculation of net force from given velocity time graph

A block of mass 12 kg is hanging with a light string in a lift as shown in the figure (a). The lift is moving in upward direction and its speed time graph is plotted as shown in the figure (b) : (take

g = 10 m / s^{2}

):
In the time interval

t = 0

t = 5

, the acceleration

a = \frac{1 0 - 0}{5 - 0} = 2 m / s^{2}

Hence

T = m (g + a)

T = 12 \times (10 + 2) = 144 N

In the time interval

t = 5

t = 10

, the acceleration

a = 0

Hence

T = m g = 12 \times 10 = 120 N

In the time interval

t = 10

t = 15

, the acceleration

a = \frac{3 0 - 1 0}{1 5 - 1 0} = 4 m / s^{2}

Hence

T = m (g + a)

T = 12 \times (10 + 4) = 168 N

Advanced Knowledge of Distance/Displacement - Time Graph

result

Distance in nth second(distance-time graph)

definition

Average velocity from a displacement-time graph

definition

Calculation of net force from given velocity time graph

REVISE WITH CONCEPTS

Advanced Knowledge of Velocity - Time Graph

Average and Instantaneous Acceleration

Advanced Knowledge of Acceleration

Average and Instantaneous Acceleration - Problem L1

Quick Summary With Stories

Displacement-Time Graph

Equation For Position-Time Relation

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

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?

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

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to $v (x) = β x^{- 2 n}$ , where $β$ and $n$ are constants and $x$ is the position of the particle. The acceleration of the particle as a function of $x$ , is given by

result

Distance in nth second(distance-time graph)

definition

Average velocity from a displacement-time graph

definition

Calculation of net force from given velocity time graph

REVISE WITH CONCEPTS

Advanced Knowledge of Velocity - Time Graph

Average and Instantaneous Acceleration

Advanced Knowledge of Acceleration

Average and Instantaneous Acceleration - Problem L1

Quick Summary With Stories

Displacement-Time Graph

Equation For Position-Time Relation

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 ?

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 body is moving with velocity 30 m/s towards east. After 10 seconds its velocity becomes 40 m/s towards north. The average acceleration of the body is

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x)=βx−2n, where β and n are constants and x is the position of the particle. The acceleration of the particle as a function of x, is given by

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

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?

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to $v (x) = β x^{- 2 n}$ , where $β$ and $n$ are constants and $x$ is the position of the particle. The acceleration of the particle as a function of $x$ , is given by