Q: 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.

Q: 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

Q: 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\$

Q: 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}\$.

Q: 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 \$

Q: 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\$

Q: 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\$

Q: 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.

Q: 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}\$

Question 1

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:-

Accepted Answer

$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.

Question 2

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?

Accepted Answer

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

Question 3

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

Accepted Answer

$t=ax^2+bx$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, $a=-2av^3$

Question 4

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

Accepted Answer

$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}$.

Question 5

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

Accepted Answer

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 $

Question 6

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 ?

Accepted Answer

(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$

Question 7

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

Accepted Answer

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   $a = \dfrac{|\Delta v|}{\Delta t} = \dfrac{50}{10} = 5 \ m/s^2$

Question 8

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

Accepted Answer

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.

Question 9

The elevator shown in figure is descending, with an acceleration of 2 $m/s^{2}$. The mass of the block A is 0.5 kg. The force exerted by the block A on the block B is -

Accepted Answer

When the elevator in descending apseudo force acts on it in the upword direction $\therefore \ $ from free body diagram$m=0.5$$g=10$$a=2$$mg-N=ma$$N=m(g-a)$$=0.5(10-2)=4\ N$so the force exerted by the block $A\ m$ the block $B$ in $4N$

Question 10

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

Accepted Answer

$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}$

Revise with Concepts

Advanced Knowledge of Distance/Displacement - Time Graph

Advanced Knowledge of Velocity - Time Graph

Average and Instantaneous Acceleration

Advanced Knowledge of Acceleration

Average and Instantaneous Acceleration - Problem L1

Quick Summary With Stories

Acceleration

Motion in Plane with Constant Acceleration

Displacement-Time Graph

Equation For Position-Time Relation

Velocity-Time Graph

Acceleration From Velocity-Time Graph

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 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 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 elevator shown in figure is descending, with an acceleration of 2 $m / s^{2}$ . The mass of the block A is 0.5 kg. The force exerted by the block A on the block B 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

Revise with Concepts

Advanced Knowledge of Distance/Displacement - Time Graph

Advanced Knowledge of Velocity - Time Graph

Average and Instantaneous Acceleration

Advanced Knowledge of Acceleration

Average and Instantaneous Acceleration - Problem L1

Quick Summary With Stories

Acceleration

Motion in Plane with Constant Acceleration

Displacement-Time Graph

Equation For Position-Time Relation

Velocity-Time Graph

Acceleration From Velocity-Time Graph

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 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 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 elevator shown in figure is descending, with an acceleration of 2 m/s2. The mass of the block A is 0.5 kg. The force exerted by the block A on the block B 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 elevator shown in figure is descending, with an acceleration of 2 $m / s^{2}$ . The mass of the block A is 0.5 kg. The force exerted by the block A on the block B 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