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 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}\$

Solve Study Textbooks Guides

Use app

>>Class 11
>>Physics
>>Motion in a Straight Line
>>Acceleration
>> $Average and Instantaneous A...$

Average and Instantaneous Acceleration

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: 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: 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}\$

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\$

Physics

definition

Average Acceleration

Average acceleration is the change in velocity divided by an elapsed time. For instance, if the velocity of a marble increases from 0 to 60 cm/s in 3 seconds, its average acceleration would be 20 cm/s

^{2}

. This means that the marble's velocity will increase by 20 cm/s every second.

definition

Define and calculate instantaneous acceleration

When the object is moving with variable acceleration, then the object possesses different acceleration at different instant. The acceleration of the object at different instant of time or at given time of motion, is called instantaneous acceleration.

REVISE WITH CONCEPTS

Advanced Knowledge of Distance/Displacement - Time Graph

ExampleDefinitionsFormulaes

Advanced Knowledge of Velocity - Time Graph

ExampleDefinitionsFormulaes

Advanced Knowledge of Acceleration

ExampleDefinitionsFormulaes