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\$

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>>Class 11
>>Physics
>>Motion in a Straight Line
>>Acceleration
>> $Uniform and Non-uniform Acc...$

Uniform and Non-uniform 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: 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}\$

Q: 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.

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}\$

Physics

definition

Uniform acceleration

A body under constant acceleration is called as uniformly accelerated motion. This happens when a constant external force is applied on the body.
Note:
Direction of acceleration plays a role in the change of velocity. However, direction of acceleration of a body may not be same as the direction of its motion.

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