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
>> $Acceleration - Time Graph$

Acceleration - Time Graph

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

Interpretation of acceleration-time graph

Acceleration vs. time graphs tell us about an object's velocity in the same way that velocity vs. time graphs tell us about an object's displacement. The change in velocity in a given time interval is equal to the area under the graph during that same time interval.

definition

Acceleration Vs Time Graph for piece-wise execution of motion

Acceleration-time graph for a particle describing motion piece-wise for separate time intervals.

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