Question

The position of a particle along the x-axis at time ‘t’ is given by . The displacement and the distance travelled in the interval to are respectively: -
A). $-2,2.16$
B) 0, 2
C) 2, 2
D) None.

Answer 1

Hint In order to solve this particular question, the first thing we need to do is to differentiate the equation of position(x) with respect to ‘t’.
Further we would do differentiation once again to check whether the body is accelerating or retarding from its position.
 Now, we will equate the velocity to zero (0), to get the time at t=0 and at time t = 1 and hence will find distance and displacement.

Complete Step – by –Step Solution
Given in the question; the position of particle;
$x=2+t-3{{t}^{2}}$ (1)
To find out it’s velocity and acceleration, we need to differentiate with respect to ‘it’ on both sides of eqn
So, differentiating both the sides of eqn
$\Rightarrow \dfrac{dx}{dt}=\dfrac{d}{dt}\left( 2+t-3{{t}^{2}} \right)$
$\Rightarrow \dfrac{dx}{dt}=\dfrac{d}{dt}\left( 2 \right)+\dfrac{d}{dt}\left( t \right)-\dfrac{d}{dt}\left( 3{{t}^{2}} \right)$
$\Rightarrow \dfrac{dx}{dt}=0+1-\left( 3\times 2 \right)t$
$\Rightarrow \dfrac{dx}{dt}=0+1-6t=1-6t$
$\Rightarrow \dfrac{dx}{dt}=1-6t$ (2)
Since, differentiation of any constant term with respect to ‘it’ is zero (0);
So, $\dfrac{d}{dt}\left( 2 \right)=0;$
Now; we know that velocity ‘v’$=\dfrac{dx}{dt};$ because it means that the velocity over a certain period of time is the instantaneous change in position $\left( dx \right)$ over the instantaneous change in time $\left( dt \right)$. This period of time is intentionally very small,. hence “instantaneous”.
Now, to get the acceleration; differentiate equation$\left( 2 \right);$
Differentiating both the sides;
$\Rightarrow \dfrac{dx}{dt}=1-6t-\left( 2 \right)$
$\Rightarrow \dfrac{{{d}^{2}}x}{d{{t}^{2}}}=\dfrac{d}{dt}\left( 1 \right)-\dfrac{d}{dt}\left( 6t \right)$
$\Rightarrow \dfrac{{{d}^{2}}x}{d{{t}^{2}}}=0-6$
$\Rightarrow \dfrac{{{d}^{2}}x}{d{{t}^{2}}}=-6$ < 0
Hence, motion is retarding here, because
$\dfrac{{{d}^{2}}x}{d{{t}^{2}}}=-6$ < 0
and,. acceleration ‘a’$=\dfrac{{{d}^{2}}x}{d{{t}^{2}}}.$
Let the velocity be zero (0) at time to;
So, $v=\dfrac{dx}{dt}$
and, from equation $\left( 2 \right),\dfrac{dx}{dt}=1-6t$
So, v$=\dfrac{dx}{dt}=1-6t=0$
$\Rightarrow 1-6t=0$
$\Rightarrow \therefore $
$\therefore $
So, at $\therefore $ $t=\dfrac{1}{6},$ velocity of partial is zero , there partite moves in back direction at $t=1\sec .$
Now , displacement = final position of the
Particle- initial position of the particle
$=x\left( 1 \right)-x\left( 0 \right)$
$={{\left( 2+t-3{{t}^{2}} \right)}_{1}}-{{\left( 2+t-3{{t}^{2}} \right)}_{0}}$
$=\left[ 2+1-3{{\left( 1 \right)}^{2}} \right]-\left[ 2+0-3{{\left( 0 \right)}^{\begin{smallmatrix}
 2 \\

\end{smallmatrix}}} \right]$
$=\left[ 3-3 \right]-\left[ 2-0 \right]$
$=0-2$
$=-2m.$
 displacement $=-2m;$
Distance $=\left| x\left( 1 \right)-x\left( \dfrac{1}{6} \right) \right|+\left| x\left( \dfrac{1}{6} \right)-x\left( 0 \right) \right|$
$=\left[ {{\left( 2+t-3{{t}^{2}} \right)}_{1}}-{{\left( 2+t-3{{t}^{2}} \right)}_{\dfrac{1}{6}}} \right]+$ $\left[ {{\left( 2+t-3{{t}^{2}} \right)}_{\dfrac{1}{6}}}-{{\left( 2+t-3{{t}^{2}} \right)}_{0}} \right]$
$=\left[ 2+1-3{{\left( 1 \right)}^{2}}-\left( 2+\dfrac{1}{6}-3{{\left( \dfrac{1}{6} \right)}^{2}} \right) \right]+$
$\left[ \left\{ 2+\dfrac{1}{6}-3\left( {{\dfrac{1}{6}}^{2}} \right) \right\}-\left\{ 2+0-3{{\left( 0 \right)}^{2}} \right\} \right]$
$=\left[ 3-3\left\{ 2+\dfrac{1}{6}-\dfrac{3}{36} \right\} \right]+\left[ \left\{ 2+\dfrac{1}{6}-\dfrac{3}{36} \right\}-2 \right]$
\[=\left[ 0-\left( \dfrac{72+6-3}{36} \right) \right]+\left[ \left( \dfrac{72+6-3}{36} \right)-2 \right]\]
$=\left[ -2-\dfrac{1}{6}+\dfrac{1}{12} \right]+\left[ 2+\dfrac{1}{6}-\dfrac{1}{12}-2 \right]$
$=\left[ -2-\dfrac{1}{12} \right]+\left[ \dfrac{1}{12} \right]$
$=2+\dfrac{1}{12}+\dfrac{1}{12}$
$=2+\dfrac{1}{6}$
$=\dfrac{13}{6}=2.16\ m$
Distance $=2.16m$
Displacement $=-2m$

Hence, option 'a' is the correct answer

Note Approach to solve this question can be done by differentiation and double differentiation as because velocity is the first derivative of position with respect to time and acceleration is the first derivative of the velocity with respect to time. Reverse this operation. Instead of differentiating position to find velocity, integrate velocity to find position.