Question

A particle moves along the curve $y=\dfrac{{{x}^{2}}}{2}$. Here $x$ varies with time as $x=\dfrac{{{t}^{2}}}{2}$. Here $x$ and $y$ are measured in terms metres and $t$ is in seconds. At $t=2s$ the speed of the particle in $m{{s}^{-1}}$will be,
$\begin{align}
  & A.2\sqrt{5}m{{s}^{-1}} \\
 & B.2m{{s}^{-1}} \\
 & C.4m{{s}^{-1}} \\
 & D.5m{{s}^{-1}} \\
\end{align}$

Answer 1

Hint: First of all let us calculate the velocity of the particle. The velocity components will be there in both the directions $x$ and $y$. so first of all find out these components. Taking the resultant of these will give us the speed of the particle. These will be helping you in solving this question.

Complete step-by-step answer:
it has been mentioned in the question that the curve is mentioned in the equation as,
$y=\dfrac{{{x}^{2}}}{2}$
And also the $x$coordinates varies with the time which is given as,
$x=\dfrac{{{t}^{2}}}{2}$
As we all know, the velocity in a particular direction is the change in position of the object in that direction. Therefore we can write that,
The velocity of the particle in the $x$-direction can be written as,
$\therefore {{v}_{x}}=\dfrac{dx}{dt}$
As already mentioned the value of the $x$ can be written as,
$x=\dfrac{{{t}^{2}}}{2}$
Substituting this in the equation will give,
${{v}_{x}}=\dfrac{dx}{dt}=\dfrac{d}{dt}\left( \dfrac{{{t}^{2}}}{2} \right)$
Performing the differentiation will give,
${{v}_{x}}=\dfrac{d}{dt}\left( \dfrac{{{t}^{2}}}{2} \right)=\dfrac{2\times t}{2}=t$
In the similar way the velocity in the $y$direction can be written as,
${{v}_{y}}=\dfrac{d}{dt}y$
We already mentioned the equation of the curve.
$y=\dfrac{{{x}^{2}}}{2}$
And also,
$x=\dfrac{{{t}^{2}}}{2}$
Substituting this in the equation of curve will give,
$y=\dfrac{{{\left( \dfrac{{{t}^{2}}}{2} \right)}^{2}}}{2}=\dfrac{{{t}^{4}}}{8}$
Substitute this in the equation of velocity.
$\begin{align}
  & {{v}_{y}}=\dfrac{d}{dt}y \\
 & \Rightarrow {{v}_{y}}=\dfrac{d}{dt}\left( \dfrac{{{t}^{4}}}{8} \right)=\dfrac{4{{t}^{3}}}{8}=\dfrac{{{t}^{3}}}{2} \\
\end{align}$
As the value of time is mentioned in the question as,
$t=2s$
Let us substitute in the velocity equations will give,
\[{{v}_{y}}=\dfrac{{{t}^{3}}}{2}=\dfrac{{{2}^{3}}}{2}=4m{{s}^{-1}}\]
And the velocity in \[x\] direction will be given as,
\[{{v}_{x}}=t=2s\]
The magnitude of the velocity will be the speed. That is the resultant of the speed can be written as,
\[v=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}\]
Substituting the values in it will give,
\[v=\sqrt{{{2}^{2}}+{{4}^{2}}}=2\sqrt{5}m{{s}^{-1}}\]
Therefore the speed of the object will be given as,
\[s=2\sqrt{5}m{{s}^{-1}}\]

So, the correct answer is “Option A”.

Note: The velocity of the object can be shown in vector form as the \[\hat{i}\] and \[\hat{j}\]components. That is we can write that,
\[\vec{v}=2\hat{i}+4\hat{j}\]
The angle between velocity vectors will be the tangent of the velocity components. That is,
\[\begin{align}
  & \tan \theta =\dfrac{4}{2}=2 \\
 & \Rightarrow \theta =63.435{}^\circ \\
\end{align}\]