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Question: A particle moves in a straight line and its position x at time t is given by \({{x}^{2}}=2+t\) . Its...

A particle moves in a straight line and its position x at time t is given by x2=2+t{{x}^{2}}=2+t . Its acceleration is given by :-

& \text{A}\text{. }\dfrac{-2}{{{x}^{3}}} \\\ & \text{B}\text{. }-\dfrac{1}{4{{x}^{3}}} \\\ & \text{C}\text{. }-\dfrac{1}{4{{x}^{2}}} \\\ & \text{D}\text{. }\dfrac{1}{{{x}^{2}}} \\\ \end{aligned}$$
Explanation

Solution

At first we need to look carefully into the question and analyze it, in the question its given a position with respect to time so we know that if we differentiate it with ‘t’ then we will get the velocity of the particle and if we again differentiate the velocity with time we will get acceleration so and that is what the question wants us to answer.

Complete answer:
So, there is a particle whose position is given with respect to time,
Its position is x2=2+t{{x}^{2}}=2+t, or we can say,
x=2+tx=\sqrt{2+t}.
Now, as the position is with respect to time so, we will differentiate it with t,
dxdt=d2+tdt\dfrac{dx}{dt}=\dfrac{d\sqrt{2+t}}{dt}
So, on differentiating,
dxdt=122+t\dfrac{dx}{dt}=\dfrac{1}{2\sqrt{2+t}},
Now we know that displacement with respect to time is, velocity therefore,
v=dxdt=122+tv=\dfrac{dx}{dt}=\dfrac{1}{2\sqrt{2+t}},
Again differentiating this with respect to ‘t’ and this is known as acceleration, as velocity with respect to time is acceleration,
a=dvdt=d122+tdta=\dfrac{dv}{dt}=\dfrac{d\dfrac{1}{2\sqrt{2+t}}}{dt}
On solving this we get,
a=dvdt=d122+tdta=\dfrac{dv}{dt}=\dfrac{d\dfrac{1}{2\sqrt{2+t}}}{dt}
⇒a=12(−12)(2+t)−12−1\Rightarrow a=\dfrac{1}{2}\left( -\dfrac{1}{2} \right){{\left( 2+t \right)}^{-\dfrac{1}{2}-1}}
⇒a=(−14)(2+t)−32\Rightarrow a=\left( -\dfrac{1}{4} \right){{\left( 2+t \right)}^{-\dfrac{3}{2}}}
⇒a=(−14)(2+t)−3\Rightarrow a=\left( -\dfrac{1}{4} \right){{\left( \sqrt{2+t} \right)}^{-3}}
⇒a=(−14(2+t)3)\Rightarrow a=\left( -\dfrac{1}{4{{\left( \sqrt{2+t} \right)}^{3}}} \right)
Now, we see that x=2+tx=\sqrt{2+t}
So,
a=(−14x3)a=\left( -\dfrac{1}{4{{x}^{3}}} \right)

So, option C that is−14x2-\dfrac{1}{4{{x}^{2}}}, is the correct option.

Note:
When we are doing differentiation one might feel problem in doing the first step that is dxdt=122+t\dfrac{dx}{dt}=\dfrac{1}{2\sqrt{2+t}} here we will have to apply chain rule to get the desired result. Analysis of a question must like the values that are given, what the question wants to tell. While differentiating never skip a step as it will lead to mistakes and steps containing marks.