Question

The distance $x$ covered by a particle varies with time $t$ as $x^{2}=2t^{2} +6i +1.$. Its acceleration varies with $x$ as

• A. $x$
• B. $x^{2}$
• C. $x^{-1}$
• D. $x^{-3}$

Answer 1

The Correct Option is D

Given, $x^{2}=2 t^{2}+6 t+1\,\,\,\,\dots(i)$
Differentiating E (i) w.r.t. $t$, we get
$2 x \frac{d x}{d t}=4 t+t $
$2 x v=4 t+6$ $\left(\because v=\frac{d x}{d t}\right) $
$x v=2 t+3\,\,\,\,\,\dots(ii)$
Now, again differentiating E (ii) w.r.t. $t$, we get
$x\, \frac{d v}{d t}+v \frac{d x}{d t}=2$
$x \cdot a+v \cdot v=2 \, \left(\because a=\frac{d v}{d t} \text { and } v=\frac{d x}{d t}\right) $
$x a+v^{2}=2 \,\,\,\,\,\dots(iii)$
Here, $v^{2}=\frac{4 t^{2}+12 t+9}{x^{2}}$
$v^{2}=\frac{2\left(2 t^{2}+6 t+1\right)+7}{x^{2}} $
$v^{2}=\frac{2 t^{2}+7}{x^{2}}\,\,\,\,\,\dots(iv)$
Put the value of $v^{2}$ in E (iii) from E (iv), we get
$x a+\frac{2 x^{2}+7}{x^{2}} =2 $
$x^{3} a+2 x^{2}+7=2 x^{2} $
$x^{3} a+7 =0 $
$x^{3} a =-7 $
$a =\frac{-7}{x^{3}}$
Hence, the acceleration varies with $x^{-3}$.

Answer 2

Given equation (i):
\(x^2 = 2t^2 + 6t + 1\)

Differentiating with respect to t:
\(2x \frac{dx}{dt} = 4t + 6\)

Solving for \(\frac{dx}{dt}\):
\(\frac{dx}{dt} = \frac{4t + 6}{2x} = \frac{2t + 3}{x}\)

This gives us equation (ii):
\(xv = 2t + 3\)

Now, Differentiating with respect to t:
\(x\frac{dv}{dt} + v\frac{dx}{dt} = 2\)

Using \(v = \frac{dx}{dt}\):
\(xa + v^2 = 2\)

Substituting \(v^2\):
\(xa + \frac{2t^2 + 7}{x^2} = 2\)

Multiplying through by \(x^2\):
\(x^3a + 2x^2 + 7 = 2x^2\)

Simplifying:
\(x^3a + 7 = 0\)
\(x^3a = -7\)
\(a=\frac{-7}{x^3}\)
Hence, the acceleration varies with x^-3.

So, the correct option is (D): x^-3.