Question

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

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

Answer 1

Answer: (D)

$x^{-3}$

Given, $x^2=2t^2+6t+1$ ...(i)
Differentiating Eq. (i) w.r.t. t, we get
$2x\frac{dx}{dt}=4t+6$
$2xv=4t+6$ $\left(\begin{array}{c}\because v=\frac{dx}{dt}\end{array}\right)$
$xv=2t+3$ ...(ii)
Now, again differentiating Eq.(ii) w.r.t. t, we get
$x\frac{dv}{dt}+v\frac{dx}{dt}=2$
$x.a+v.v=2$ $\left(\begin{array}{c}\because a=\frac{dv}{dt}andv=\frac{dx}{dt}\end{array}\right)$
$xa+v^2=2$ ...(iii)
Here, $v^2=\frac{4t^2+12t+9}{x^2}$
$v^2=\frac{2(2t^2+6t+1)+7}{x^2}$
$v^2=\frac{2x^2+7}{x^2}$ ...(iv)
Put the value of $v^2$ in Eq. (iii) from Eq. (iv), we get
$xa+\frac{2x^2+7}{x^2}=2$
$x^{3}a+2x^2+7=2x^2$
$x^{3}a+7=0$
$x^{3}a=-7$
$a=\frac{-7}{x^3}$
Hence, the acceleration varies with $x^{-3}$