Question

In the one-dimensional motion of a particle, the relation between position $x$ and time $t$ is given by $x^2+2x=t$ , where $x>0$. Choose the correct statement.

• A. The retardation of the particle is $\frac{1}{4(x+1{)}^3}$.
• B. The velocity of the particle is $\frac{1}{(x+1{)}^3}$.
• C. Both are correct
• D. Both are wrong

Answer 1

The correct option is A The retardation of the particle is $\frac{1}{4(x+1{)}^3}$.

Given, $t=x^2+2x(x>0)$

Differentiating with respect to ${}^{\prime }{x}^{\prime }$ on both sides,
$\frac{dt}{dx}=2x+2=2(x+1)$

$v=\frac{dx}{dt}=\frac{1}{2(x+1)}.....\left(1\right)$

Differentiating with respect to ${}^{\prime }{t}^{\prime }$ on both sides,
$a=\frac{dv}{dt}=-\frac{1}{2(x+1{)}^2}.\frac{dx}{dt}$

Using $\left(1\right)$ in the above equation , we get
$a=-\frac{1}{4(x+1{)}^3}$

Hence, option (a) is the correct answer.