Question

A point moves such that its displacement as a function of time is given by $x^3=t^3+1$. Its acceleration as a function of time $t$ will be:

• A. $\frac{2}{x^5}$
• B. $\frac{2t}{x^3}$
• C. $\frac{2t}{x^4}$
• D. $\frac{{2t}^2}{x^5}$

Answer 1

The correct option is B $\frac{2t}{x^3}$

$x^3=t^3+1$

$x={(t^3+1)}^{1/3}$

The velocity is given by $v=\frac{dx}{dt}=\frac{d}{dt}{(t^3+1)}^{1/3}$

$=\frac{1}{3}{(t^3+1)}^{-2/3}.{3t}^2$

$v=t^2.{(t^3+1)}^{-2/3}$

The acceleration is given by $a=\frac{dv}{dt}=\frac{d}{dt}\left(t^2{\left(t^3+1\right)}^{-2/3}\right)$

$=2t.{(t^3+1)}^{-2/3}+t^2.\frac{-2}{3}{(t^3+1)}^{-5/3}.{3t}^2$

$=2t{\left(x^3\right)}^{-2/3}-2t^4{\left(x^3\right)}^{-5/3}$

$=2t{x}^{-2}-{2t}^4{x}^{-5}$

$=2t(x^{-2}-t^3.x^{-5})$

$=2t(\frac{1}{x^2}-\frac{x^3-1}{x^5})\because t^3=x^3-1$

On solving

$a=\frac{2t}{x^3}$