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Q.

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

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a

$\frac{2 t^{2}}{x^{5}}$

b

$\frac{2 t}{x^{4}}$

c

$\frac{2 t}{x^{5}}$

d

$\frac{2}{x^{5}}$

answer is B.

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Detailed Solution

$x^{3} = t^{3} + 1 \Rightarrow 3 x^{2} \frac{dx}{dt} = 3 t^{2} \Rightarrow \frac{dx}{dt} = \frac{t^{2}}{x^{2}}$ ……………………..(1)
$\begin{array}{l} \frac{d^{2} x}{{dt}^{2}} = \frac{2 {tx}^{2} - 2 x \frac{dx}{dt} t^{2}}{x^{4}} \\ = \frac{1}{x^{4}} [2 {tx}^{2} - 2 {xt}^{2} \frac{t^{2}}{x^{2}}] \\ = \frac{2 {tx}^{2}}{x^{4}} [1 - \frac{t^{3}}{x^{3}}] = \frac{2 {tx}^{2}}{x^{4}} [1 - \{\frac{x^{3} - 1}{x^{3}}\}] = \frac{2 t}{x^{5}} \end{array}$

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