Single Choice

A point moves in a straight line so that its displacement $$x$$ metre at time t second is given by $$x^2=1+t^2$$. Its acceleration in $$ms^{-2}$$ at time t second is

A$$\dfrac { 1 }{ x^{ 3 } } $$
Correct Answer
B$$\dfrac { -t }{ x^{ 3 } } $$
C$$\dfrac{1}{x}-\dfrac{t^2}{x^3}$$
D$$\dfrac{1}{x}-\dfrac{1}{x^2}$$

Solution

$$x^2=t^2+1$$
taking its differentiation,
$$2x\times (\dfrac{dx}{dt})=2t$$

$$x\times \left(\dfrac{dx}{dt}\right)=t$$

again taking differentiation for acceleration,
$$x\times (\dfrac{d^{2}x}{dt^{2}}) + \left(\dfrac{dx}{dt}\right)^2 = 1$$

$$x\times\left (\dfrac{d^{2}x}{dt^2}\right) = 1 – (\dfrac{dx}{dt})^2$$

$$x\times\left(\dfrac{d^{2}x}{dt^{2}}\right) = 1 – \left(\dfrac{t}{x}\right)^2 $$ $$ (because \ x\times\left(\dfrac{dx}{dt}\right)=t)$$

$$\left(\dfrac{d^{2}x}{dt^2}\right) =\dfrac{ \left(1 – \left(\dfrac{t}{x}\right)^2\right)}{x}$$

$$\left(\dfrac{d^{2}x}{dt^{2}}\right) = \dfrac{1}{x} –\dfrac{ t^2}{x^3}$$

put $$t^2 = x^2 -1$$

$$\left(\dfrac{d^2x}{dt^2}\right) = \dfrac{1}{x^3}$$

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