Question

If the time and displacement of particles along the positive X-axis are related as t = √(x²-1), then find the acceleration in terms of x.

Answer 1

Final Answer :
$a(x) = \frac{1}{ {x}^{3} }$
Steps and Understanding :
1)
$t = \sqrt{ {x}^{2} - 1} \\ = > {t}^{2} = {x}^{2} - 1 \\ differentiating \: wrt \: t \\ = > 2t = 2x \frac{dx}{dt} - 0 \\ = > t = xv - - - - (1) \\ \: differentiating \: wrt \: t \: \\ = > 1 = xa + {v}^{2} \\ = > a = \frac{1 - {v}^{2} }{x} \\ = > a = \frac{1 - {( \frac{t}{x}) }^{2} }{x} \\ = > a = \frac{ {x}^{2} - {t}^{2} }{ {x}^{3} } \\ = > a \: \: = \frac{ {x}^{2} - ( {x}^{2} - 1) }{ {x}^{3} } \\ = > a \: = \frac{1}{ {x}^{3} }$

Answer 2

Answer:

The acceleration in term of x is $a = \dfrac{1}{x^3}$.

Explanation:

Given that,

$t = \sqrt{x^2-1}$

$t^2=x^2-1$...(I)

Where, t shows the time and x shows the displacement.

On differentiating of equation w.r.t. t

$2t=2x\dfrac{dx}{dt}-0$

$t=x\dfrac{dx}{dt}$

$t= xv$....(II)

Here,$\dfrac{dx}{dt}=v$

On differentiating  equation (II) w.r.t. t

$1=x\times\dfrac{dv}{dt}+v\times\dfrac{dx}{dt}$

Here, $\dfrac{dv}{dt}=a$

a = acceleration

$1=xa+v^2$

$1-v^2=xa$

$a = \dfrac{1-v^2}{x}$

$a = \dfrac{1-(\dfrac{t}{x})^2}{x}$

$a= \dfrac{x^2-t^2}{x^3}$

$a =\dfrac{x^2-x^2-1}{x^3}$

$a = \dfrac{1}{x^3}$

Hence, The acceleration in term of x is $a = \dfrac{1}{x^3}$.