Question

a particle moves in a straight line and its position x at time t is given by x^2=2+t. its acceleration is given by (1)2/x^3 (2)-1/4x^3 (3)-1/4x^2 (4)1/x^2

Answer 1

Its yoir ans
Can you understand
Ans=-1/4x^3

Answer 2

Answer:

The acceleration is $-\dfrac{1}{4x^3}$

(2) is correct .

Explanation:

Given that,

A particle moves in a straight line and its position x at time t.

The position is

$x^2 =2+t$

Differentiating with respect to time

$2x\dfrac{dx}{dt}=0+1$

$\dfrac{dx}{dt}=\dfrac{1}{2x}$

The velocity is

$v=\dfrac{dx}{dt}=\dfrac{1}{2x}$

Now, The acceleration is

The acceleration is the rate of change of velocity

$a = \dfrac{dv}{dt}=\dfrac{d}{dt}(\dfrac{1}{2x})$

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

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

Put the value of v

$a = -\dfrac{1}{4x^3}$

Hence, The acceleration is $-\dfrac{1}{4x^3}$