Question

A displacement x of a particle moving along x axis at time t is given by x^2 = 2t^2 + 6t find velocity at time t

Answer 1

We have,
1)
${x}^{2} = 2 {t}^{2} + 6t - - - (1) \\ = > x = \sqrt{2 {t}^{2} + 6t } \: or \: - \sqrt{2 {t}^{2} + 6t}$
Signs represent directions.
Let us assume particle is moving in + x -axis.
So,
$x = \sqrt{2 {t}^{2} + 6t } \:$

2)Implicilty Differentiating eq. (1),

$2x \frac{dx}{dt} = 4t + 6 \\ = > \frac{dx}{dt} = \frac{2t + 3}{x} \\ = > v = \frac{2t + 3}{ \sqrt{2 {t}^{2} + 6t} }$
Hence, Velocity as a function of time t, is given by
$\boxed {v = \frac{2t + 3}{ \sqrt{2 {t}^{2} + 6t } } }$

Answer 2

So,

The velocity and displacement relation is given by;

v= (dx)/dt

Now,

The relation between displacement and time is given in the question as,

x² = 2t² + 6t

So,

x= √(2t² + 6t )......................1)

Differentiating this equation w.r.t t ,

2x(dx / dt) = 4t + 6

dx / dt= (4t + 6) /2x

dx / dt= (2t + 3)/x

v= (2t + 3)/x

But from equation 1)

v=(2t + 3)/√(2t² + 6t )

So,

This is the required velocity time relation