Question

Relation 2t equal to root x + 5 describe the displacement of a particle in one direction where x is in metres and t seconds the displacement when velocity is zero is

Answer 1

Answer:

solved

Explanation:

2t equal to root x + 5

2t = √ x + 5

x+5 = 4t²

x = 4t² - 5

v = 8t = 0 , t = 0

x = - 5 m.

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

Given relation:-

$\large{\bold{ \tt 2t = \sqrt{x + 5}}}$

$\huge{\bold{\underline{Explanation:-}}}$

$\large{ \tt 2t = \sqrt{x + 5}}$

Squaring on both sides,

$\large{(2t)^2 = x + 5}$

$\large{4t^2 = x + 5}$

$\large{x = 4t^2 - 5}$

$\large{x = 4t^2 - 5 \: ----(1)}$

Differentiating, it w.r.t time to get velocity,

$\large{ \dfrac{dx}{dt} = \dfrac{4t^2}{dt} - \dfrac{5}{dt}}$

Now,

$\large{ v = 8t - 0 }$

As differentiating a constant gives a value of zero.

$\large{\boxed{ \tt v = 8t}}$

Now, the velocity is Zero.

Substituting,

$\large{0 = 8t}$

$\large{t = \dfrac{0}{8}}$

$\large{\boxed{ \tt t = 0 \: seconds}}$

Substituting in the given equation (1).

$\large{x = 4t^2 - 5}$

$\large{x = 4 \times (0)^2 - 5}$

$\large{x = 4 \times 0 - 5}$

$\large{x = 0 - 5}$

$\huge{\boxed{\boxed{ \tt x = - 5 \: meters}}}$

So, the displacement covered by the body when velocity is zero is - 5 meters.