Question

If the distance covered by a particle is piven
by relation x= at^2 The particle is moving with
(where a is constant) :
(1) constant acceleration
2) zero acceleration
(3) variable acceleration
(4) none of these
please answer reliable with reason......​

Answer 1

Answer:

• The particle is moving with Constant Acceleration.

Given:

1. Given relation is x = at².

Explanation:

$\rule{300}{1.5}$

From The Question We Know,

$\large\bigstar \; {\boxed{\tt x = at^2}}$

$\bold{Here}\begin{cases}\text{x Denotes Displacement} \\ \text{t Denotes Time taken} \\ \text{a Denotes Constant}\end{cases}$

Now,

$\longmapsto{\large \tt x = at^2}$

Differentiating the Equation to get Velocity.

$\longmapsto \large\boxed{\tt v = \dfrac{dx}{dt}}$

$\longmapsto \large{\tt v = \dfrac{d(a \; t^2)}{dt}}$

As " a " is a constant and cannot be differentiated

$\longmapsto \large{\tt v = a \; \dfrac{d(t^2)}{dt}}$

$\longmapsto \large{\tt v = a \times 2 \; t}$

$\longmapsto \large{\underline{\boxed{\tt v = 2 \; at \; m/s}}}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

$\longmapsto \large{\tt v = 2 \; at}$

Again, Differentiating the Equation to get Acceleration.

$\longmapsto \large\boxed{\tt a = \dfrac{dv}{dt}}$

$\longmapsto \large{\tt a = \dfrac{d(2 \; a \; t)}{dt}}$

As a & 2 are constant and cannot be differentiated,

$\longmapsto \large{\tt a = 2 \; a \; \dfrac{d(t)}{dt}}$

$\longmapsto \large{\tt a = 2 \times a \times 1}$

$\longmapsto \large{\underline{\boxed{\tt a = 2 \; a \; m/s^2}}}$

From this we can Conclude that,

$\longmapsto \large{\underline{\boxed{\red{\tt a \propto t^0}}}}$ ∵ [t⁰ = 1]

∵ Acceleration is Directly Proportional to the zeroth power of time period which is 1.

Hence, it doesn't changes with time.

∴ The magnitude of acceleration of the body is constant with time (Option - 1).

$\rule{300}{1.5}$

Answer 2

Answer

Acceleration is constant

Given

Position of the particle is given by :

$\tt \: x = a {t}^{2}$

To finD

Type of acceleration the particle is executing

Differentiating x w.r.t to t,we get velocity of the particle :

$\sf \: v = \dfrac{dx}{dt} \\ \\ \leadsto \: \sf \: v = \dfrac{d(a {t}^{2} )}{dt} \\ \\ \leadsto \: \sf \: v = a \times \dfrac{d( {t}^{2} )}{dt} \\ \\ \leadsto \: \boxed{ \boxed{ \sf \: v = 2at \: {ms}^{ - 1}}}$

Differentiating v w.r.t to t,we get acceleration of the particle :

$\sf \: a = \dfrac{dv}{dt} \\ \\ \leadsto \: \sf \: a = \dfrac{d(2at)}{dt} \\ \\ \leadsto \: \: \boxed{ \boxed{ \sf \: a \: = 2a \: {ms}^{ - 2} }}$

Since,

$\huge{ \boxed{ \boxed{ \sf \: a \: \propto \: {t}^{0} }}}$

Acceleration is independent of time here and shows no remarkable change

Option (A) is correct