Question

The acceleration of a particle is given by a=x where x is a constant. If the particle start at origin from rest,its distance from origin after time t is given by

Answer 1

S = ut + 1/2 at^2
but ut-> 0 as initial velocity(u) is 0.
Therefore S(distance)= 1/2at^2.

Answer 2

Answer:

After time t, the distance from origin $s =\dfrac{xt^2}{2}$

Explanation:

Given that,

Acceleration a = x

The acceleration is the first derivative of the velocity of the particle.

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

The velocity is

$\int_{0}^{v}{dv}=\int_{0}^{x}({x})dt$

On integrating w.r.to t

$v = xt+C$....(I)

When, t=0,v=0

So, C =0

Therefore,

$v = xt$....(II)

The velocity is the first derivative of the position of the particle.

$v=\dfrac{ds}{dt}$

Now, $\dfrac{ds}{dt} = xt$

$ds=xtdt$

$\int_{0}^{s}ds=\int_{0}^{x}{xt}dt$

On integrating w.r.to t

$s=\dfrac{xt^2}{2}+C$

When t = 0, s= 0

Then, C = 0

$s =\dfrac{xt^2}{2}$

Hence, After time t, the distance from origin $s =\dfrac{xt^2}{2}$