Question

If the displacement of an object is proportional to square of time then object moves with.? ​

Answer 1

The object moves with uniform or constant acceleration.

Given,

$\longrightarrow\sf{s\propto t^2}$

Let,

$\longrightarrow\sf{s=kt^2\quad[k=a\ constant]}$

Differentiating with respect to time,

$\longrightarrow\sf{\dfrac{ds}{dt}=\dfrac{d}{dt}[kt^2]}$

We know first derivative of displacement is velocity.

$\longrightarrow\sf{v=k\cdot\dfrac{d}{dt}[t^2]}$

$\longrightarrow\sf{v=2kt}$

Again differentiating with respect to time,

$\longrightarrow\sf{\dfrac{dv}{dt}=\dfrac{d}{dt}[2kt]}$

We know first derivative of velocity is acceleration.

$\longrightarrow\sf{a=2k\cdot\dfrac{dt}{dt}}$

$\longrightarrow\sf{a=2k}$

Since k is a constant,

$\Longrightarrow\sf{\underline{\underline{a=constant}}}$

Answer 2

$\large \green{ \fcolorbox{gray}{black}{ ☑ \: \textbf{Verified \: answer}}}$

Object moves with Uniform Acceleration.

EXPLANATION

Displacement is proportional to square of time

⠀⠀⠀⠀⠀⠀⠀⠀⠀s ∝ t²

⠀⠀⠀⠀⠀⠀⠀⠀⠀s = kt²

⠀⠀⠀ , where k is constant of proportionality.

Differentiating Displacement , we get Velocity (v)

⠀⠀⠀⠀⠀⠀⠀ ds/dt = 2kt

Differentiating Velocity , we get Acceleration (a)

⠀⠀⠀⠀⠀⠀ dv/dt = 2k = constant.

ᗇ So, object moves with Constant Acceleration.