A particle moves in $xy-$plane according to the equation $x=\alpha t$ and $y=\alpha t(1-\beta t)$ where $\alpha$ and $\beta$ are positive constants and $t$ is time.

Choose the correct option in regard to the speed of the particle.

The correct option is A The speed of the particle at time $t=\frac{1}{4\beta }$ is $\frac{\sqrt{5}}{2}\alpha$.
Given:

$x=\alpha t\Rightarrow \frac{dx}{dt}=\alpha =v_x...\left(i\right)$, and,

$y=\alpha t(1-\beta t)\Rightarrow \frac{dy}{dt}=\alpha -2\alpha \beta t=v_y...\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$,

The velocity of the particle is,
$\overrightarrow{v}=\alpha \hat{i}+(\alpha -2\alpha \beta t)\hat{j}$

So, at $t=\frac{1}{4\beta }$,

$\overrightarrow{v}=\alpha \hat{i}+(\alpha -2\alpha \beta \times \frac{1}{4\beta })\hat{j}=\alpha \hat{i}+\frac{\alpha }{2}\hat{j}\ne 0$

Further, its magnitude,

$v=\sqrt{{\alpha }^2+(\alpha /2{)}^2}=\frac{\sqrt{5}}{2}\alpha$

Also, at $t=\frac{1}{2\beta }$,

$\overrightarrow{v}=\alpha \hat{i}+(\alpha -2\alpha \beta \times \frac{1}{2\beta })\hat{j}=\alpha \hat{i}\ne 0$

Hence, option $\left(A\right)$ is correct.