Correct Option is (A)

Given $\frac{dv}{dt}=-2.5\sqrt{v}$

$\Rightarrow \frac{dv}{\sqrt{v}}=-2.5dt$

On integrating, ${\int }_{6.25}^0{v}^{-1/2}dv=-2.5{\int }_0^{t}dt$

$\Rightarrow {\left[\frac{v^{+1/2}}{\left(1/2\right)}\right]}_{6.25}^0=-2.5{\left[t\right]}_0^t$

$\Rightarrow -2{\left(6.25\right)}^{1/2}=-2.5t$

$\Rightarrow t=2sec$

Correct Option is (C)

Given that, v = v₀ + gt + ft²

We know that, $v=\frac{dx}{dt}$

$\Rightarrow dx=vdt$

Integrating, $\underset{0}{\overset{x}{\int }}dx=\underset{0}{\overset{t}{\int }}vdt$

or $x=\underset{0}{\overset{t}{\int }}\left(v_0+gt+f{t}^2\right)dt$

$={\left[v_{0}t+\frac{g{t}^2}{2}+\frac{f{t}^3}{3}\right]}_0^t$

or $x=v_{0}t+\frac{g{t}^2}{2}+\frac{f{t}^3}{3}$

At $t=1,x=v_0+\frac{g}{2}+\frac{f}{3}.$

Correct Option is (A)

Given $v=\alpha \sqrt{x}$

$\therefore$ $\frac{dx}{dt}=\alpha \sqrt{x}$

$\Rightarrow \frac{dx}{\sqrt{x}}=\alpha dt$

On integrating we get,

$\underset{0}{\overset{x}{\int }}\frac{dx}{\sqrt{x}}=\alpha \underset{0}{\overset{t}{\int }}dt$

$\Rightarrow {\left[\frac{2\sqrt{x}}{1}\right]}_0^x=\alpha {\left[t\right]}_0^t$

$\Rightarrow 2\sqrt{x}=\alpha t$

$\Rightarrow x=\frac{{\alpha }^2}{4}{t}^2$

So $x\propto t^2$

Correct Option is (D)

Given $t=a{x}^2+bx;$

Differentiate with respect to time $(t)$

$\frac{d}{dt}\left(t\right)=a\frac{d}{dt}\left(x^2\right)+b\frac{dx}{dt}$

$=a.2x\frac{dx}{dt}+b.\frac{dx}{dt}$

$1=2axv+bv=v\left(2ax+b\right)$

[as $\frac{dx}{dt}=v$]

$\Rightarrow 2ax+b=\frac{1}{v}.$

Again differentiating,

$2a\frac{dx}{dt}+0=-\frac{1}{v^2}\frac{dv}{dt}$

$\Rightarrow \frac{dv}{dt}=f=-2a{v}^3$

(as $\frac{dv}{dt}=f$ = Acceleration )