Question

The acceleration vector along x-axis of a particle having initial velocity $v_0$ changes with distance x as $a=\sqrt{x}$.The distance covered by the particles, When its speed becomes twice that of initial speed is:-

• A. ${\left(\frac{9}{4}{v}_0\right)}^{4/3}$
• B. ${\left(\frac{3}{2}{v}_0\right)}^{4/3}$
• C. ${\left(\frac{2}{3}{v}_0\right)}^{4/3}$
• D. $2v_0$

Answer 1

The correct option is C ${\left(\frac{2}{3}{v}_0\right)}^{4/3}$

$a=\sqrt{x}$

$\frac{vdv}{dx}=\sqrt{x}$

$vdv=\sqrt{x}dx$

${\int }_{v_0}^{2v_0}vdv={\int }_0^x\sqrt{x}dx$

$\frac{{4v_0}^2-v_o^2}{2}=\frac{2}{3}{x}^{3/2}$

$\frac{3}{2}{v}_0^2=\frac{2}{3}{x}^{3/2}$

$x^{3/2}=(\frac{2}{3}{v}_0{)}^2$

$x=(\frac{2}{3}{v}_0{)}^{4/3}$