A particle is projected with velocity _0 along x - axis. The dece… - Vidyadip

MCQ

A particle is projected with velocity ${\upsilon _0}$ along $x - axis$. The deceleration on the particle is proportional to the square of the distance from the origin i.e., $a = \alpha {x^2}.$ The distance at which the particle stops is

A
$\sqrt {\frac{{3{\upsilon _0}}}{{2\alpha }}} $
B
${\left( {\frac{{3{v_o}}}{{2\alpha }}} \right)^{\frac{1}{3}}}$
C
$\sqrt {\frac{{3\upsilon {}_0^2}}{{2\alpha }}} $
✓
${\left( {\frac{{3\upsilon {}_0^2}}{{2\alpha }}} \right)^{\frac{1}{3}}}$

✓

Answer

Correct option: D.

${\left( {\frac{{3\upsilon {}_0^2}}{{2\alpha }}} \right)^{\frac{1}{3}}}$

d
(d) $a = \frac{{dv}}{{dt}} = \frac{{dv}}{{dx}}\frac{{dx}}{{dt}}$ $ = v\frac{{dv}}{{dx}} = - \alpha {x^2}$ (given)

==> $\int\limits_{{v_0}}^0 {vdv = - \alpha \int\limits_0^S {{x^2}dx} } $ ==> $\left[ {\frac{{{v^2}}}{2}} \right]_{{v_0}}^0 = - \alpha \left[ {\frac{{{x^3}}}{3}} \right]_0^S$

==> $\frac{{v_0^2}}{2} = \frac{{\alpha \,{S^3}}}{3}$==> $S = {\left( {\frac{{3v_0^2}}{{2\alpha }}} \right)^{\frac{1}{3}}}$

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