NEET UG PHYSICS MCQ: MOTION IN A STRAIGHT LINE

1 Hard

1 of 1

Application +4 | -1

A particle is projected with velocity

{v_0}

along x-axis.The deceleration on the particle is proportional to the square of the distance from the origin i.e.,

a = - a{x^2}.

The distance at which the particle stops is

View answer and explanation

Correct option: D

{\left( {\frac{{3v_0^2}}{{2\alpha }}} \right)^{\frac{1}{3}}}

Option D (

{\left( {\frac{{3v_0^2}}{{2\alpha }}} \right)^{\frac{1}{3}}}

) is correct because it is the distance at which the particle stops, derived by integrating the given deceleration function.

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8 \mathrm{~m} / \mathrm{s}

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150 \mathrm{~m}

2 \mathrm{~ms}^{-2}

{V_1} = 10\,kmph

{V_2} = 20\,kmph

{V_3} = 60\,kmph