NEET UG PHYSICS MCQ: MOTION IN PLANE

1 Hard

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Application +4 | -1

A particle moves in the x-y plane with velocity

{v_x} = 8t - 2

and

{v_y} = 2

. If it passes through the point  x=14 and y=4 at t=2s, find the equation (x-y relation) of the path

View answer and explanation

Correct option: A

x = {y^2} - y + 2

Option A (

x = {y^2} - y + 2

) is correct because it represents the path of the particle derived by integrating the velocity components and using the given initial conditions.

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\mathrm{\vec a}\,and\,\mathrm{\vec b}

\mathrm{\vec a}\,and\,\mathrm{\vec b}

\overrightarrow P \,\,.\,\,\overrightarrow Q = 0\,and\,\overrightarrow P \,\, \times \,\,\overrightarrow Q = 0

\overrightarrow Q \,and\,\overrightarrow R

\vec{A}=3 \hat{i}+4 \hat{j}

\vec{B}=6 \hat{i}+8 \hat{j}

A

B

\overrightarrow{\mathrm{A}}

\overrightarrow{\mathrm{B}}

{\cot ^{ - 1}}\left( 2 \right)