An object moves at a constant speed along a circular path in a horizontal plane with centre at origi

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3-2.Motion in Plane

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An object moves at a constant speed along a circular path in a horizontal plane with centre at the origin. When the object is at $x =+2\,m$, its velocity is $-4 \hat{ j }\, m / s$. The object's velocity $(v)$ and acceleration $(a)$ at $x =-2\,m$ will be

A$v =4 \hat{ i }\,m / s , a =8 \hat{ j }\,m / s ^2$

B$v =4 \hat{ j }\,m / s , a =8 \hat{ i }\,m / s ^2$

C$v =-4 \hat{ j }\,m / s , a =8 \hat{ i }\,m / s ^2$

D$v =-4 \hat{ i }\,m / s , a =-8 \hat{ j }\,m / s ^2$

(JEE MAIN-2023)

Solution

$a _{ c }=\frac{ V ^2}{ r }=\frac{4^2}{2}=\frac{16}{2}=8\,m / s ^2$
$\overrightarrow{ V }=4 \hat{ j }$
$\rightarrow a _{ c }=8 \hat{ i }$

Standard 11

Physics

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