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JEE Mains · Physics · STD 11 - 3.2 motion in plane
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 \(v =-4 \hat{ i }\,m / s , a =-8 \hat{ j }\,m / s ^2\)
Answer & Solution
Correct Answer
(B) \(v =4 \hat{ j }\,m / s , a =8 \hat{ i }\,m / s ^2\)
Step-by-step Solution
Detailed explanation
\(a _{ c }=\frac{ V ^2}{ r }=\frac{4^2}{2}=\frac{16}{2}=8\,m / s ^2\) \(\overrightarrow{ V }=4 \hat{ j }\) \(\rightarrow a _{ c }=8 \hat{ i }\)
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