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

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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 horizontal $XY$ plane with centre at origin. When the object is at $x = -2\,m$ , its velocity is $-(4\,m/ s)\hat j$ . What is object's acceleration when it is at $y = 2\,m$ ?

A

$-(8\,m/ s^2)\hat j$

B

$-(8\,m/ s^2)\hat i$

C

$-(4\,m/ s^2)\hat j$

D

$(4\,m/ s^2)\hat i$

Solution

at $B$

$a_{B}=\frac{V^{2}}{R}=\frac{(4)^{2}}{2}=\frac{16}{2}=8 \mathrm{m} / \mathrm{s}^{2}$

$\overrightarrow{\mathrm{a}}_{\mathrm{B}}=-\left(8 \mathrm{m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}$

Standard 11

Physics

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