MCQ
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 $y = 2\,m$ ?
  • $-(8\,m/ s^2)\hat j$
  • B
    $-(8\,m/ s^2)\hat i$
  • C
    $-(4\,m/ s^2)\hat j$
  • D
    $(4\,m/ s^2)\hat i$

Answer

Correct option: A.
$-(8\,m/ s^2)\hat j$
a
at $B$

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

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

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