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  • An object moves at a constant speed along a circular path in a horizontal plane with center at the origin. When the object is at <img alt="x=+2 \mathrm{~m}" src="https://entrancecorner.oncodecogs.c

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

Option: 1

v=-4 \hat{\imath} \frac{\mathrm{m}}{\mathrm{s}}, a=-8 \hat{\jmath} \mathrm{m} / \mathrm{s}^2


Option: 2

v=4 \hat{\imath} \frac{\mathrm{m}}{\mathrm{s}}, a=8 \hat{\jmath} \mathrm{m} / \mathrm{s}^2


Option: 3

v=4 \hat{\jmath} \frac{\mathrm{m}}{\mathrm{s}}, a=8 \hat{i} \mathrm{m} / \mathrm{s}^2


Option: 4

v=-4 \hat{\jmath} \frac{\mathrm{m}}{\mathrm{s}}, a=8 \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}^2


Answers (1)

best_answer

\begin{aligned} & a_c=\frac{v^2}{r}=\frac{4^2}{2}=8 \mathrm{~ms}^{-2} \\ & \vec{v}=4 \hat{j} \\ & \overrightarrow{a_c}=8 i \end{aligned}

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