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Q.

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 $x = + 2 m$ , its velocity is $- 4 \hat{j} m / s$ . The object’s velocity (v) and acceleration (a) at $x = - 2 m$ will be

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a

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

b

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

c

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

d

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

answer is A.

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Detailed Solution

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at x=+2m, the velocity is tangential and directed along +y-axis direction.

$a_{r} = \frac{v^{2}}{R} = \frac{4^{2}}{2} = 8$ , directed towards the centre(+x-direction).

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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 x=+2m, its velocity is −4j^ m/s. The object’s velocity (v) and acceleration (a) at x=−2m will be