Find the angle between the point's velocity and acceleration vectors.

\frac{5 π}{2}

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π

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\frac{3 π}{2}

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\frac{π}{2}

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Solution

The correct option is D $\frac{π}{2}$
$x = a sin w t y = a (1 - cos w t)$
position vector $\to r = a sin w t^i + a (1 - cos w t)^j$
velocity $\to v = a w cos w t^i + a w sin w t^j$
acceleration $\to a = - a w^{2} sin w t^i + a w^{2} cos w t^j$
$\to a \cdot \to v = - a^{2} w^{3} sin w t cos w t + a^{2} w^{3} sin w t cos w t = 0$ , as the dot product is zero angle between them is $\frac{π}{2}$ .

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Find the angle between the point's velocity and acceleration vectors.

The correct option is D π2x=asinwty=a(1−coswt)position vector →r=asinwt^i+a(1−coswt)^jvelocity →v=awcoswt^i+awsinwt^jacceleration →a=−aw2sinwt^i+aw2coswt^j→a⋅→v=−a2w3sinwtcoswt+a2w3sinwtcoswt=0, as the dot product is zero angle between them is π2.