A body is moving on a circle of radius 80 , m | vedclass

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(C) The centripetal acceleration $a_c$ is given by $a_c = \frac{v^2}{r} = \frac{20^2}{80} = \frac{400}{80} = 5 \, m/s^2$. Since the speed is decreasin

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$135$

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(C) The centripetal acceleration $a_c$ is given by $a_c = \frac{v^2}{r} = \frac{20^2}{80} = \frac{400}{80} = 5 \, m/s^2$. Since the speed is decreasing,the tangential acceleration $a_t$ is directed opposite to the velocity vector,so $a_t = -5 \, m/s^2$. The total acceleration vector $\vec{a}$ is the vector sum of centripetal acceleration $\vec{a}_c$ (directed towards the center) and tangential acceleration $\vec{a}_t$ (directed opposite to velocity). The angle $\phi$ between the total acceleration $\vec{a}$ and the tangential acceleration $\vec{a}_t$ is given by $	an \phi = \frac{a_c}{|a_t|} = \frac{5}{5} = 1$,which means $\phi = 45^{\circ}$. Since the tangential acceleration $\vec{a}_t$ is directed opposite to the velocity $\vec{v}$,the angle between $\vec{a}_t$ and $\vec{v}$ is $180^{\circ}$. The angle $	heta$ between the total acceleration $\vec{a}$ and the velocity $\vec{v}$ is $	heta = 180^{\circ} - 45^{\circ} = 135^{\circ}$.

$A$ body is moving on a circle of radius $80 \, m$ with a speed $20 \, m/s$ which is decreasing at the rate of $5 \, m/s^2$ at an instant. The angle made by its acceleration with its velocity is .......... (in $^{\circ}$)

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