A slider moving outward on a rod OR with a velocity of 3 m/s and acceleration of 9 $$m/s^2$$. The rod has an angular velocity of 7 rad/s counter clockwise about O and an angular acceleration of 12 rad/$$s^2$$ clockwise. The magnitude of absolute acceleration of the slider at location A in $$m/s^2$$ is 

Correct option is B

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$$\text{For radius } r, \text{ radial speed } \dot{r} \text{ and radial acceleration } \ddot{r}, \text{ angular speed } \omega \text{ and angular acceleration } \alpha,\\[6pt]\vec{a} = (\ddot{r} - r\omega^2)\, \hat{e}_r \; + \; (r\alpha + 2\dot{r}\omega)\, \hat{e}_\theta.\\[8pt]\text{Given: } r = 1~\text{m}, \; \dot{r} = +3~\text{m/s (outward)}, \; \ddot{r} = +9~\text{m/s}^2, \; \omega = +7~\text{rad/s (CCW)}, \; \alpha = -12~\text{rad/s}^2 \\ \text{(clockwise → negative).}\\[8pt]\text{Compute components:}\\[4pt]a_r = \ddot{r} - r\omega^2 = 9 - 1(7^2) = 9 - 49 = -40~\text{m/s}^2\\[6pt]a_\theta = r\alpha + 2\dot{r}\omega = 1(-12) + 2(3)(7) = -12 + 42 = +30~\text{m/s}^2\\[8pt]\text{Magnitude:}\\[4pt]|\vec{a}| = \sqrt{a_r^2 + a_\theta^2} = \sqrt{(-40)^2 + (30)^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50~\text{m/s}^2$$​​