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(B) Given: Initial velocity $u = 0$,tangential acceleration $a_t = 12 \ m/s^2$,and radius $r = 3 \ m$.The total acceleration vector $\vec{a}$ is the v

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$(1/2) \ s$

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(B) Given: Initial velocity $u = 0$,tangential acceleration $a_t = 12 \ m/s^2$,and radius $r = 3 \ m$.The total acceleration vector $\vec{a}$ is the vector sum of radial acceleration $\vec{a}_c$ and tangential acceleration $\vec{a}_t$.The angle $	heta$ between the total acceleration and the radial acceleration is given by $	an 	heta = \frac{a_t}{a_c}$.Given $	heta = 45^{\circ}$,so $	an 45^{\circ} = 1$,which implies $a_t = a_c$.We know $a_c = \frac{v^2}{r}$,where $v$ is the instantaneous velocity.Since $a_t = 12 \ m/s^2$ and $a_c = \frac{v^2}{3}$,we have $12 = \frac{v^2}{3} \Rightarrow v^2 = 36 \Rightarrow v = 6 \ m/s$.Using the equation of motion $v = u + a_t t$,we get $6 = 0 + 12 	imes t$.Therefore,$t = \frac{6}{12} = 0.5 \ s$ or $(1/2) \ s$.

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