A point P moves in counter-clockwise direction on a circular path as shown in the figure. The movement of 'P' is such that it sweeps out a length $s = t ^ { 3 } + 5$, where $s$ is in metres and $t$ is in seconds. The radius of the path is 20 m. The acceleration of 'P' when $t = 2 \mathrm {~s}$ is nearly
(1) $13 \mathrm {~m} / \mathrm { s } ^ { 2 }$
(2) $12 \mathrm {~m} / \mathrm { s } ^ { 2 }$
(3) $7.2 \mathrm {~m} / \mathrm { s } ^ { 2 }$
(4) $14 \mathrm {~m} / \mathrm { s } ^ { 2 }$

A particle is moving in a circle of radius 50 cm in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at $\mathrm { t } = 0$ is $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the time taken to complete the first revolution will be $\frac { 1 } { \alpha } \left[ 1 - \mathrm { e } ^ { - 2 \pi } \right] \mathrm { s }$, where $\alpha =$ $\_\_\_\_$ .