Answer
The speed $v$ and the tangential acceleration ${a_{\bf{T}}}$ at $P$ :
$v \simeq 32.57$ m/min
and
${a_{\bf{T}}} = 25\sqrt 2 $ $m/{\min ^2}$
Work Step by Step
We have at the point $P$ the acceleration vector ${\bf{a}} = \left( {0, - 50} \right)$ $m/{\min ^2}$ pointing down. So, $a = ||{\bf{a}}|| = 50$.
From Figure 12, we get the tangential and normal accelerations:
${a_{\bf{T}}} = a\cos 45^\circ = 25\sqrt 2 $, ${\ \ }$ ${a_{\bf{N}}} = a\sin 45^\circ = 25\sqrt 2 $
Since ${a_{\bf{N}}} = \kappa {v^2} = \frac{{{v^2}}}{R}$, so the speed $v$ at $P$ is
$v = \sqrt {{a_{\bf{N}}}R} = \sqrt {25\sqrt 2 \cdot30} \simeq 32.57$ m/min.
