I was trying yo solve it by:

$*$ Calculating the "angular acceleration" knowing that the acceleration that the truck applies in the reel at point $($let$\text{'}$s call it A$)$ is $2$ ft$/$s$^2.$

$*$ Then, by knowing that the linear acceleration is$...$ a$=$v$($dv$/$dx$),$ I integrated that expression $($again$,$ at point A$)$ to know the velocity at that point. I believe this was correct bc the mention of the displacement of $6$ ft$.$

$*$ Knowing the velocity y use the equation... v$=$rw$,$ and then solve for angular velocity $($w$).$

$*$ with the acquired values I solve for each acceleration $($tangential and normal$)$ at point P$.$

The issue is that my answer is different from the one mentioned in the textbook. Please tell me what I did wrong.

The answer is a$\_$p $=11\mathrm{.}14$ ft$/$s$^2$