Answer
The linear speed as it reaches the end of the string is $~~0.56~m/s$
Work Step by Step
In part (a), we found that the magnitude of the linear acceleration is $~~0.13~m/s^2$
In part (b), we found that it takes $~~4.3~s~~$ to reach the end of the string.
We can find the linear speed as it reaches the end of the string:
$v = v_0+at$
$v = 0+(0.13~m/s^2)(4.3~s)$
$v = 0.56~m/s$
The linear speed as it reaches the end of the string is $~~0.56~m/s$.
