Answer
The linear speed as it reaches the end of the string is $~~1.4~m/s$
Work Step by Step
In part (a), we found that the magnitude of the linear acceleration is $~~0.13~m/s^2$
In part (a), we found that it takes $~~0.88~s~~$ to reach the end of the string.
We can find the linear speed as it reaches the end of the string:
$v = v_0+at$
$v = 1.3~m/s+(0.13~m/s^2)(0.88~s)$
$v = 1.4~m/s$
The linear speed as it reaches the end of the string is $~~1.4~m/s$.
