Answer
a) $v_x =\dfrac{ TF_0}{m} \bigg[1-e^{-t/T} \bigg]$
b) $ v_x =\dfrac{ TF_0}{m} $
Work Step by Step
a)
According to Newton's second law, acceleration is given by
$$a_x=\dfrac{\sum F_x}{m}$$
And in this case,
$$a_x=\dfrac{ F_x}{m}=\dfrac{F_0e^{-t/T}}{m}\tag 1$$
We can see at $t=0$, the initial force is $F_0$ which is constant and hence the initial acceleration is $F_0/m$.
To find the expression of velocity, we need to integrate the acceleration expression relative to $t$;
$$v_x=\int_0^t a_xdt=\dfrac{F_0}{m}\int_0^t e^{-t/T}$$
$$v_x =\dfrac{-TF_0}{m} \bigg[e^{-t/T} \bigg]_0^t $$
$$v_x =\dfrac{-TF_0}{m} \bigg[e^{-t/T}- e^{-0/T}\bigg] $$
$$v_x =\dfrac{-TF_0}{m} \bigg[e^{-t/T}- 1\bigg] $$
$$\boxed{ v_x =\dfrac{ TF_0}{m} \bigg[1-e^{-t/T} \bigg] } $$
b)
Thus, at $t=0$,
$$v_x =\dfrac{ TF_0}{m} \bigg[1-1 \bigg]=0$$
And after a long time $t=\infty$, $e^{-\infty}=0$, and hence,
$$v_x =\dfrac{ TF_0}{m} \bigg[1-0 \bigg]$$
$$\boxed{v_x =\dfrac{ TF_0}{m} }$$
