The electrons that produce the picture in a TV set are accelerated by a very large electric force as they pass through a small region in the neck of the picture tube. This region is 1.9 cm in length, and the electrons enter with a speed of 1 × 105 m/s and leave with a speed of 2.5 × 106 m/s.
Is it the average acceleration that you are after?
If so, use v2 = u2 + 2as where v = final velocity, u = initial velocity a = acceleration and s = distance
Rearrange as: a = (v2 - u2)/(2s)
$${\mathtt{a}} = {\frac{\left({\left({\mathtt{2.5}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{6}}}\right)}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\left({\mathtt{1}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{5}}}\right)}^{{\mathtt{2}}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{0.019}}\right)}} \Rightarrow {\mathtt{a}} = {\mathtt{164\,210\,526\,315\,789.473\: \!684\: \!210\: \!526\: \!315\: \!8}}$$
This is in m/sec2
.
Is it the average acceleration that you are after?
If so, use v2 = u2 + 2as where v = final velocity, u = initial velocity a = acceleration and s = distance
Rearrange as: a = (v2 - u2)/(2s)
$${\mathtt{a}} = {\frac{\left({\left({\mathtt{2.5}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{6}}}\right)}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\left({\mathtt{1}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{5}}}\right)}^{{\mathtt{2}}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{0.019}}\right)}} \Rightarrow {\mathtt{a}} = {\mathtt{164\,210\,526\,315\,789.473\: \!684\: \!210\: \!526\: \!315\: \!8}}$$
This is in m/sec2
.