May 10, 2018

See below:

#### Explanation:

when $t = 40$, $x \approx 3 c m$

Each square in the vertical direction represents 0.4 cm (from my point of view), so the bar stretches approximately across 0.5 cm.

So the percentage uncertainty is, therefore:

0.5/3.0 approx 0.167 times 100=17% (two sig.figs)