A particle moves in a straight line with a constant acceleration. It changes its velocity from $10{\text{ms}}^{-1}$ to $20{\text{ms}}^{-1}$ while passing through a distance of $135\text{m}$ in $t$ seconds. What is the value of $t$?

Given:Initial velocity, $u=10{\text{ms}}^{-1}$ Final velocity, $v=20{\text{ms}}^{-1}$ Distance, $s=135\text{m}$ Using the third equation of motion to find acceleration ($a$):
$$v^2-u^2=2as$$
$$(20{)}^2-(10{)}^2=2\cdot a\cdot 135$$
$$400-100=270a$$
$$300=270a⟹a=\frac{300}{270}=\frac{10}{9}{\text{ms}}^{-2}$$
Now, using the first equation of motion to find time ($t$):
$$v=u+at$$
$$20=10+\left(\frac{10}{9}\right)t$$
$$10=\left(\frac{10}{9}\right)t$$
$$t=\frac{10\times 9}{10}=9\text{s}$$
Alternatively, using the average velocity formula:
$$s=\left(\frac{u+v}{2}\right)t$$
$$135=\left(\frac{10+20}{2}\right)t$$
$$135=15t⟹t=\frac{135}{15}=9\text{s}$$