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

let u and v be the first and final velocities of particle and a and s be the constant acceleration and distance covered by it.

Using ${\mathrm{v}}^2 = {\mathrm{u}}^2+2\mathrm{as}$

$\Rightarrow {\left(20\right)}^2 = {\left(10\right)}^2+2\mathrm{a}\times 135$

or a = $\frac{300}{2\times 135} = \frac{10}{9} {\mathrm{ms}}^{-2}$

Now using, v = u+at

t = $\frac{\mathrm{v}-\mathrm{u}}{\mathrm{a}} = \frac{20-10}{\left({\displaystyle \raisebox{1ex}{$10$}\!\left/ \!\raisebox{-1ex}{$9$}\right.}\right)} = \frac{10\times 9}{10} = 9\mathrm{s}$