A particle travels 10m in first 5 sec and 10m in next 3 sec. Assuming constant acceleration what is the distance travelled in next 2 sec:

Let initial (t = 0) velocity of particle = u For, first 5 sec of motion S₅ = 10 metre,

So, by using $s=ut+\frac{1}{2}a{t}^2$

$\begin{array}{l}10=5\mathrm{u}+\frac{1}{2}\mathrm{a}(5{)}^2\\ \Rightarrow 2\mathrm{u}+5\mathrm{a}=4 . . . \left(\mathrm{i}\right)\end{array}$

For, first 8 sec of motion S₈ = 20 metre

$\begin{array}{l}20=8\mathrm{u}+\frac{1}{2}\mathrm{a}(8{)}^2\\ \Rightarrow 2\mathrm{u}+8\mathrm{a}=5 . . . \left(\mathrm{ii}\right)\end{array}$

By solving (i) and (ii)

$u=\frac{7}{6}m/sa=\frac{1}{3}m/s^2$

Now distance travelled by particle in total 10 sec.

$s_{10}=u\times 10+\frac{1}{2}a(10{)}^2$

by substituting the value of u and a, we will get

$s_{10}=28.3\mathrm{m}$

So, the distance in last $2\sec =s_{10}-s_8=28.3-20=8.3\mathrm{m}$