A particle covers 10m in first 5s and 10m in next 3s. Assuming constant acceleration. Find initial speed, acceleration and distance covered in next 2s.

Let u be the initial velocity

Distance travelled in 5 s
$$\begin{gathered} 10=\mathrm{u}\left(5\right)+\frac{1}{2}\mathrm{a}{\left(5\right)}^2 \\ 10=5\mathrm{u}+\frac{1}{2}\mathrm{a}\times 25 \\ 20=10 \mathrm{u}+25 \mathrm{a} \\ 4=2\mathrm{u}+5\mathrm{a}\to \left(1\right) \end{gathered}$$
Distance travelled in 8 s
$$\begin{gathered} 20=8\mathrm{u}+\frac{1}{2}\mathrm{a} 64 \\ 20=8\mathrm{u}+32 \mathrm{a}\to \left(2\right) \\ \left(1\right)\times 4 8u+20a=16 \left(2\right)\times 1 8u+32a=20 - - - -12a=4 \\ \mathrm{a}=\frac{1}{3} {\mathrm{ms}}^{-2} \\ \mathrm{Substituting} \mathrm{a} \mathrm{value} \mathrm{in} \mathrm{eq} \left(1\right) \\ 4=2\mathrm{u}+\frac{5}{3} \\ 2\mathrm{u}=4-\frac{5}{3} \\ 2\mathrm{u}=\frac{12-5}{3} \\ 2\mathrm{u}=\frac{7}{3} \\ \mathrm{u}=\frac{7}{6} {\mathrm{ms}}^{-1} \\ \mathrm{Distance} \mathrm{travelled} \mathrm{in} 10 \mathrm{s} \\ \mathrm{s}\left(10\right)=10\left(\frac{7}{6}\right)+\frac{1}{2}\frac{1}{3}\times 100 \\ \mathrm{s}\left(10\right)=\frac{70+100}{6} \\ =\frac{170}{6} \\ \mathrm{Distance} \mathrm{travelled} \mathrm{in} \mathrm{next} 2 \mathrm{s} \mathrm{is} \\ =\frac{170}{6}-20 \\ =\frac{170-120}{6} \\ =\frac{50}{6} \\ =8.33 \mathrm{m} \end{gathered}$$