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

The rectilinear motion of the body is a factor in the difficulty mentioned above. Therefore, we can readily solve this issue using the equations of motion. Rectilinear motion is the name given to a body's motion when it moves in a straight line. Translational motion is exemplified by this action.

If acceleration is a and beginning velocity is u, the distance in the first five seconds will be as follows:$\mathrm{S}=ut+{\mathrm{at}}^2/2$ i.e $10=5\mathrm{u}+25\mathrm{a}/2\dots \left(1\right)$
Moreover, the distance in 8 seconds will be$\mathrm{S}=8\mathrm{u}+64\mathrm{a}/2=8\mathrm{u}+32\mathrm{a}$
as a result, separation $t=5$ and $t=8$ will be $S-10=8 u+32 a-10$
It is stated to be $10 \mathrm{m}$ so $10=8\mathrm{u}+32\mathrm{a}-10$ or 20=8 $\mathrm{u}+32\mathrm{a}\dots .$. (2)
Equation 1 and equation 2 are solved to provide $\mathrm{a}=1/3$ and $\mathrm{u}=35/30$
The current distance in 10 seconds is ${\mathrm{S}}_1=10\mathrm{u}+100\mathrm{a}/2=10\mathrm{u}+50\mathrm{a}$
as a result, separation $t=8 \mathrm{s}$ and $t=10$ s i.e. the last two seconds' distance will be
${\mathrm{S}}_1-\mathrm{S}=10\mathrm{u}+50\mathrm{a}-(8\mathrm{u}+32\mathrm{a})=2\mathrm{u}+18\mathrm{a}=\frac{70}{30}+\frac{18}{3}=\frac{25}{3}$ which is approximately one meter
8.3meter

Hence, the correct answer is option A.