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

The correct option is 8.3 m.

Let's solve the problem step-by-step using the equations of motion for constant acceleration. Let the initial velocity of the particle be u and its constant acceleration be a.

Step 1: Analyze the motion in the first 5 seconds.
The distance travelled in the first $t_1=5\text{s}$ is $s_1=10\text{m}$.
Using the second equation of motion, $s=ut+\frac{1}{2}a{t}^2$:
$10=u\left(5\right)+\frac{1}{2}a{\left(5\right)}^2$
$10=5u+12.5a$
Dividing the entire equation by 5 gives:
$2=u+2.5a$ --- (Equation 1)

Step 2: Analyze the motion in the first 8 seconds.
The particle travels 10 m in the first 5 seconds and another 10 m in the next 3 seconds.
So, the total distance travelled in the first $t_2=5+3=8\text{s}$ is $s_2=10+10=20\text{m}$.
Applying the equation of motion again:
$20=u\left(8\right)+\frac{1}{2}a{\left(8\right)}^2$
$20=8u+32a$
Dividing the entire equation by 4 gives:
$5=2u+8a$ --- (Equation 2)

Step 3: Solve Equation 1 and Equation 2 for u and a.
Multiply Equation 1 by 2:
$4=2u+5a$ --- (Equation 3)
Subtract Equation 3 from Equation 2:
$(2u+8a)-(2u+5a)=5-4$
$3a=1$
$a=\frac{1}{3}{\text{m/s}}^2\approx 0.333{\text{m/s}}^2$
Now, substitute the value of a back into Equation 1 to find u:
$2=u+2.5\left(\frac{1}{3}\right)$
$2=u+\frac{5}{6}$
$u=2-\frac{5}{6}=\frac{7}{6}\text{m/s}\approx 1.167\text{m/s}$

Step 4: Find the distance travelled in the next 2 seconds.
The next 2 seconds corresponds to the time interval from $t=8\text{s}$ to $t=10\text{s}$.
Let's first find the total distance travelled in the first $t_3=10\text{s}$:
$s_3=u\left(10\right)+\frac{1}{2}a{\left(10\right)}^2$
Substitute the values of $u=\frac{7}{6}$ and $a=\frac{1}{3}$:
$s_3=\left(\frac{7}{6}\right)\left(10\right)+\frac{1}{2}\left(\frac{1}{3}\right)\left(100\right)$
$s_3=\frac{70}{6}+\frac{100}{6}=\frac{170}{6}=\frac{85}{3}\text{m}\approx 28.33\text{m}$

The distance travelled in the next 2 seconds (from 8 s to 10 s) is the difference between the distance travelled in 10 seconds and the distance travelled in 8 seconds:
$d=s_3-s_2$
$d=28.33\text{m}-20\text{m}=8.33\text{m}$

Rounding to one decimal place, the distance travelled in the next 2 seconds is 8.3 m.