A body travels for 15 sec starting from rest with constant acceleration. If it travels distances S1, S2 and S3 in the first five seconds, second five seconds and next five seconds respectively the relation between S1,S2 and S3 is

To find the relation between the distances $S_1$, $S_2$, and $S_3$ traveled by the body in successive 5-second intervals, we can use the equations of motion for constant acceleration.

Let the body start from rest, so its initial velocity is:
$u=0$
Let the constant acceleration of the body be $a$.

The distance $S$ covered in time $t$ starting from rest is given by the second equation of motion:
$S=ut+\frac{1}{2}a{t}^2$
Since $u=0$, this simplifies to:
$S=\frac{1}{2}a{t}^2$

Step 1: Distance traveled in the first 5 seconds ($S_1$)
For $t=5\text{s}$:
$S_1=\frac{1}{2}a\left(5^2\right)=\frac{25}{2}a$

Step 2: Distance traveled in the next 5 seconds ($S_2$)
The total distance traveled in the first 10 seconds is:
$S_{\text{total, 10}}=\frac{1}{2}a\left({10}^2\right)=\frac{100}{2}a=50a$
Therefore, the distance traveled during the second 5-second interval (from $t=5$ to $t=10$) is:
$S_2=S_{\text{total, 10}}-S_1=50a-\frac{25}{2}a=\frac{75}{2}a$

Note that we can write $S_2$ in terms of $S_1$:
$S_2=3\left(\frac{25}{2}a\right)=3S_1$
Which gives:
$S_1=\frac{1}{3}$

Step 3: Distance traveled in the next 5 seconds ($S_3$)
The total distance traveled in the first 15 seconds is:
$S_{\text{total, 15}}=\frac{1}{2}a\left({15}^2\right)=\frac{225}{2}a$
The distance traveled during the third 5-second interval (from $t=10$ to $t=15$) is:
$S_3=S_{\text{total, 15}}-S_{\text{total, 10}}=\frac{225}{2}a-50a=\frac{125}{2}a$

We can also write $S_3$ in terms of $S_1$:
$S_3=5\left(\frac{25}{2}a\right)=5S_1$
Which gives:
$S_1=\frac{1}{5}$

Step 4: Finding the relation
Combining the relations for $S_1$:
$S_1=\frac{1}{3}$

Thus, the correct option is:
S1 = (1/3)S2 = (1/5)S3