The average velocity of a body moving with uniform acceleration travelling a distance of 3.06 m is 0.34 m/s¹.If the change in velocity of the body is 0.18m/s¹ during this time, its uniform acceleration is

The correct option is 0.02 m/s².

To find the uniform acceleration of the body, we can break the problem down into two steps: finding the total time taken for the travel, and then calculating the acceleration using the change in velocity over that time.

Step 1: Find the time duration ($t$) of the motion.
We are given the total distance ($s$) and the average velocity ($v_{\text{avg}}$). The relationship between distance, average velocity, and time is given by:
$v_{\text{avg}}=\frac{s}{t}$
Rearranging the formula to solve for time ($t$):
$t=\frac{s}{v_{\text{avg}}}$
Substitute the given values ($s=3.06\text{m}$ and $v_{\text{avg}}=0.34\text{m/s}$):
$t=\frac{3.06}{0.34}=9\text{s}$

Step 2: Calculate the uniform acceleration ($a$).
Acceleration is defined as the rate of change of velocity over time:
$a=\frac{\Delta v}{t}$
Here, the change in velocity ($\Delta v$) is given as $0.18\text{m/s}$, and we calculated the time duration ($t$) to be $9\text{s}$.
Substitute these values into the acceleration formula:
$a=\frac{0.18}{9}=0.02{\text{m/s}}^2$
Therefore, the uniform acceleration of the body is $0.02{\text{m/s}}^2$.