Question

A body starts from rest and moves with uniform acceleration of 6m/s^2. What distance does it cover in the third second?

Answer 1

To find the distance covered by the body in the third second, we can use the equation of motion:

$\text{Distance}=\text{Initial Velocity}\times \text{Time}+\frac{1}{2}\times \text{Acceleration}\times {\text{Time}}^2$

In this case, the body starts from rest, so the initial velocity ($v_0$) is 0 m/s. The acceleration ($a$) is given as 6 m/s^2, and the time ($t$) is 3 seconds.

Substituting these values into the equation, we get:

$\text{Distance}=0\times 3+\frac{1}{2}\times 6\times 3^2$

Simplifying the equation:

$\text{Distance}=\frac{1}{2}\times 6\times 9$

Distance=3×9 ext{Distance} = 3 imes 9Distance<span class="mspace" style="margi