A body moving in a straight line with an acceleration of 12ms² has a velocity of 80ms after 5 seconds. find the initial velocity of the body and the distance covered in the 9th second

We are given the acceleration ($a$), final velocity ($v$), and time ($t$). We can use the first equation of motion:

$$v = u + at$$

Given:

• $a = 12\,m/s^2$
• $v = 80\,m/s$
• $t = 5\,s$

Substitute the values into the equation:

$$80\,m/s = u + (12\,m/s^2)(5\,s)$$ $$80 = u + 60$$

Subtract 60 from both sides to solve for $u$:

$$u = 80 - 60$$ $$u = 20\,m/s$$

The initial velocity of the body is $\boxed{\text{20 m/s}}$.

The distance covered in the $n^{th}$ second is given by the formula:

$$S_n = u + \frac{a}{2}(2n - 1)$$

Here, $n = 9$ (for the 9th second).
We use the initial velocity $u = 20\,m/s$ (found in Step 1) and the acceleration $a = 12\,m/s^2$.

Substitute the values into the formula:

$$S_9 = 20\,m/s + \frac{12\,m/s^2}{2}(2 \times 9 - 1)$$ $$S_9 = 20 + 6(18 - 1)$$ $$S_9 = 20 + 6(17)$$ $$S_9 = 20 + 102$$ $$S_9 = 122\,m$$

The distance covered in the 9th second is $\boxed{\text{122 m}}$.