Answer :
To find the distance travelled by a body in the fifth second when it starts from rest and moves with uniform acceleration, we use the formula for distance travelled in a specific second:
[tex]s_n = u + \frac{1}{2}a(2n-1)[/tex]
where:
- [tex]s_n[/tex] is the distance travelled in the [tex]n[/tex]-th second,
- [tex]u[/tex] is the initial velocity,
- [tex]a[/tex] is the uniform acceleration,
- [tex]n[/tex] is the specific second.
Given that:
- Initial velocity, [tex]u = 0[/tex] m/s (since it starts from rest),
- Uniform acceleration, [tex]a = 8[/tex] m/s²,
- We need the distance travelled in the fifth second, so [tex]n = 5[/tex].
Substituting these values into the equation:
[tex]s_5 = 0 + \frac{1}{2} \times 8 \times (2 \times 5 - 1)[/tex]
Simplify the expression inside the parenthesis:
[tex]2 \times 5 - 1 = 10 - 1 = 9[/tex]
Now substitute back:
[tex]s_5 = \frac{1}{2} \times 8 \times 9 = 4 \times 9 = 36~\text{metres}[/tex]
Thus, the distance travelled by the body in the fifth second is [tex]36[/tex] metres.
Therefore, the correct option is [tex]\mathbf{A.}[/tex] 36 metres.
