91Ó°ÊÓ

To do that, we need to remember the formula for distance traveled with uniform acceleration during any interval of time: \[S = v_0t + \frac{1}{2}at^2\] where \(S\) is the distance traveled, \(v_0\) is the initial velocity, \(t\) is the time, and \(a\) is the acceleration. However, since we know that the body starts from rest, we can set \(v_0=0\), and our equation simplifies to: \[S = \frac{1}{2}at^2\] The body traveled \(8 \mathrm{~m}\) during the 2nd second, so we can also write that the total distance during the first 2 seconds is: \[S_{total} = \frac{1}{2}a(2)^2\] We'll find the distance traveled in the first second, \(S_1\), by: \[S_1 = \frac{1}{2}a(1)^2\] Now, let's find the distance during the 2nd second, \(S_2\): \[S_2 = S_{total} - S_1\]

Plug in the given distance, \(8 \mathrm{~m}\), for \(S_2\) and solve for the acceleration, \(a\): \[8 \mathrm{~m} = \frac{1}{2}a(2)^2 - \frac{1}{2}a(1)^2\] Now, solve for a: \[8 = 2a - \frac{1}{2}a\] \[8 = \frac{3}{2}a\] \[a = \frac{16}{3} \mathrm{ms^{-2}}\]

First, we need to find the distance the body travels during the first 5 seconds, \(S_{5s}\): \[S_{5s} = \frac{1}{2}\times\frac{16}{3}\times(5)^2 = \frac{200}{3} \mathrm{~m}\] Next, find the distance traveled during the first 4 seconds, \(S_{4s}\): \[S_{4s} = \frac{1}{2}\times\frac{16}{3}\times(4)^2 = \frac{128}{3} \mathrm{~m}\] Now, we can calculate the distance traveled during the 5th second, \(S_5\): \[S_5 = S_{5s} - S_{4s} = \frac{200}{3} - \frac{128}{3} = \frac{72}{3} = 24 \mathrm{~m}\] So, the correct answer is: (B) \(24 \mathrm{~m}\)