Explanation:
Distance travelled by the particle is
\(x = 40 + 12t - {t^3}\)
We know that, speed is rate of change of distance i.e., \(v = \frac{{dx}}{{dt}}\)
\(\therefore v = \frac{d}{{dt}}\left( {40 + 12t - {t^3}} \right) = 0 + 12 - 3{t^2}\)
But final velocity \(v = 0\)
\(\therefore \;12 - 3{t^2} = 0\)
\( \Rightarrow {t^2} = \frac{{12}}{3} = 4 \Rightarrow t = 2s\)
Hence, distance travelled by the particle before coming to rest is given by
\(x = 40 + 12\left( 2 \right) - {\left( 2 \right)^3}\)
\( = 40 + 24 - 8 = 64 - 8 = 56 \, m\)
At, \(t = 0,\;{x_1} = 40 \, m\)
\(t = 2s,\;{x_2} = 56 \, m\)
\(\therefore S = {x_2} - {x_1} = 16 \, m\)