A particle starts from rest and moves through a distance \(S = 12t^{2} – 2t^{3}\) metres in time t seconds. Find its acceleration in 1 second.
- 24\(ms^{-2}\)
- 18\(ms^{-2}\)
-
12\(ms^{-2}\)
- 10\(ms^{-2}\)
The correct answer is: C
Explanation
\(\frac{\mathrm d s(t)}{\mathrm d t} = v(t)\) and \(\frac{\mathrm d v(t)}{\mathrm d t} = a(t)\)
\(\therefore v(t) = \frac{\mathrm d (12t^{2} - 2t^{3})}{\mathrm d t} = 24t - 6t^{2}\)
\(\frac{\mathrm d (24t - 6t^{2})}{\mathrm d t} = 24 - 12t = a(t)\)
\(a(1) = 24 - 12(1) = 24 - 12 = 12ms^{-2}\)