A particle starts from rest and moves through a distance \(S = 12t^{2} - 2t^{3}\)
FURTHER MATHEMATICS
WAEC 2018
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.
- A. 24\(ms^{-2}\)
- B. 18\(ms^{-2}\)
- C. 12\(ms^{-2}\)
- D. 10\(ms^{-2}\)
Correct Answer: C. 12\(ms^{-2}\)
Explanation
\(\frac{\mathrm d s(t)}{\mathrm d t} = v(t)\) and \(\frac{\mathrm d v(t)}{\mathrm d t} = a(t)\)
\(\therefore v(t) = \frac{\mathrm d (12t^{2} - 2t^{3})}{\mathrm d t} = 24t - 6t^{2}\)
\(\frac{\mathrm d (24t - 6t^{2})}{\mathrm d t} = 24 - 12t = a(t)\)
\(a(1) = 24 - 12(1) = 24 - 12 = 12ms^{-2}\)
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