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: Option 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})