The distance s metres of a particle from a fixed point at time t seconds is given by (s = 7 + pt^{3} + t^{2}), where p is a constant. If the acceleration at t = 3 secs is (8 ms^{-2}), find the value of p.
-
A.
(frac{1}{3}) -
B.
(frac{4}{9}) -
C.
(frac{5}{9}) -
D.
(1)
Correct Answer: Option A
Explanation
Differentiate distance twice to get the acceleration and then equate to get p.
(s = 7 + pt^{3} + t^{2})
(frac{mathrm d s}{mathrm d t} = v(t) = 3pt^{2} + 2t)
(frac{mathrm d v}{mathrm d t} = a(t) = 6pt + 2)
(a(3) = 6p(3) + 2 = 8 implies 18p = 8 – 2 = 6)
(p = frac{6}{18} = frac{1}{3})