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: A
Explanation
Differentiate distance twice to get the acceleration and then equate to get p. \(s = 7 + pt^{3} + t^{2}\) \(\frac{\mathrm d s}{\mathrm d t} = v(t) = 3pt^{2} + 2t\) \(\frac{\mathrm d v}{\mathrm d t} = a(t) = 6pt + 2\) \(a(3) = 6p(3) + 2 = 8 \implies 18p = 8 - 2 = 6\) \(p = \frac{6}{18} = \frac{1}{3}\)
