WAEC - Further Mathematics (2010 - No. 21)
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.
\(\frac{1}{3}\)
\(\frac{4}{9}\)
\(\frac{5}{9}\)
\(1\)
Explanation
Differentiate distance twice to get the acceleration and then equate to get p.
\(s = 7 + pt^{3} + t^{2}\)
\(\frac{\mathrm d s}{\mathrm d t} = v(t) = 3pt^{2} + 2t\)
\(\frac{\mathrm d v}{\mathrm d t} = a(t) = 6pt + 2\)
\(a(3) = 6p(3) + 2 = 8 \implies 18p = 8 - 2 = 6\)
\(p = \frac{6}{18} = \frac{1}{3}\)
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