The velocity \(v ms^{-1}\) of a particle moving in a straight line is given by \(v = 3t^{2} - 2t + 1\) at time t secs. Find the acceleration of the particle after 3 seconds.
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option D
Explanation:
\(v(t) = 3t^{2} - 2t + 1\)
\(\frac{\mathrm d v}{\mathrm d t} = a(t) = 6t - 2\)
\(a(3) = 6(3) - 2 = 18 - 2 = 16 ms^{-2}\)
\(v(t) = 3t^{2} - 2t + 1\)
\(\frac{\mathrm d v}{\mathrm d t} = a(t) = 6t - 2\)
\(a(3) = 6(3) - 2 = 18 - 2 = 16 ms^{-2}\)