The distance s in metres covered by a particle in t seconds is \(s = \frac{3}{2}t^{2} - 3t\). Find its acceleration.
\(1 ms^{-2}\)
\(2 ms^{-2}\)
\(3 ms^{-2}\)
\(4 ms^{-2}\)
Correct answer is C
The two time differentiation of distance with respect to time will give the acceleration.
\(s = \frac{3}{2}t^{2} - 3t\)
\(\frac{\mathrm d s}{\mathrm d t} = v = 3t - 3\)
\(\frac{\mathrm d v}{\mathrm d t} = a = 3\)