The distance S metres moved by a body in t seconds is given by \(S = 5t^3 – \frac{19}{2} t^2 + 6t – 4\). Calculate the acceleration of the body after 2 seconds
- 19 \(ms ^{-2}\)
- 21 \(ms ^{-2}\)
-
41 \(ms ^{-2}\)
- 31 \(ms ^{-2}\)
The correct answer is: C
Explanation
\(S = 5t^3 - \frac{19}{2} t^2 + 6t - 4\)
\(v(t)=\frac{dS}{dt}=15t^2-19t+6\)
\(a(t)=\frac{dv}{dt}=30t-19\)
∴a(2)=30(2)-19=60-19=41 \(ms ^{-2}\)