Acceleration from Velocity Expression

If the velocity of a particle is given by \(v = (180 - 16x)^{1/2}\text{ ms}^{-1}\), then its acceleration will be

zero

\(16\text{ ms}^{-2}\)

\(-8\text{ ms}^{-2}\)

\(4\text{ ms}^{-2}\)

We have \(v^2 = 180 - 16x\). Comparing with \(v^2 = u^2 + 2ax\), we get \(2a = -16 ⇒ a = -8\text{ ms}^{-2}\).