Acceleration from Velocity Expression – Rankers Physics

Calculus Based Questions: Practice Problem & Solution

If the velocity of a particle is given by \(v = (180 - 16x)^{1/2}\text{ ms}^{-1}\), then its acceleration will be
zero
\(16\text{ ms}^{-2}\)
\(-8\text{ ms}^{-2}\)
\(4\text{ ms}^{-2}\)

Solution Explained:

To solve this problem, we apply the core principles of Calculus Based Questions. Understanding the underlying formula is key to arriving at the correct answer below:

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

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