Acceleration from Time-Distance Relation – Rankers Physics

Calculus Based Questions: Practice Problem & Solution

The relation between times t and distance x is \(t = ax^2 + bx\), where a and b are constants. The acceleration is
\(-2abv^2\)
\(2bv^3\)
\(-2av^3\)
\(2av^2\)

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:

Differentiating \(t = ax^2 + bx\) with respect to \(t\) gives \(1 = (2ax + b)v\). Differentiating again with respect to \(t\) gives \(a = \frac{dv}{dt} = -2av^3\).

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