Time-dependent kinematics: A particle starts from rest and moves on a straight line with x(t) = t^3 - 3 t^2 + 5. What is the ratio of accelerations a(5 s) : a(3 s)?

Introduction / Context:

This exercise reinforces the relationship between position, velocity, and acceleration through differentiation. With position given as a time polynomial, differentiating once gives velocity, and differentiating twice gives acceleration. Ratios at specific times test accurate substitution and algebra.

Given Data / Assumptions:

• Position: x(t) = t^3 - 3 t^2 + 5.
• We need accelerations at t = 5 s and t = 3 s.
• One-dimensional motion; units are consistent.

Concept / Approach:

Velocity v(t) = dx/dt; acceleration a(t) = dv/dt = d^2x/dt^2. Use power-rule differentiation term-by-term.

Step-by-Step Solution:

Verification / Alternative check:

Second derivative check: differentiating x twice should remove the constant term and reduce polynomial degree properly; the result 6 t - 6 satisfies this.

Why Other Options Are Wrong:

• 3, 4, 5 do not match the evaluated ratio.
• 1 would imply equal accelerations at 5 s and 3 s, which is false for a(t) linear in t.

Common Pitfalls:

• Stopping after the first derivative (giving velocity instead of acceleration).
• Arithmetic mistakes when substituting t values.

Final Answer:

2