Kinematics on a line: For a particle with position x(t) = t^2 * (t + 1) = t^3 + t^2, what is the instantaneous acceleration a(t)?

Introduction / Context:

This is a direct application of single-variable calculus to particle motion. Position x(t) is given; velocity is the first derivative dx/dt, and acceleration is the time derivative of velocity, d^2x/dt^2. Such questions are common in basic dynamics and control of motion along a line.

Given Data / Assumptions:

• Position: x(t) = t^3 + t^2 (units consistent but unspecified).
• Differentiation rules apply; motion is along a straight line.

Concept / Approach:

Velocity v(t) = dx/dt; Acceleration a(t) = dv/dt. Differentiate term-by-term using power rules.

Step-by-Step Solution:

Verification / Alternative check:

As a quick check, units are consistent: if x is in m and t in s, then v is m/s and a is m/s^2; nothing contradictory arises from the differentiation.

Why Other Options Are Wrong:

• 3t^2 + 2t is the velocity, not the acceleration.
• 6t - 2 and 3t - 2 alter constant terms incorrectly.
• 3t^3 - 2t is unrelated to correct differentiation steps.

Common Pitfalls:

• Differentiating once and mistaking v(t) for a(t).

Final Answer:

6t + 2