Uniformly accelerated motion (retardation case): A particle starts with speed u = 2 m/s along a straight line and experiences a constant negative acceleration a = −0.1 m/s^2. How much time t is required to traverse a distance S = 1.5 m from the start?

Introduction / Context:
Translational kinematics with constant acceleration allows direct computation of travel time when the initial velocity, uniform acceleration (including deceleration), and displacement are known. Care must be taken with signs and with multiple mathematical roots of the quadratic equation.

Given Data / Assumptions:

• Initial speed u = 2 m/s.
• Constant acceleration a = −0.1 m/s^2 (retardation).
• Required displacement S = 1.5 m in the initial direction.
• Straight-line motion; no turning points considered within the shortest travel time.

Concept / Approach:
Use the displacement relation for constant acceleration:
S = u * t + (1/2) * a * t^2.
This yields a quadratic in t. The physically meaningful time is the smaller positive root (the later larger root corresponds to a mathematically valid but physically subsequent position after slowing, stopping, and reversing if applicable).

Step-by-Step Solution:

Verification / Alternative check:
Compute velocity at t = 0.765 s: v = u + a t ≈ 2 − 0.1 * 0.765 ≈ 1.9235 m/s (still positive), confirming the first passage before the particle meaningfully slows down.

Why Other Options Are Wrong:
Values like 10–20 s are far beyond the first physical crossing; 1.50 s is a distractor from using S ≈ u t only (ignoring acceleration).

Common Pitfalls:
Dropping the acceleration term; selecting the larger quadratic root without interpreting motion; sign mistakes with negative acceleration.

Final Answer:
t ≈ 0.76 s