A 3 kg particle decelerates uniformly along a straight line from 40 m/s to 20 m/s over a distance of 300 m. Assuming the same constant deceleration continues, how much additional distance will it travel before coming to rest?

Introduction / Context:
Uniformly accelerated (or decelerated) rectilinear motion problems are common in dynamics. The kinematic relations between speed, distance, and acceleration allow us to determine stopping distances once the deceleration is known.

Given Data / Assumptions:

• Initial segment: u = 40 m/s down to v = 20 m/s over s = 300 m.
• Motion is uniformly decelerated (constant a).
• Mass is irrelevant for purely kinematic calculations.

Concept / Approach:
Use the equation v^2 = u^2 + 2 a s to determine the constant deceleration. Then apply the same deceleration from 20 m/s to 0 to compute the additional stopping distance.

Step-by-Step Solution:
Find a from 20^2 = 40^2 + 2*a*300 → 400 = 1600 + 600a → a = -2 m/s^2.From 20 m/s to rest with a = -2 m/s^2: use 0^2 = 20^2 + 2*a*s₂.Solve for s₂: s₂ = (0 - 400) / (2 * -2) = 100 m.Thus, the particle travels an additional 100 m before stopping.

Verification / Alternative check:
Time-based check: deceleration 2 m/s^2; time from 20 m/s to 0 is 10 s; average speed during this phase is 10 m/s → distance = 10 * 10 = 100 m.

Why Other Options Are Wrong:

• 10 m, 20 m, 50 m underestimate stopping distance for the given deceleration.
• 150 m overestimates; it would require a smaller magnitude of deceleration.

Common Pitfalls:
Sign errors in v^2 = u^2 + 2 a s; mixing time and distance formulas without consistency checks.

Final Answer:
100 m