One-dimensional kinematics (with a turning point): A particle moves along a straight line with position x(t) = t^2 − 10 t + 30 (x in metres, t in seconds). What is the total distance travelled between t = 0 s and t = 10 s?

Introduction / Context:
Total distance travelled differs from net displacement when the motion includes reversals (turning points). For polynomial position-time functions, identifying turning instants via velocity analysis is essential to sum absolute path segments correctly.

Given Data / Assumptions:

• Position: x(t) = t^2 − 10 t + 30 (metres).
• Time interval: t from 0 s to 10 s.
• Straight-line motion; continuous differentiability allows critical-point analysis using velocity.

Concept / Approach:
Compute velocity v(t) = dx/dt. Turning points occur where v(t) = 0 (and acceleration indicates direction change). Evaluate positions at the start, at each turning instant within the interval, and at the end. Sum absolute differences to get total distance; net displacement would be the algebraic difference x(10) − x(0).

Step-by-Step Solution:

Verification / Alternative check:
Acceleration a(t) = dv/dt = 2 m/s^2 > 0 confirms a single minimum at t = 5 s. Net displacement x(10) − x(0) = 30 − 30 = 0 m, while total distance is 50 m, consistent with a “down then up” path.

Why Other Options Are Wrong:

• 0 m: This is the net displacement, not the total distance.
• 30 m or 60 m: Do not match the sum of absolute segment lengths determined by the turning point.
• None of these: Incorrect because 50 m is attainable and correct.

Common Pitfalls:
Confusing displacement with distance; forgetting to split the time interval at v = 0; sign errors in evaluating absolute differences.

Final Answer:
50 m