Calculate the total distance covered by a train before coming to rest if its initial speed is 30$ms^{-1}$ with a constant retardation of 0.1m-2

Analyze the question: A train moves at an initial speed u = 30 m/s and slows uniformly (retardation a = 0.1 m/s², written as "m-2" in the question, which is a typo for m/s²) until it stops (v = 0). We need the distance covered.

Concept: For constant acceleration, use the equation of motion that links speed, acceleration and distance without time:

v² = u² − 2as

(Retardation is negative acceleration, so we subtract.)

Solution:

• u = 30 m/s
• v = 0
• a = 0.1 m/s²

0 = 30² − 2(0.1)s

0 = 900 − 0.2s

s = 900 / 0.2 = 4500 m

Check using time:

• t = (u − v)/a = 30/0.1 = 300 s
• s = average speed × time = (30 + 0)/2 × 300 = 15 × 300 = 4500 m ✔

Tip: When a question gives speeds and acceleration but no time, go straight to v² = u² + 2as. For "coming to rest", set v = 0 and use a negative sign for retardation, so s = u²/(2a). Also watch for unit typos like "m-2": retardation must be in m/s².