Answer:
The distance the train travels before coming to a (complete) stop = 40/81 km which is approximately 493.83 meters
Explanation:
The initial speed of the train u = 80 km/h = 22 2/9 m/s = 22. m/s
The magnitude of the constant acceleration with which the train slows, a = 0.5 m/s²
Therefore, we have the following suitable kinematic equation of motion;
v² = u² - 2 × a × s
Where;
v = The final velocity = 0 (The train comes to a stop)
s = The distance the train travels before coming to a stop
Substituting the values gives;
0² = 22.² - 2 × 0.5 × s
2 × 0.5 × s = 22.²
s = 22.²/1 = 493 67/81 m = 40/81 km
The distance the train travels before coming to a (complete) stop = 40/81 km ≈ 493.83 m.