Explanation:
a) The train first accelerates from 0 m/s to a velocity v in 5 minutes.
It then maintains that velocity v for 20 minutes.
Finally, it decelerates from v to 0 m/s at twice the acceleration rate, or in half the time (2.5 minutes).
So the total time the train travels is 5 + 20 + 2.5 = 27.5 minutes.
b) The train travels 4.5 km during acceleration, so the acceleration is:
Δx = v₀ t + ½ at²
4500 m = (0 m/s) (300 s) + ½ a (300 s)²
a = 1/10 m/s²
The distance traveled equals the area under the speed-time graph.
d = ½ (v) (300 s) + v (1200 s) + ½ (v) (150 s)
d = (1425 s) v
Use the acceleration to find the velocity reached.
v = at = (1/10 m/s²) (300 s) = 30 m/s
The distance is therefore:
d = (1425 s) (30 m/s)
d = 42,750 m