Someone help pleeeaaaaseee

Step-by-step 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

1= 16 counters

2=28 counters

3=34 counters

4= 20 counters

5=6 counters

6=19 counters

Step-by-step explanation: