the train accelerates at a rate of 2ms^-2 for 10s when it leaves station A and decelerates at a rate of 4ms^-2 when it approaches station B. the whole journey takes 100s. Find the distance between the two stations.

The distance between the stations is equal to the area under the velocity vs. time curve. In this case, it's the area of the two triangles plus the area of the rectangle.

First, we find the velocity reached after acceleration (the height of the triangles and rectangle).

v = at + v₀

v = (2 m/s²) (10 s) + 0 m/s

v = 20 m/s

Next, we find the time it takes to decelerate (the width of the second triangle).

v = at + v₀

0 m/s = (-4 m/s²) t + 20 m/s

t = 5 s

The whole journey takes 100 s, so the time the train traveled at its top speed is 100 s − 10 s − 5 s = 85 s. This is the width of the rectangle.

Therefore, the distance is:

d = ½ (10 s) (20 m/s) + (85 s) (20 m/s) + ½ (5 s) (20 m/s)

d = 1850 m

the train accelerates at a rate of 2ms^-2 for 10s when it leaves station A and decelerates at a rate of 4ms^-2 when it approaches station B. the whole journey takes 100s. Find the distance between the two stations.​

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the train accelerates at a rate of 2ms^-2 for 10s when it leaves station A and decelerates at a rate of 4ms^-2 when it approaches station B. the whole journey takes 100s. Find the distance between the two stations.