Cy) = 3y - 4 d(y) = 3 – 2y Find (c. d)(y).
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Given that the subway stations are 930 m apart, the train have to be accerelated for half the distance and then decerelated for the rest of the distance.
Recall that the distance travelled by an object with an initial velocity, u, for a period of time, t, at an accereration, a, is given by
But, we assume that the train accelerates from rest, thus
The maximum speed is attained at half the center of the distance between subway stations (i.e. at distance = 465 m).
Thus, maximum speed = distance / time = 465 / 23.12 = 20.11 m/s.
Recall that the distance travelled by an object with an initial velocity, u, for a period of time, t, at an accereration, a, is given by
But, we assume that the train accelerates from rest, thus
The maximum speed is attained at half the center of the distance between subway stations (i.e. at distance = 465 m).
Thus, maximum speed = distance / time = 465 / 23.12 = 20.11 m/s.