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
$s=ut+ \frac{1}{2} at^2$

But, we assume that the train accelerates from rest, thus
$s=\frac{1}{2} at^2 \\ \\ \Rightarrow465=\frac{1}{2}(1.74)t^2 \\ \\ \Rightarrow t^2=534.48 \\ \\ \Rightarrow t=\sqrt{534.48}=23.12$

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.

6 0

3 years ago

Are all parallelograms squares?

pav-90 [236]

No at least two side need to be parallel to be that so almost anything will be your answer

6 0

3 years ago

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