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1 answer:
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Explanation:
At point B, the velocity speed of the train is as follows.
=
= 26.34 m/s
Now, we will calculate the first derivative of the equation of train.
y =
Now, second derivative of the train is calculated as follows.
Radius of curvature of the train is as follows.
=
= 3808.96 m
Now, we will calculate the normal component of the train as follows.
=
= 0.1822
The magnitude of acceleration of train is calculated as follows.
a =
=
=
Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is .