Answer:
A) OA, AB, BC
B) 25m/s^2
C) see explanation
D) 25
E) Rest
Explanation:
From the Velocity time graph shown:
The positive slope = OA ; This is positive because, it is the point of uniform acceleration on the graph.
Constant slope = AB, the slope here is constant because, AB on the graph is the point of constant velocity.
-ve slope = BC
B) Acceleration of body in path OA.
Acceleration = change in Velocity / time
Acceleration = (150 - 0) / 6
Acceleration = 150/6 = 25m/s^2
C) Path AB is Parallel to the because it marks the period of constant velocity (that is Velocity does not increase or decrease during the time interval).
D) Length of BC
BC corresponds to the distance moved, that velocity / time
Velocity = 150 ; time = 6
Therefore Distance (BC) = 150/6 = 25
E.) Velocity =0 ; Hence body is at rest
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.
![\rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}](https://tex.z-dn.net/?f=%5Crho%20%3D%20%5Cfrac%7B%5B1%20%2B%20%28%5Cfrac%7Bdy%7D%7Bdx%7D%29%5E%7B2%7D%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%7B%5Cfrac%7Bd%5E%7B2%7Dy%7D%7Bdx%5E%7B2%7D%7D%7D)
= ![\frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B1%20%2B%200.2e%5E%7B%5Cfrac%7B400%7D%7B1000%7D%7D%5E%7B2%7D%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%7B0.2%2810%5E%7B-3%7D%29e%5E%7B%5Cfrac%7B400%7D%7B1000%7D%7D%7D)
= 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
.
The elements which have similar behavior are Barium, strontium and beryllium.
Explanation: