Respuesta :
Explanation:
At point B, the velocity speed of the train is as follows.
[tex]\nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})[/tex]
= [tex](30)^{2} + 2(-0.25(412 - 0))[/tex]
= 26.34 m/s
Now, we will calculate the first derivative of the equation of train.
y = [tex]200 e^{\frac{x}{1000}}[/tex]
[tex]\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}[/tex]
Now, second derivative of the train is calculated as follows.
[tex]\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}[/tex]
[tex]\frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}[/tex]
Radius of curvature of the train is as follows.
[tex]\rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}[/tex]
= [tex]\frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}[/tex]
= 3808.96 m
Now, we will calculate the normal component of the train as follows.
[tex]a_{n} = \frac{\nu^{2}_{B}}{\rho}[/tex]
= [tex]\frac{(26.34)^{2}}{3808.96}[/tex]
= 0.1822 [tex]m/s^{2}[/tex]
The magnitude of acceleration of train is calculated as follows.
a = [tex]\sqrt{(a_{t})^{2} + (a_{n})^{2}}[/tex]
= [tex]\sqrt{(-0.25)^{2} + (0.1822)^{2}}[/tex]
= [tex]0.309 m/s^{2}[/tex]
Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is [tex]0.309 m/s^{2}[/tex].
The acceleration of the train when it reaches point B is -0.016 m/s².
The given parameters;
- distance between point A and B = 412 m
- speed of train, v = 30 m/s
- acceleration of the train at point A = -0.25 m/s²
The final velocity of the train when it reaches point B is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\v^2 = (30)^2 + 2(-0.25)(412)\\\\v^2 = 694\\\\v = \sqrt{694} \\\\v = 26.34 \ m/s[/tex]
The time to travel from point A to point B at the given constant velocity;
[tex]412 = 30t + (0.5)(-0.25)t^2\\\\412 = 30t - 0.125t^2\\\\0.125t^2 - 30t + 412 = 0\\\\a = 0.125, \ b = -30, \ c = 412\\\\t = \frac{-b \ \ + \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-(-30) \ \ + \ \sqrt{(-30)^2 - 4(0.125\times 412)} }{2(0.125)} \\\\t = 225.38 \ s[/tex]
The acceleration of the train when it reaches point B is calculated as;
[tex]a = \frac{\Delta V}{t} \\\\a = \frac{26.34 - 30}{225.38} \\\\a = -0.016 \ m/s^2[/tex]
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