Question

The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determine the magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m

Answer 1

Explanation:

At point B, the velocity speed of the train is as follows.

          $\nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})$

                           = $(30)^{2} + 2(-0.25(412 - 0))$

                           = 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

          y = $200 e^{\frac{x}{1000}}$

      $\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

Now, second derivative of the train is calculated as follows.

         $\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$      

       $\frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}$    

Radius of curvature of the train is as follows.  

   $\rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}$

               = $\frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}$

              = 3808.96 m

Now, we will calculate the normal component of the train as follows.

            $a_{n} = \frac{\nu^{2}_{B}}{\rho}$

                        = $\frac{(26.34)^{2}}{3808.96}$

                        = 0.1822 $m/s^{2}$

The magnitude of acceleration of train is calculated as follows.

            a = $\sqrt{(a_{t})^{2} + (a_{n})^{2}}$

               = $\sqrt{(-0.25)^{2} + (0.1822)^{2}}$

              = $0.309 m/s^{2}$

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is $0.309 m/s^{2}$.