Question

A train slows down as it rounds a sharp horizontal turn, going from 100 km/h to 60.0 km/h in the 14.0 s it takes to round the bend. The radius of the circle is 160 m. Compute the total acceleration and the angle at the moment the train speed reaches 60.0 km/h.

Answer 1

Answer:

The magnitude of the total acceleration of the train is approximately 1.909 meters per square second.

The direction of total acceleration with respect to radial acceleration is approximately -24.578º.

Please note that negative sign indicates that train is decelerating.

Explanation:

From Physics, we know that magnitude of the acceleration experimented by the train ($a$), measured in meters per square second, while rounding the bend is the Pythagorean identity involving radial and tangential accelerations ($a_{r}$, $a_{t}$), measured in meters per square second:

$a = \sqrt{a_{t}^{2}+a_{r}^{2}}$ (Eq. 1)

If we assuming that train decelerates at constant rate, then each component is determined by following expressions:

$a_{t} = \frac{v_{f}-v_{o}}{t}$ (Eq. 2)

$a_{r} = \frac{v_{f}^{2}}{R}$ (Eq. 3)

Where:

$v_{o}$, $v_{f}$ - Initial and final speeds, measured in meters per second.

$t$ - Deceleration time, measured in seconds.

$R$ - Bend radius, measured in meters.

If we know that $v_{o} = 27.778\,\frac{m}{s}$, $v_{f} = 16.667\,\frac{m}{s}$, $t = 14\,s$ and $R = 160\,m$, then each acceleration component is calculated:

$a_{t} = \frac{16.667\,\frac{m}{s}-27.778\,\frac{m}{s} }{14\,s}$

$a_{t} = -0.794\,\frac{m}{s^{2}}$

$a_{r} = \frac{\left(16.667\,\frac{m}{s} \right)^{2}}{160\,m}$

$a_{r} = 1.736\,\frac{m}{s^{2}}$

Then, the magnitude of the total acceleration of the train is:

$a = \sqrt{\left(-0.794\,\frac{m}{s^{2}} \right)^{2}+\left(1.736\,\frac{m}{s^{2}} \right)^{2}}$

$a \approx 1.909\,\frac{m}{s^{2}}$

The magnitude of the total acceleration of the train is approximately 1.909 meters per square second.

After that, we obtain the angle of acceleration by using the following trigonometric expression:

$\theta = \tan^{-1} \frac{a_{t}}{a_{n}}$ (Eq. 4)

Where $\theta$ is the direction of total acceleration with respect to radial acceleration, measured in sexagesimal degrees.

If we get that $a_{t} = -0.794\,\frac{m}{s^{2}}$ and $a_{r} = 1.736\,\frac{m}{s^{2}}$, the direction of total acceleration experimented by the train is:

$\theta = \tan^{-1}\left(\frac{-0.794\,\frac{m}{s^{2}} }{1.736\,\frac{m}{s^{2}} } \right)$

$\theta \approx -24.578^{\circ}$

Please note that negative sign indicates that train is decelerating.

The direction of total acceleration with respect to radial acceleration is approximately -24.578º.