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

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.

Step-by-step 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) = (v_(f)-v_(o))/(t)` (Eq. 2)

`a_(r) = (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\,(m)/(s)`,
`v_(f) = 16.667\,(m)/(s)`,
`t = 14\,s` and
`R = 160\,m`, then each acceleration component is calculated:

`a_(t) = (16.667\,(m)/(s)-27.778\,(m)/(s) )/(14\,s)`

`a_(t) = -0.794\,(m)/(s^(2))`

`a_(r) = (\left(16.667\,(m)/(s) \right)^(2))/(160\,m)`

`a_(r) = 1.736\,(m)/(s^(2))`

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

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

`a \approx 1.909\,(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) (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\,(m)/(s^(2))` and
`a_(r) = 1.736\,(m)/(s^(2))`, the direction of total acceleration experimented by the train is:

`\theta = \tan^(-1)\left((-0.794\,(m)/(s^(2)) )/(1.736\,(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.