Respuesta :
Answer:
(a) The train took 100 seconds (1.667 minutes) to reach its top speed.
(b) The train travelled a distance of 270000 meters (270 kilometers) at top speed.
(c) The train decelerates at a rate of 2.5 meters per square second.
Explanation:
Statement is incomplete. Complete description is presented below:
A train started from a station and accelerated at [tex]0.5\,\frac{m}{s^{2}}[/tex] to reach its top speed of [tex]50\,\frac{m}{s}[/tex] and maintained this speed for 90 minutes.
When the train approached the next station the driver applied the brakes uniformly to bring the train to a stop in a distance of 500 meters.
(a) Calculate how long it took the train to reach its top speed.
(b) Calculate how far it travelled at its top speed.
(c) Calculate the acceleration experienced by the train when the brakes were applied.
(a) Let suppose that train accelerates uniformly. Then, the time needed by the vehicle ([tex]t[/tex]), measured in seconds, is described by the following equation of motion:
[tex]t = \frac{v-v_{o}}{a}[/tex] (1)
Where:
[tex]a[/tex] - Acceleration, measured in meters per square second.
[tex]v_{o}[/tex], [tex]v[/tex] - Initial and final speed, measured in meters per second.
If we know that [tex]a = 0.5\,\frac{m}{s^{2}}[/tex], [tex]v_{o} = 0\,\frac{m}{s}[/tex] and [tex]v = 50\,\frac{m}{s}[/tex], then the time needed by the train is:
[tex]t = \frac{50\,\frac{m}{s}-0\,\frac{m}{s}}{0.5\,\frac{m}{s^{2}} }[/tex]
[tex]t = 100\,s[/tex]
The train took 100 seconds (1.667 minutes) to reach its top speed.
(b) Let suppose that train travels at constant speed, the distance travelled by the train ([tex]s[/tex]), measured in meters, is:
[tex]s = v\cdot t[/tex] (2)
Where:
[tex]v[/tex] - Top speed, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
If we know that [tex]v = 50\,\frac{m}{s}[/tex] and [tex]t = 5400\,s[/tex], then the distance travelled by the train at top speed is:
[tex]s = \left(50\,\frac{m}{s} \right)\cdot (5400\,s)[/tex]
[tex]s = 270000\,m[/tex]
The train travelled a distance of 270000 meters (270 kilometers) at top speed.
(c) Let suppose that train decelerates at constant rate. We can calculate the acceleration from the following expression:
[tex]a = \frac{v^{2}-v_{o}^{2}}{2\cdot s}[/tex] (3)
Where:
[tex]v_{o}[/tex], [tex]v[/tex] - Initial and final speed, measured in meters per second.
[tex]s[/tex] - Travelled distance, measured in meters.
If we know that [tex]v_{o} = 50\,\frac{m}{s}[/tex], [tex]v = 0\,\frac{m}{s}[/tex] and [tex]s = 500\,m[/tex], then the acceleration of the train is:
[tex]a = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(50\,\frac{m}{s} \right)^{2}}{2\cdot (500\,m)}[/tex]
[tex]a = -2.5\,\frac{m}{s^{2}}[/tex]
The train decelerates at a rate of 2.5 meters per square second.
