Answer:
The train will travel 493.817 meters until it stops completely.
Step-by-step explanation:
In this case, we know that train applies the brakes and decelerates at constant rate until rest is reached after travelling an unknown distance. Travelled distance ([tex]\Delta s[/tex]), measured in meters, can be found by using this kinematic formula:
[tex]v^{2} = v_{o}^{2}+2\cdot a \cdot \Delta s[/tex] (Eq. 1)
Where:
[tex]v_{o}[/tex] - Initial speed, measured in meters per second.
[tex]v[/tex] - Final speed, measured in meters per second.
[tex]a[/tex] - Acceleration, measured in meters per square second.
Now travelled distance is cleared within the formula:
[tex]\Delta s = \frac{v^{2}-v_{o}^{2}}{2\cdot a}[/tex]
If we know that [tex]v_{o} = 22.222\,\frac{m}{s}[/tex], [tex]v = 0\,\frac{m}{s}[/tex] and [tex]a = -0.5\,\frac{m}{s^{2}}[/tex], then the distance travelled by the train is:
[tex]\Delta s = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(22.222\,\frac{m}{s} \right)^{2}}{2\cdot \left(-0.5\,\frac{m}{s^{2}}\right) }[/tex]
[tex]\Delta s = 493.817\,m[/tex]
The train will travel 493.817 meters until it stops completely.
