Answer:
t = 25.0 s
Explanation:
- Assuming that the engineer applies the brakes just over the crossing, the train moves exactly 350 m at a constant acceleration, with a final speed (when the last car of the train leaves the crossing) of 15.8km/h.
- Since we know the initial and final speeds, and the horizontal distance traveled (the length of the train) we can use the following kinematic equation to get the acceleration:
[tex]v_{f}^{2} - v_{o}^{2} = 2*a* \Delta x (1)[/tex]
- Since we need to find the time in seconds, it is advisable to convert vf and vo to m/s first, as follows:
[tex]v_{o} = 84.1 km/h*\frac{1h}{3600s} *\frac{1000m}{1km} = 23.4 m/s (2)[/tex]
[tex]v_{f} = 15.8 km/h*\frac{1h}{3600s} *\frac{1000m}{1km} = 4.4 m/s (3)[/tex]
- Replacing (2) and (3) in (1), since Δx =350m, we can solving for a:
[tex]a = \frac{(4.4m/s)^{2} - (23.4m/s)^{2}}{2*350m} = -0.76 m/s2 (4)[/tex]
- In order to get the time, we can simply use the definition of acceleration, and rearrange terms:
[tex]t =\frac{v_{f}-v_{o}}{a} = \frac{(4.4m/s)-(23.4m/s)}{-0.76m/s2} = 25.0 s (5)[/tex]
