Question

A train 350 m long is moving on a straight track with a speed of 84.1 km/h. The engineer applies the brakes at a crossing, and later the last car passes the crossing with a speed of 15.8 km/h. Assuming constant acceleration, determine how long the train blocked the crossing. Disregard the width of the crossing.

Answer 1

t = 25.0 s

Step-by-step 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:

`v_(f)^(2) - v_(o)^(2) = 2*a* \Delta x (1)`

• Since we need to find the time in seconds, it is advisable to convert vf and vo to m/s first, as follows:

`v_(o) = 84.1 km/h*(1h)/(3600s) *(1000m)/(1km) = 23.4 m/s (2)`

`v_(f) = 15.8 km/h*(1h)/(3600s) *(1000m)/(1km) = 4.4 m/s (3)`

• Replacing (2) and (3) in (1), since Δx =350m, we can solving for a:

`a = ((4.4m/s)^(2) - (23.4m/s)^(2))/(2*350m) = -0.76 m/s2 (4)`

• In order to get the time, we can simply use the definition of acceleration, and rearrange terms:

`t =(v_(f)-v_(o))/(a) = ((4.4m/s)-(23.4m/s))/(-0.76m/s2) = 25.0 s (5)`