In: Physics
A train has a mass of 4.82E+6 kg and is moving at 69.9 km/hr. The engineer applies the brakes, which results in a net backward force of 1.91E+6 N on the train. The brakes are held on for 28.2 s. What is the new speed of the train?
How far does it travel during this period?
Speed of the train = 69.9 km/hr = 19.42 m/s
Mass of the train = 4.82 x 106 kg
Backward acceleration of the train = (1.91 x 106)/(4.82 x 106) = 0.4 m/s2
So, velocity of train after 28.2 s is
Now we know that
Putting all the values in this equation, we get
Hence, the train will travel nearly 386 meters in this time period.
For any doubt please comment.