Question

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?

Answer 1

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

                         

                   or,    

Hence, the train will travel nearly 386 meters in this time period.

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