A train is travelling at a speed of 90 km per hour. Brakes are applied so as to produce a uniform acceleration of –0.5 ms^-2. Find how far the train will go before it is brought to rest?

Solution

Here comma space space space space Initial space velocity space of space train comma space straight u space equals space 90 space km divided by straight h space equals space fraction numerator 90 space cross times space 1000 space straight m over denominator 3600 space straight s end fraction space equals space 25 space straight m divided by straight s Uniform space acceleration space produce comma space straight a space equals space minus 0.5 space straight m divided by straight s squared Final space velocity comma space straight v space equals space 0 Now comma space using space the space formula comma space space space straight v squared space minus space straight u squared space equals space 2 as therefore space space space space 0 squared space minus space 25 squared space equals space 2 space cross times space left parenthesis negative 0.5 right parenthesis space straight s space space space space space space space space space space space space space space straight s space equals space fraction numerator 625 over denominator 2 space cross times space 0.5 end fraction space equals space 625 space straight m therefore space space Distance space covered space by space the space train space before space it space stops space equals space 625 space straight m

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