Question

A train is travelling at a speed of 90 km/h. Brakers are applied so as to produce a uniform acceleration of -0.5 m/s^2. Find how far the train will go before it is brought to rest

Answer 1

Answer:

625 m

Explanation:

Here , we have given Acceleration =   - 0.5 m / sec^2

           and speed , which is initial speed = 90 KM / H

we first convert       KM/H     to    M/SEC     in order to match unit

    so       when we convert KM/H to M/SEC we simple multiply KM/H value to 5/18

                            so , 90KM/H =   90 x 5/18 m/sec

                                                      = 25 m/sec

so we have now        Acceleration (a) = -0.5 m/sec^2

                                    Initial Velocity (u) = 25 m/sec

                                    Final Velocity   (v) = 0 (as train finally came to rest)

                                    Distance (s)      = we have to calculate

we have equation of motion relating U, V A and S

  i,e                                        v^2 - u^2 = 2as

 putting values                     (0)^2  -  25^2  = 2 x (-0.5) x s

                                                          - 625     = -1 x s

                         - and - will get cancel and we have   Distance (s) = 625m

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