Question

The speed of a train is reduced from 60km/hr to 15km/hr while it travels a distance of 450m. If the retardation is uniform find how much further it will travel before coming to rest.

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Answer 1

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Answer 2

The initial velocity of train [u] = 60 km/hr

The final velocity of train [v] = 15 km/hr

The initial distance travelled [s] = 450 m

We have 1 km = 1000 m.

               1000 m = 1 km

               $1\ m=\frac{1}{1000}\ km$

               $450\ m=\frac{1}{1000} *450\ km$

                $450\ m=\ 0.45\ km$

Let the acceleration of the particle is  a .

From equation of kinematics we know that -

                                     $v^2-u^2=2as$

                                     $15^2-60^2\ =\ 2*a*0.45$

                                     $225-3600=0.90a$

                                     $\ -3375=0.90a$

                                     $a=\frac{-3375}{0.90}\ km/hr^2$

                                     $a=\ -3750\ km/hr^2$

Here, negative sign is due to the fact that it is retardation.

Now, we are asked to to calculate the stopping distance.

For stopping distance, the final velocity [v] = 0

                                      The initial velocity [u] = 15 km/hr

From equation of kinematics we know that-

                          $v^2-u^2=2as$

                          $0-15^2=2*[-3750]*s$

                          $-225=-7500*s$

                          $s=\frac{225}{7500}\ km$

                                 $=0.03\ km$    [ans]