Question

a yard engine shunts a freight car along a level siding. if the car stops in 50.0 seconds, 250 m from the point where it was released by the engine, calculate the velocity of the car when it was released.

Answer 1

We have the equations of motion

                                                                       .............................................. (1)

10+n= 2

                        and                                       .............................................(2)

= u? + 2as

Where u is initial velocity of the object

            v is final velocity

            a is acceleration

            t is time of motion

For the given problem,

                           S = 250 m

                            v = 0 m/s

                           t = 50 s

Putting the values in equation (1), we get

                         

0= u + 50a

                     or,

= D

Now putting the values in equation (2), we get

                         $0^{2} = u^{2} + 2\times \left ( \frac{-u}{50} \right )\times 250$

                    or,

u- 10u = 0

                    or,  

u= 10 m/s

Hence, the velocity of car, when it was released, was 10 m/s.

For any doubt please comment.