Question

You are at a train station, standing next to the train at the front of the first car. The train starts moving with constant acceleration, and 9.0 s later the back of the first car passes you.How long does it take after the train starts moving until the back of the seventh car passes you? All cars are the same length.t=?

Answer 1

Answer:

The answer is t = 23.81 s

Explanation:

A car in this question means those partition that make up the train

       Let denote the the of each car as Z

 The equation of motion that has to do with distance is given as

                 S = ut + (1/2)at²

          Where u = initial velocity = 0

                       t = time = 9 s

 Substituting these values we have

                       Z = (1/2) a(9)²

                          = 40.5 a

making acceleration the subject we have that a = Z / 40.5

                    Let us denote the length of the 7th car as  =7Z

Considering the distance at the 7th car we have

                    S₇ = ut₇ + (1/2)at²₇

          Where  u = initial velocity =0

     Substituting values we have

                    7Z = (1/2)(Z/40.5)(t₇)²

                      7  = {1/2 ×40.5}(t₇)²

                    t₇²  = 7 × 2 × 40.5

                   t₇²    =  567

                   t₇     =  23.81 s