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 5.0 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.

Express your answer with the appropriate units.

The solution HAS to include the Kinematic equations with constant acceleration

Please include a graph.

Answer 1

The time needed for the 7th car to pass is 13.2 s

Explanation:

The motion of the train is a uniformly accelerated motion, therefore we can use suvat equations.

We start by analzying the motion of the first car, by using the equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered by the first car in a time t, which corresponds to the length of one car

u = 0 is the initial velocity

a is the acceleration

t = 5.0 s is the time

The equation can be rewritten as

$a=\frac{2s}{t^2}=\frac{2L}{(5.0)^2}=0.08L[m/s^2]$

where L is the length of one car.

The same equation can be written considering the first 7 cars:

$7L = ut+\frac{1}{2}at'^2$

where

7L is the distance covered by the 7 cars

t' is the time needed

We still have

u = 0

And the acceleration is constant so it is

$a=0.08L$

Substituting into the equation, we can find t':

$7L = \frac{1}{2}(0.08L)t'^2\\7=0.04t'^2\\t'=\sqrt{\frac{7}{0.04}}=13.2 s$

In attachment the graph of the distance covered versus the time taken: since the motion is uniformly accelerated, the relationship between the two variables is quadratical.

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