- #1

eltavo809

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## Homework Statement

A train pulls away from a station with a constant

acceleration of 0.40 m/s2. A passenger arrives at the

track 6.0 s after the end of the train has passed the

very same point. What is the slowest constant speed at

which she can run and catch the train?

## Homework Equations

X = X

_{0}+ V

_{0}* t + 0.5 * a * t

^{2}

V = V

_{0}+ a * t

## The Attempt at a Solution

The passenger arrieves 6 sec. latter, so: t

_{p}= t

_{t}- 6sec

__Passenger__

X = V

_{0}* ( t - 6 sec)

__Train__

X = 0.5 * 0.4 m/s

^{2}* t

^{2}

Both, train and passenger, have to be at the same place at the same time in order to achieve an encounter, so:

x

_{train}= x

_{passenger}

V

_{0}* ( t - 6 sec) = 0.2 m/s

^{2}* t

^{2}So here is the problem. I know I have to find out V

_{0}, but I have 3 variables and 2 equations. Any ideas? This is supose to be easy... :(

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